What is the Schrödinger Equation? A basic introduction to Quantum Mechanics

In this video i would like to tell you about one of the most famous and important equations in physics the schrodinger equation it's an equation that lies at the heart of quantum mechanics and it forms the basis of much of our modern understanding of the universe but what exactly is the schrodinger equation and how can we use it to perform quantum mechanical calculations In this video i will provide an introductory overview of schrodinger's equation with my aim being that by the end you should hopefully be ready to pick up an undergraduate textbook on quantum mechanics

and actually enjoy the experience so if you're up for the challenge stay tuned our story begins in vienna when on the 12th of august 1887 irwin rudolph joseph alexander schrodinger was born As was customary at the time schrodinger was educated at home for the first 11 years of his life before attending the local gymnasium school almost from the very first day until he left eight years later schrodinger excelled in mathematics and physics when schrodinger eventually arrived at the university of vienna much was expected of him and he did not disappoint consistently coming top of his

class His efforts were rewarded in 1910 when he received his doctorate and after a short stint of military service schrodinger joined his old university as an assistant of experimental physics in 1914 the first world war intervened and soon after the war ended schrodinger took up a position at the university of jenner the next 10 years saw showing a move between numerous institutions in europe before he finally settled in zurich Switzerland at the age of 34. as the next few years passed schrodinger began to wonder if he would ever make a major contribution that would establish

him among the first rank of contemporary physicists at the beginning of 1925 he was 37 years old long past his 30th birthday which was believed by many to be the watershed in the creative life of a theoretical physicist however by the end of the same year he Would make a breakthrough that would ensure his place in the pantheon of physics so what was it that suddenly changed the tide for schrodinger in october 1925 schrodinger read a paper that einstein had written earlier in the year relating to the latest developments in atomic physics a footnote in

einstein's paper mentioned the thesis of a young french physicist named louis de bruy who had recently published his thesis on the Theory of wave particle duality a few years earlier this rather innocent footnote caught the attention of schrodinger and he set about acquiring a copy of de bruy's thesis what was particularly exciting for schrodinger was that de bruy had dared to ask the question that if light waves can behave like particles as has been shown by einstein's explanation of the photoelectric effect Is it then possible that particles such as electrons can behave like waves the

boy's answer to this question was an unequivocal yes in summary de bruy proposed that for a matter particle with mass m moving at velocity v you can associate a plane wave of wavelength lambda equals h over p or h over mv where p equals nv is the momentum of the particle and h is planck's constant furthermore if the particle has energy e Then you can assign a frequency given by f equals e over h and these relations connect the particle concepts of energy and momentum e and p with the wave concepts of frequency and wavelength

f and lambda now what de bruy did was simply assume that these two relationships hold for all matter particles including electrons although there was no experimental evidence at the time to support de bruy's hypothesis he was encouraged by The fact that he was able to provide an explanation for boar's model of the atom although other physicists had taken a passing interest into boy's work the lack of experimental evidence dissuaded most from spending too much time thinking about the ideas however when schrodinger read du bois paper he immediately wrote to einstein to share his thoughts about

the implications of wave particle duality a couple of weeks later schrodinger was Asked by the physicist peter dubai to present a talk on dubroy's work at a physics colloquium in zurich schrodinger gave a beautifully clear account of de bruy's work however not everyone in the audience was convinced in particular peter dubai did not like the fact that according to de bruy's theory there appeared to be mata waves but no wave equation to describe these waves how can you have a wave without a wave Equation according to the well-established classical laws of wave motion any wave

whether it's a sound wave electromagnetic wave or wave on a vibrating violin string has an equation that describes its motion however in de bruy's theory there was no wave equation de bruy had never attempted to derive one and neither had einstein schrodinger knew at once that dubai was correct you cannot have waves without a wave equation and almost at once he decided To set himself the task of finding the missing equation and in the space of only a few months schrodinger was able to take de bruy's tentative ideas surrounding wave particle duality and develop them

into a fully blown theory of quantum mechanics in the remainder of this video i would like to provide a simplified overview of schrodinger's theory we will begin our investigation with a simple review of the properties of Classical waves a simple progressive wave which is moving towards the right can be described using a cosine function where y of x t is the displacement at position x and time t and a is the amplitude or maximum displacement of the wave in this expression we have defined the wave number k as 2 pi over lambda where lambda is

the wavelength and we've also defined the angular frequency omega as 2 pi f where f is the frequency of Oscillation or vibration of the wave now this displacement function is a solution to a general wave equation which takes the following form it's worth noting that this equation involves partial derivatives which are represented by the curly d symbol in the wave equation the displacement function y is a function of two variables x and t and so it's appropriate to use the partial derivative symbol when differentiating with respect to one of The variables we can confirm that

the cosine wave function is indeed a solution to this wave equation by substituting it into the equation if we take the first and second derivative of the wave displacement relation with respect to time and space we get the following relations and then if we sub these back into the wave equation we find the following green expression If we then divide both sides by y we find that v squared equals omega squared over k squared and therefore we see that this displacement function is indeed a solution to the wave equation provided that v equals omega over

k if we then substitute in the definitions of omega and k we see that this constraint is none other than the famous velocity frequency and wavelength relation v equals f lambda Now it's worth emphasizing a few points at this stage we see that the structure of the wave equation has ensured that the expected wave function is indeed a solution and it's also ensured that we arrive at the correct relationship between velocity frequency and wavelength this can ultimately be traced back to the fact that when we differentiated twice with respect to space we got a factor

of k squared and when we differentiated twice With respect to time we got a factor of omega squared and therefore we found that v squared equals omega squared over k squared from which it followed that v equals f lambda and so to summarize we've seen that a wave function is a solution to a wave equation provided that a certain relationship between frequency and wavelength holds true and for the classical wave we've considered this relationship between wavelength and frequency is precisely What we expect for a traveling wave and it was this relationship between wave function and

wave equation that schrodinger started thinking about in 1925. if de bruy was correct and particles behave like waves then these de broglie wave functions should be solutions to a wave equation if so what form should this wave equation take would it have the same form as the classical wave equation we've just seen And if not why not now if a freely moving particle is to be described by an associated wave then a reasonable first guess would be to suggest that the wave is described by a progressive traveling wave which as we've already seen can be

described by the following function this makes sense as a first guess since this function describes a wave with fixed frequency and wavelength and according to the einstein de bruy Relations if the wavelength and frequency is constant then so was the momentum and energy as would be expected for a freely moving particle however if a force were to act upon the particle then according to newton's second law the particle's momentum would change and therefore according to the de bruy postulate lambda equals h over p the wavelength associated with the particle will also change however the concept

of wavelength is not Well defined if it changes very rapidly it's difficult to define even a variable wavelength since the separation between adjacent maxima is not equal to the separation between adjacent minima to put this another way if the linear momentum of a particle is not of constant magnitude because the particles acted on by a force then functions which are more complicated than a cos kx minus omega t are required to describe the associated wave The challenge facing schrodinger was to develop a wave equation that could determine the wave function for any given situation in

other words schrodinger wanted to develop a wave equation that tells us the form of the wave function if we tell it information about the force acting on the associated particle and the way that we do this is by specifying the potential energy corresponding to the force now as we've Already seen with classical wave equation the most common type of equation which has a function for a solution is a differential equation and so it's a differential equation that we shall seek to help us construct a suitable quantum mechanical wave equation we're going to make a few

reasonable assumptions firstly the wave equation must be consistent with the de broy einstein postulates which connect the wavelength Lambda of the wave function with the linear momentum p of the associated particle and the frequency f of the wave function with the total energy e of the particle secondly it must be consistent with the energy equation for a particle which states that the total energy should be the sum of the kinetic and potential energies now we can write the kinetic energy half mv squared in terms of the momentum as p Squared over 2m and we can

write the potential energy as v of x t as you may remember from classical mechanics by specifying the potential energy we are providing information about the forces acting on the particle since the force is simply the derivative of the potential energy function and therefore we can write the total energy of the particle as e equals p squared over 2m plus v where v is our Potential energy function thirdly our equation must be linear in the wave function psi that is if psi 1 and psi 2 are two different solutions to the equation for a given

potential energy v then any arbitrary linear combination of these solutions should also be a solution this linearity requirement ensures that we should be able to add together wave functions to produce constructive and Destructive interference patterns that are characteristic of waves and the final assumption is that when the potential energy is constant the desired differential equation should have a sinusoidal traveling wave solution of constant wavelength and frequency this follows from the fact that if the potential energy is constant then the force acting on the particle moving in this potential will be equal to zero Since the

force is equal to the derivative of the potential energy function and if there is no force acting on the particle then newton's second law of motion tells us that the linear momentum p and the total energy e of the particle will be constant as is the case for a freely moving particle and we've already seen from the de bruy einstein relations that if the momentum and energy are constant then the Associated wavelength and frequency are constant we therefore assume that in this case the desired differential equation will have sinusoidal traveling wave solutions of constant wavelength

and frequency as already discussed so let's now use these assumptions to arrive at an expression for the schrodinger wave equation firstly if we rewrite the energy relation using the de bruy einstein postulates we find the Following yellow expression now before proceeding any further it will prove convenient to rewrite this expression in terms of the angular frequency omega and the wave number k as well as the reduced planck constant h bar these relations are useful because they keep variables out of the denominators of expressions and they absorb a factor of two pi that would otherwise appear

every time we write a sinusoidal wave Function subbing these relations into our energy expression we find the following result notice that we have a factor of k squared on the left hand side and a factor of omega on the right hand side of this expression now if you recall our fourth assumption suggested that when the potential energy is constant we have a freely moving particle and therefore the desired differential equation should have sinusoidal traveling wave solutions And so we will begin by assuming that our wave function has the following form next notice that when we

differentiate this wave function twice with respect to x we obtain a factor of k squared and the effect of taking the first time derivative is to introduce a factor of omega and since the differential equation we seek must be consistent with the expression in the red box which contains a factor of k squared in one term and a Factor of omega in another these facts suggest that the differential equation that we seek should contain a second derivative with respect to x and a first derivative with respect to time in addition there must be a term

containing the potential energy function and in order to ensure linearity this term must contain a factor of the wave function putting all these ideas together we are led to a differential equation of the Following form where we've introduced two constants alpha and beta to provide a bit of flexibility our job is to fix these constants by ensuring that our proposed wave function and wave equation is consistent with the de bruy einstein relations now since we are currently assuming a freely moving particle we will set the potential energy to a constant value if we then sub

in our proposed wave function into our proposed wave equation We find the following result which we can then write in the following form now an immediate problem with this result is that we have a mixture of sine and cosine terms and so in order for this equation to be satisfied for all values of x and t requires that the coefficient of each term is separately equal to zero in which case we're forced to set beta equal to zero and k squared equals v naught over alpha and we clearly see That there is no way of

recovering the required de broy einstein relation so what's gone wrong well part of the difficulty can be traced back to the fact that differentiation changes cosines into sines and vice versa and since we are differentiating once with respect to time and twice with respect to space we end up with a mix of sine and cosine terms furthermore we've been quite restrictive with our initial choice of wave function if you recall we Simply borrowed the wave function result from classical physics so perhaps we should try using a combination of cosine and sine functions to specify our

free particle wave function with this in mind let's consider a more general wave function of the following form here gamma is a constant which again provides some additional flexibility let's now see if this more general wave function helps us get a bit closer Towards our goal if we differentiate this function twice with respect to x we find the following equation and if we differentiate once with respect to time we find the following green expression then if we sub these results into our proposed wave equation we find the following ghastly yellow result which we can rewrite

in the following form now hopefully you can see that the two Terms in square brackets on the left hand side of the equation have exactly the same form as our initial proposed wave function we also notice that the term in square brackets on the right hand side also very nearly has the same form however we see that the minus gamma squared term is preventing it from being exactly the same as the other terms if we could choose gamma appropriately then we would be able to cancel these Terms from the equation and drastically simplify it however

in order to do this we require that minus gamma squared equals one and therefore gamma squared equals minus one and it follows that gamma must therefore be equal to plus or minus the square root of minus one now you may at this stage panic and think that something's gone terribly wrong surely this doesn't make sense after all You remember being taught at school that it's not possible to take the square root of a negative number however let's take this possibility seriously for a moment and define the square root of -1 as being equal to i

where i is referred to as an imaginary number which is simply a number with the property that i squared equals minus 1. now if we look back at our original equation and set gamma equal to i then we see that we can cancel square bracket Terms from both sides of the equation and we then arrive at the following simplified expression and if we then compare this with the energy momentum relationship that we derived using the de broy einstein relation we see that we can match the two expressions if we identify alpha equals minus h bar

squared over 2m and beta equals i h bar if we then sub these expressions back into our proposed wave equation we Arrive at the following beautiful equation this differential equation satisfies all four of our assumptions concerning the quantum mechanical wave equation now it's worth emphasizing that we've been led to this equation by considering the special case of a free particle with a constant potential we will however postulate that the quantum mechanical wave equation has the same form in the general case when the Potential energy can vary as a function of space and time but we

cannot prove this to be true we can however postulate it to be true and the validity of the postulate will ultimately be judged by comparing its implications with experiment and so it seems that we've finally arrived at the desired equation a quantum mechanical wave equation whose solutions give us the wave function which is to be associated with the Motion of a particle under the influence of forces which are described by the potential energy function v and it was this exact equation that was first obtained by schrodinger in 1926 and which now carries his name now

it's worth noting that schrodinger was led to his equation by a more esoteric line of reasoning than the one that we've adopted here however he was strongly influenced by the dubois postulate in the same way that we have Been so what is the significance of the schrodinger equation and what can we use it for well the basic idea is as follows if you wish to determine the structure of the wave function associated with a particle of mass m using the schrodinger equation the first thing you do is to specify the potential energy function which tells

you something about the forces acting on the particle once you've specified the potential you Then need to solve the corresponding partial differential equation to find the wave function but what exactly does the wave function represent and how does it relate to the particle that it's associated with let's now turn to this question in order to answer this question it will prove illuminating to look again at the free particle wave function that we used in our previous analysis If you recall we had the following expression but we also found that this function is only a solution

to the schrodinger equation if gamma equals i and if we sub this into the wavefunction expression we find the following result we see that our wave function is a complex function meaning that it contains the imaginary number i and if you recall this behavior was forced upon us to ensure consistency with the de bruy einstein postulates furthermore we Found that an imaginary number also appears in the schrodinger equation itself and this implies that in the general case of varying potential the wave functions which are solutions to the schrodinger equation will also be complex now since

a general wave function of quantum mechanics is a complex function it's worth digressing momentarily to explore complex numbers in a bit more detail Any complex number may be written in the form z equals a plus i b where as we've already seen i squared equals minus one the number a is referred to as the real component of the complex number and the number b is referred to as the imaginary component of the complex number and it's worth noting that both a and b are real numbers notice that the complex wave function that we've encountered is

of the form a Plus ib and therefore it specifies simultaneously two real functions its real part and its imaginary part this is in contrast to a wave function of classical mechanics for instance a wave on a string can be specified by one real function which gives the displacement of various elements of the string at various times this classical wave function is not complex because the classical wave equation does not contain a factor of i Since it relates a second time derivative to a second space derivative now one way in which we can represent a complex

number is to plot its position on a graph in which the y-axis specifies the imaginary component of the complex number and the x-axis specifies the real component we can then plot the complex number by marking the positions a and b note that if we draw an arrow starting at the origin and ending at the point we Plotted then we can specify the angle theta which is the angle the arrow makes with the real axis if r is the length of this arrow then we can use some simple high school trigonometry to see that the distance

from the origin along the real axis is given by r cos theta and this is simply a and the distance along the y axis is given by r sine theta which is simply b if we then sub these expressions for a and b back into the expression z equals A plus ib we get the following green result next if we set r equals 1 and theta equals kx minus omega t then we recover the complex wave function that we had before now if you have a complex number of the form z equals a plus ib

then it's possible to perform an operation on this number which is referred to as taking the complex conjugate this operation simply involves replacing I with minus i and it's represented by the asterisk symbol armed with the complex conjugate operation we can then calculate the modulus squared of a complex number which involves multiplying a complex number with its complex conjugate in our example we see that the modulus squared of z can be written as a plus ib times a minus ib which simplifies to a squared plus b squared Now the important thing to note about this

expression is that the modulus squared of a complex number is a real number that can never be negative this will have profound consequences for our interpretation of the wave function in what follows now the fact that wave functions are complex functions should not be considered a weak point of the quantum mechanical theory actually it's a desirable feature Because it makes it immediately apparent that we should not attempt to assign a physical existence to wave functions in the same sense that water waves have a physical existence the reason is that a complex quantity cannot be measured

by any actual physical instrument and therefore the quantum wave function is not a measurable quantity so what does this complex wave function represent if it corresponds to no Physical measurable quantity after all a particle by its nature is localized at a point whereas the wave function as its name suggests is spread out in space well a hint to answering this question lies in the fact that as we've just seen the modulus squared of a complex number represents a real number so perhaps the modulus squared of the wave function corresponds to something real that can be

determined experimentally in 1926 the physicist max born suggested That the modulus squared of the wave function represents the probability of finding a particle at a particular location in space in other words if at some instant in time t a measurement is made to locate the particle associated with the wave function psi then the probability that the particle would be found at a coordinate between x and x plus dx is given by the following green expression Where as we've just seen the modulus squared of the wave function is calculated by multiplying the wave function by its

complex conjugate if we then want to find the probability that the particle is located within a given region we simply need to integrate the modulus squared so for example the probability of finding a particle located between two points a and b is given by the integral of the modulus squared evaluated with a and b as the Limits of the integral the justification of bourne's postulate can be found in the following considerations since the motion of a particle is connected with the propagation of an associated wave function these two entities must be associated in space that

is the particle must be at some location where the waves have an appreciable amplitude and therefore the probability density must have an Appreciable value where the wave function has an appreciable value furthermore since the probability density is a measurable quantity it should be a real positive valued number whereas the wave function is in general complex and therefore it's obviously not possible to equate the probability simply with the wave function itself however since the modular squared of the wave function is always real and Non-negative bourne had reason to believe that it referred to a measurable quantity

so what's this all telling us bourne's interpretation introduces a kind of indeterminacy into quantum mechanics for even if you know everything the theory has to tell you about the particle in other words you know the particle's way function you still cannot predict with certainty the outcome of a simple experiment to Measure its position all that quantum mechanics can offer is statistical information about the possible result of measurements this indeterminacy has been deeply disturbing to physicists and philosophers alike and it's natural to wonder whether it's a fact of nature or a defect of the theory itself

now i think you might agree that this discussion so far has been rather abstract and so i think it would prove Useful if we ground the remainder of this video in a series of concrete examples which explicitly show how the schrodinger equation can be used to calculate various useful properties and so our first step is to figure out how to actually solve the schrodinger equation now the standard technique we will use is referred to as the separation of variables and involves searching for solutions in which the wave function can Be written as a product of

two terms where the first term is a function of x alone and the second term is a function of t alone now it's worth noting that this method is only valid if the potential energy is independent of time but since most physical systems of interest have potential energies of this form the condition is not a very serious restriction now if we substitute our expression for Psi into the time-dependent schrodinger equation we find the following blue equation if we then act with the derivatives on the terms inside the square brackets we find the following purple expression

next we want to rearrange this expression so that all the terms involving x are on the left hand side and all the terms involving t are on the right hand side a simple rearrangement then leads to the following yellow Expression since the right hand side does not depend on x while the left-hand side does not depend on t their common value cannot depend on x or t in other words the common value must be a constant which we should call k and so we can write two equations involving k a space dependent equation and a

time dependent equation the constant k is Called the separation constant and we see that the effect of employing this technique has been to convert the single partial differential equation involving two independent variables x and t into a pair of ordinary differential equations one involving x alone and the other involving t alone let's now focus on this time equation if we rearrange this expression we find the following blue equation this differential equation tells us that The function phi of t which is its solution has the property that its first derivative is proportional to the function itself

now anyone familiar with calculus will recognize that the derivative of an exponential function gives another exponential function and therefore it follows that phi of t can be written as an exponential function therefore let's assume that the solution to the differential equation has the Following form where alpha is a constant that will be determined shortly we can test whether this is indeed a solution by plugging it back into the equation in which case we find the following expression if we then cancel phi of t from both sides we see that alpha equals minus i k over

h bar and so if we sub this back into our trial solution we find that phi of t is equal to e to the minus i k t over h bar Note that this substitution has caused the exponential to become complex since it contains a factor of i in the exponential now at this stage it will prove illustrative to rewrite this expression in an alternative but identical form by making use of euler's famous equation which links complex exponentials with trigonometric functions applying euler's equation to our specific example we find the following Expression and if we

reintroduce the factors of 2 pi we find the following now the benefits of rewriting phi in this way is that we see explicitly that phi of t is an oscillatory function in time and we can read off the angular frequency immediately and we see that omega equals 2 pi k over h we also know that the angular frequency omega is related to the frequency of Oscillation f by the equation omega equals 2 pi f and therefore we see that the constant k is related to the frequency f by the relation k equals hf now this

might all seem a bit tedious and pointless until you remember that according to the de broy einstein relation the energy of a quantum particle is related to its frequency by the relationship e equals hf and so we see that the separation Constant k must in fact be equal to the energy e now i don't know about you but i think that that's a very neat result and so we can finally write that phi of t is equal to e to the minus i times e times t divided by h bar okay so now that we've

figured out a solution for the time dependent part of the equation let's sub this back into our general expression we started with our next task is to solve the space Dependent equation and find out what psi of x is to begin with let's sub in k equals e into the space equation we obtained earlier next if we rearrange the terms slightly we find the following expression this equation is referred to as the time independent schrodinger equation because the time variable t does not enter the equation the solutions of this equation determine The spatial dependence of

the general wave function and are referred to as eigenfunctions which in german means characteristic function once the eigenfunctions are determined we can then reintroduce the time dependence by multiplying by the time factor e to the minus iet over h bar so in summary the separation of variables approach has led us to the conclusion that the eigen function psi of x which specifies the space Dependence of the wave function is a solution to the time-independent schrodinger equation furthermore we have seen that the time dependence of the wave function is governed by the term phi of t

equals e to the minus i e t over h bar where e is the total energy of the particle in the system now that we've sketched the general procedure let's apply this approach to a specific example to begin with we're going to consider a Situation in which a particle is confined to a one-dimensional box of width a such that the particle is free to move around inside the box but it's unable to escape this example is often referred to as the infinite square well now although this may seem like a slightly strange and contrived example

it's worth considering since it serves as a wonderfully accessible test case For all the fancy quantum mechanical machinery that comes in a more advanced course in quantum mechanics we can represent this situation by specifying the potential energy in the following manner the first thing to notice is that outside the box the wave function must equal zero since the probability of finding the particle outside the box is zero however inside the box where v of x Equals zero the particle is completely free to move except that the two ends x equals zero and x equals a

where an infinite force prevents the particle from escaping so the natural question arises what form does the wave function take inside the box to answer this question we simply need to solve the time independent schrodinger equation inside the box since we know that the potential energy Is zero inside the box the time independent schrodinger equation simplifies to the following yellow equation we can then rearrange this expression and write it suggestively in the following form where we have defined k as the square root of 2 m e over h bar now the general solution to this

equation takes the following form where a and b are arbitrary constants as can Be verified by substituting the expression back into the equation in which case we have that d psi by dx equals the following yellow expression and if we differentiate again with respect to x we find d2 psi by dx squared equals the following purple expression which is simply equal to minus k squared psi as required we can fix the constants a and b by considering the boundary conditions of The problem in the case of the infinite square well potential we know that psi

of x equals zero outside the box and we also know that psi of x is not equal to zero inside the box in order to meaningfully interpret the wave function as a probability density function we require that the wave function is continuous in other words we require that the function changes smoothly from where Zero to where it's non-zero therefore continuity of the wave function at the edges of the box requires that psi of zero equals psi of a equals zero so what does this tell us about a and b firstly when x equals zero we

find that a sine zero plus b cos zero equals b and therefore b must be equal to zero and so finally we can write psi of x equals a sine kx next we impose the continuity constraint At x equals a and therefore we have that psi of a equals a sine ka and this must be equal to zero and so we see that either a equals 0 in which case we're left with the trivial solution psi of x equals 0 or else we have that sine of k a equals 0. now we know that the

sine function is equal to zero whenever the argument of the function is equal to an integer multiple of pi radians in other words we Can write ka equals plus or minus n times pi in which case we can define k equals n pi over a for integer values of n note we do not need to include the negative values of k since sine of minus x equals minus sine of x and we can simply absorb the minus sign into the constant a also note that we should not have included n equals zero because that would

again imply that psi of x equals Zero for all values of x so armed with this boundary constraint on k we can then combine this with the fact that k equals the square root of 2 m e over h bar to write n pi over a equals the square root of 2 m e over h bar and if we rearrange this for the energy e we find the following expression so what exactly is this telling us well the first thing that we notice is that the particle energy has become Quantized and can only take certain

discrete values depending on the value of n with each different value of n corresponding to a different energy and in turn this corresponds to a different eigen function we also notice that the lowest possible energy state is not zero as would be the case in classical physics rather the ground state energy is given by the following expression This is a general feature of quantum mechanical systems and can be understood in terms of heisenberg's uncertainty principle which roughly speaking says that whenever you confine a particle to a region of space there is an inherent uncertainty in

the particle's momentum and therefore its energy it is this quantum uncertainty that gives rise to the non-zero ground state energy when a particle is confined to a region of space We are now in a position to combine all of the results that we've derived so far and bring everything together if you recall we derived an expression for the eigenfunction psi of x when we solved the time independent schrodinger equation if we now sub in the boundary condition constraint equation into our eigenfunction expression we find that the nth eigenfunction is given by the following expression If

we sketch the first few eigenfunctions we see that they look exactly the same as the standing waves in a string of length a the lowest energy state corresponding to psi 1 of x is referred to as the ground state and those states whose energy increase in proportion to n squared are called excited states once we have specified the eigen functions we can construct the full wave function solution of the time dependent Rodinger equation for any value of n and as we've already seen we can write this in the following form where the time dependence is

now manifest and so for our particular particle in a box example we find that the wave function of our particle is given by the following blue expression where here e n refers to the energy of the nth eigenstate we see that we've managed to determine an expression for the wave function up To an overall constant a but that leaves an obvious question what value does a take and can we determine this well thankfully the answer is yes and to do this we simply use bourne's probability rule which if you recall says that the probability of

finding a particle in a region is given by the integral with respect to x of the modulus squared of the wave function now in the example of a particle Confined to the box we would expect the probability of finding the particle inside the box to be exactly equal to one and therefore we can write the following green expression for the probability our task is to now evaluate this integral to do so we recall that the modulus squared of the wave function involves multiplying the original wave function by its complex conjugate and if we do this

we find the following result Where we have used the rule of exponents to simplify our expression and so we see that our requirement that the probability of finding the particle in the box must be equal to one simplifies to the following constraint in order to evaluate this integral we use the trigonometric identity sine squared of x equals 1 minus cos of 2x all divided by 2 and this allows us to rewrite the integrand in terms of cosine as follows If we then write this as two separate terms we find the following blue expression the first

term can be integrated immediately in which case we find the following simple result but what about the second term for this slightly trickier term we let u equal two n pi x over a in which case d u by d x equals two n pi over a and so we can write d u equals two n pi over a times dx It then follows that we can rewrite our integral in the following form in terms of u and if we evaluate this we find the following yellow expression when we insert the limits we see that

we get two terms and the second term is zero for all values of n since sine of zero equals zero and so we are left with just the first term which is also equal to zero for all integer values of n And so if we go back to our original expression we see that after the dust is settled we have the following constraint and therefore a equals the square root of 2 over little a and therefore inside the box the wave function for the nth quantum state is given by the following expression and this is

our final answer the process that we've just gone through of determining the wave function Coefficient to ensure that the total probability of finding the particle somewhere in space is one is referred to as normalizing the wave function now there are a couple of key points that are worth mentioning at this stage firstly a wave function of this form describes what is known as a stationary state because the modulus squared of the wave function does not depend on time which means that the probability does not change with time We also note that such a stationary state

has a sharply defined energy corresponding to each value of n and we'll have more to say about this point later if we plot the modulus squared of the wave function for each value of n then we can visualize where we are most likely to find the particle within the box for each different eigenstate for example we see that in the ground state our particle in a box is most Likely to be found in the middle of the box since this is where the modulus squared of the wave function peaks and therefore this is where the

probability of finding the particle peaks whereas in the n equals 2 state we see that there is actually now zero probability to find the particle in the middle of the box we see that as n increases the quantum mechanical probability density oscillates more and more and in the Limit that n approaches infinity the oscillations are so compressed that no experiment could possibly have the resolution to observe the oscillations and in this limit the particle will have equal probability of being found at any location within the box furthermore the fractional separation of the eigenvalues approaches zero

as n approaches infinity so in that limit the discreteness of energy cannot be resolved and it appears as if the energy Of the particle varies continuously and so we see that the quantum mechanical predictions approach the predictions of classical mechanics in the limit of large quantum number or high energy now once we fully normalize the wave function we can calculate the probability of finding a particle in a given energy state in a particular region of space to do this we simply need to integrate The modular squared of the corresponding wave function over the region of

space that we're interested in for example suppose i wanted to calculate the probability of finding a particle in the ground state in the following interval in this case i would simply need to calculate the integral of the modulus squared of the normalized wave function between a over two and three a over four in order to evaluate this integral we Will change variables and let u equal pi x over a in which case we have that d u equals pi over a dx if we then sub this back into our integral expression we find the following

result next we use our favorite trigonometric identity to re-express our integral if we then carry out the integral and sub within limits we find an answer of a quarter plus one over two pi which is roughly equal to 0.41 And so we see that the probability of finding our particle within this region of the box has been successfully determined by calculating the integral of the modulus squared of the normalized wave function and so we see that once we've normalized the wave function we can calculate the probability of finding a particle in any particular region of

space okay but what else can we do well as we've Already seen according to bourne's probabilistic interpretation of the wave function the outcome of a measurement in quantum mechanics is a random variable with many possible values because statistics plays such a fundamental role in quantum mechanics it's worth briefly digressing to introduce some probability theory and notation and i would do so in the context of a simple example imagine a group of 12 people whose ages Are as follows if we let n of x represent the number of people of age x then we can write

the following list of information the total number of people in the room is then given by the sum of these terms now if you were to select one individual at random from this group then the probability that this person's age would be for example 9 would obviously be 1 in 12 because there are 12 possible choices All equally likely of whom only one individual has that particular age follows that if p of x is the probability of getting age x then in general we can write p of x equals n of x divided by n

for example the probability of selecting a person whose age is 12 would be p of 12 equals 2 over 12 which is equal to 1 6. it should hopefully be obvious that the sum of all the probabilities must be Equal to 1 since the person you select must have some age and so we can write the sum of p of x from x equals to zero to infinity is equal to one now if we wanted to calculate the mean or average age we would simply add together all the ages and divide by the total number

of people and we can write this in the following way which if we evaluate returns the number fifteen it follows that in general the average Value of x is then given by the following expression notice that there need not be anyone with the average age in our example no one happens to be 15 years old now in quantum mechanics the average is usually the quantity of interest and in this context is referred to as the expectation value and we'll have more to say about this later now in general the expectation value of some function of

x is given by the Following expression so for example if we wanted to calculate the average of the square of the ages in the group then we would write this in the following form now it's perfectly possible that two sets of data may have the same mean value but in one set the data could be clustered around the mean value and in the other set the data may be more spread out consider the following two histograms The first is sharply peaked around the average value whereas the second is broad and flat now it would be

useful therefore to have a numerical measure of the amount of spread in a distribution with respect to the average now it might seem like an obvious way of doing this would be to see how far each individual is from the average in other words calculate delta x equals x minus the expectation value of x and then Compute the average or expectation value of the difference however if we do this then we see that the expectation value is equal to zero now the conventional way to get around this problem is to square delta x before averaging

in which case we have the following expression here the quantity sigma squared is referred to as the variance of the distribution and sigma itself is called The standard deviation which is the customary measure of the spread of the distribution around the average value now before we move back on to quantum mechanics there is a useful theorem on variances which we'll need later so let's take a moment to digress if we take the expression for the variance and write it explicitly in terms of delta x and then expand out the bracket we see that we can

rewrite each term explicitly in terms of the Probability p of x and then we see that everything simplifies and we find that the variance can be written as the difference between the average of the square of x and the square of the average of x then if we take the square root we obtain an expression for the standard deviation itself and we note that since sigma squared is clearly non-negative from its definition then this result implies that the Expectation value of the square of x is greater than or equal to the expectation value of x

all squared and the two are equal only when sigma equals zero and the distributions have no spread at all it's worth noting that the standard deviation in the variable x is often referred to as the uncertainty in x and we will return to the importance of this concept later in the video it's worth noting that in our example we Specified the age of each individual as an integer and this therefore represented a discrete variable but it's simple enough to generalize this analysis to continuous distributions when applying this to our example of age the sensible thing

would be to specify the probability that an individual's age lies within some interval say for example the probability that their age lies between 16 and 17. It follows that the probability that some individual chosen at random has an age that lies between x and x plus dx is given by rho of x dx where r of x is called the probability density it follows that in general the probability that some variable x lies between a and b is given by the integral of the probability density and the rules that we obtain for discrete variables carry

over in an obvious way For example the expectation value of a continuous variable is given by the following expression now we can see immediately how these results in probability theory translate into quantum mechanics we've already seen that in quantum mechanics the probability of finding a particle in a region of space is equal to the integral of the modulus squared of the wave function and so we see that the modulus squared of the wave function Can be interpreted as the probability density function which is how it's often referred to likewise the expectation value of a particle's

position in quantum mechanics is given by the following integral let's now consider how this discussion of probability theory applies to the quantum mechanical example of the position of a particle in the ground state of the infinite square well potential In principle the expectation value of the position of the particle can be found by taking the average result of measurements on an infinite ensemble of identically prepared systems alternatively as we've just seen we can calculate the expectation value using the probability density function which is simply the modulus squared of the wave function if we look at

the plot of the modulus squared as a function of position it Certainly suggests that the most likely location to find the particle is in the center of the box but let's check this explicitly with a calculation using our newly developed probability theory to begin with we write an expression for the expectation value of position in terms of the ground state wave function if we then write this in terms of the energy eigenstates we see that the time dependence drops out and we're left with The following purple expression this is telling us that the expectation value

of the position does not depend on time but is a constant value if you recall from our earlier analysis the ground state eigen function is given by the following yellow expression and if we sub this into our expectation value equation we find the following blue equation if we then make use of the same trigonometric identity that we used Earlier we can get rid of the sine squared term and we find the following yellow expression next we will evaluate each integral in turn the first integral is simple enough and we find that it evaluates to a

over 2. the second integral is a little bit trickier and involves integrating by parts now if we set f equal to x then we immediately see that df equals dx And if we identify dg with cos 2 pi x over a and integrate we find that g equals a sine 2 pi x over a all divided by 2 pi if we then plug these results back into our integral we find the following blue expression now we immediately see that the first term in this expression vanishes and we're left with just the second term we can

integrate this expression by changing variables and letting u equal 2 Pi x over a in which case d u equals 2 pi over a dx if we then perform the integration we find the following result which evaluates to zero and so when the dust is settled we see that the expectation value of the position of our particle in a box is simply equal to a over two and this is precisely the result that we might have expected based on the plot of the modular squared of the wave function Furthermore we see that the expectation value

of the position is a constant that does not depend on time this is a characteristic feature of stationary states as we've already mentioned now another feature of so-called stationary states is that they have a definite energy and it's worth taking the time to show this explicitly if you recall the spatial shape of the wave function and therefore the spatial Shape of the modulus squared of the wave function is governed by the time-independent schrodinger equation which you will often see written in the following form where h is referred to as the hamiltonian operator this equation is

called the energy eigenvalue equation and as we've already mentioned the function psi of x is referred to as the eigenfunction of the hamiltonian with energy eigenvalue e Now anyone who has studied eigenvalue equations in linear algebra will be familiar with this terminology in practice as we have seen with the infinite square world potential there are many eigenvalues and many eigenfunctions an eigenfunction of the hamiltonian operator is a very special mathematical function when the complicated hamiltonian operator acts upon a general function You might expect a total mess however when the hamiltonian operator acts upon an eigenfunction

it gives the exact same eigenfunction back only this time it's multiplied by the energy of the quantum state let's now use some of the probability theory we developed earlier to show that a wave function of this form represents a state with a sharply defined energy so how do we do that well in general when the energy of a quantum system is Measured the outcome is uncertain the degree of uncertainty in the energy can be specified by calculating the standard deviation of the energy which as we've already seen takes the following form now the expectation value

of the energy for a particle with a normalized wave function is given by the following integral where we have sandwiched the hamiltonian operator between the complex conjugate Of the wave function and the wave function itself if we then simply use the energy eigenvalue equation we see that the expectation value expression simplifies and is equal to e where we've used the fact that the integral of the modulus squared inside the box is simply equal to 1 and so we see that the expectation value of the energy is simply equal to e now if psi of x

is an eigenfunction of The hamiltonian it's also an eigenfunction of the product of the hamiltonian with itself and so it follows that the following relationship is true we can then use this to find the expectation value of the energy squared which is given by the following equation and if we use the eigenvalue equation we see that this simplifies to e squared and so if we combine our results we see that The standard deviation in the energy aka the uncertainty in the energy is exactly zero so what does this mean this is telling us that the

result of an energy measurement is certain to be e when the wave function is an eigen function of the hamiltonian operator with eigenvalue e and so we conclude that an eigen function of the hamiltonian always describes a state of definite energy and These states of definite energy are exactly the stationary states that we mentioned earlier when we said that the probability density does not change with time now at this stage it's probably worth taking a moment and summarizing some of the key points we've covered so far we began by writing down the general time dependent

schrodinger equation which describes a particular quantum mechanical system once we specified the Potential energy function we found that if the potential energy is time independent then we can use the separation of variables approach and write the general wave function as a product of two terms one involving space and one involving time this then allowed us to rewrite the schrodinger equation as two separate differential equations one involving only spatial coordinates and one involving only time coordinates by solving the time equation we found That the time dependence can be described by an exponential function of the following

form where e represents the corresponding energy of the state on the other hand the spatial equation which is aptly named the time schrodinger equation was found to take the following form next we saw that the time-independent schrodinger equation can be written as an energy eigenvalue equation involving The hamiltonian operator and that the eigenfunction solutions of this equation describe stationary states in the sense that all probabilities and expectation values are independent of time and the energy uncertainty of these states is zero so that the energy of each state is well defined we also found that acceptable

solutions to the time independent schrodinger Equation exist only for certain discrete values of energy and these energies are precisely the energy eigenvalues that we've already referred to and corresponding to each eigenvalue is an energy eigenfunction which is a solution to the time-independent schrodinger equation for a given potential energy function for each eigenvalue we can construct the corresponding wave function by simply Tacking on the time dependence the index n which takes on successive integral values and which is employed to designate a particular eigenvalue eigenfunction and wave function is called the quantum number if this system is

described by the wave function psi subscript n it's said to be in the quantum state n and each of these quantum state wave functions is a particular solution to the general schrodinger equation for a Given potential energy function now since the schrodinger equation is linear in the wave function we expect that any linear combination of these wave functions will also be a solution to the schrodinger equation in other words we should be able to write an expression of the following form and this should also be a solution to the schrodinger equation in fact this expression

gives the most general form of the solution to the equation for a Given potential function and this is often referred to as a linear superposition of quantum states its generality can be appreciated by noting that it is a function which is composed of a very large number of different eigen functions combined in proportions governed by the constant c so once we've found the separable eigenfunction solutions of the time independent schrodinger equation we can immediately construct a more general Solution of the following form now you may be wondering what the coefficients c subscript n represent physically

to find out we're going to have to work through a few calculations so buckle up let's begin by normalizing the general wavefunction expression which as you now know requires that the integral of the modular squared of the wavefunction is equal to one for our general wavefunction we need to Then evaluate the following integral now hopefully you can see by staring at this expression long enough that when we multiply out the square brackets this integral will contain two types of term firstly there will be cross terms involving different energy eigenstates and these terms will have the

following form in this particular example we see that we have the complex conjugate of the n equals two eigenstate being multiplied By the n equals one eigenstate secondly they'll be terms of the following form which involve the same eigen state so how do we deal with these integrals well the second type of integral is easy to evaluate since we can use the normalization condition which tells us that the integral of the modulus squared of an energy eigenstate is simply equal to one and so we can write this particular term As simply the modulus squared of

the coefficient c1 and therefore in general we will find the following result but what about the first type of integral which involves cross terms how do we evaluate this well we can in fact show that all of these cross-term integrals will vanish and to do that let's consider a general cross-term integral of the following form where m and n take on different integer values Next if we sub in the expressions we derived earlier for the energy eigenstates for a particle in a box then we find the following yellow expression we can simplify the integral by

rewriting the product of sine terms using the following well-known trigonometric identity in which case we find that our integral takes the following form if we then evaluate this integral we find the following slightly intimidating Expression but if we then sum in the limits of the integral we find something slightly more compact and we can in fact see that this is equal to zero when m is not equal to n since in this case we will have an integer multiple of pi inside each of the sine terms and therefore the whole integral will be equal to

zero this condition is often referred to as orthogonality and the two different Eigenstates represented by m and n are said to be orthogonal to each other now it's worth noting that this argument does not work when m equals n in that case we have to use the normalization condition which tells us that when m equals n the integral will be equal to 1. in fact it's customary to combine both orthogonality and normalization into a single statement which takes the following form Here delta m n is referred to as the kronecker delta and is defined by

the fact that when m equals n it's equal to one and when m is not equal to m the kronecker delta is equal to zero and finally we say that the eigenstates satisfying this condition are orthonormal and so after that slight detail let's return to the job at hand which was to normalize our general wave function and we now see that armed with the Orthogonality relation that the cross terms will vanish because the eigenstates are orthogonal to each other and terms involving the same eigenstate will not vanish due to the normalization condition and so it

turns out that in order to normalize our general wave function the coefficients must satisfy the following condition okay but this still doesn't tell us what the coefficients actually represent To gain a bit more insight let's calculate the expectation values for the energy and energy squared corresponding to this general wave function as we've already seen we can calculate the expectation value of the energy using the following green equation and we can calculate the expectation value of the energy squared using the following yellow equation now in order to evaluate these integrals when the wave function is a

Superposition of eigenfunctions we use the energy eigenvalue equations we encountered earlier as well as the orthonormal condition if we then make use of these relations to evaluate the expectation value integrals we find that the expectation value of the energy is given by the following green expression and the expectation value of the energy squared is given by the following yellow expression And so we see that we're finally in a position to answer the question we originally posed namely what is the physical meaning of the complex coefficients if you recall from our discussion on probability theory the

expectation value of some variable x is given by the following expression and if we compare this with our expressions for the expectation values of the energy and energy squared we see That for a given value of n the modulus squared of the coefficient represents the probability that a measurement of the energy will return the value e subscript n and this fits perfectly with the normalization condition that we found earlier which said that the sum of the modulus squared of the coefficients must be equal to one this is simply the statement that the sum of the

probabilities of measuring Each energy state must be equal to one and so to summarize we found that when the wave function corresponding to a particle in a one-dimensional infinite square well potential is expressed in its most general form as a linear superposition of energy eigenstates then this linear superposition describes a particle with uncertain energy and the probability of measuring a particular energy eigenvalue is given by the modulus squared of the coefficient of The corresponding term in the linear superposition now i must admit this has all been quite heavy going and abstract so far so let's

consider a specific concrete example of a linear superposition of states which will highlight how this whole formalism works imagine that a particle in the infinite square well potential has an initial wave function comprising an even mixture of the first two stationary states If we sub in the expressions that we derived earlier for the normalized eigenstates of the ground state and the first excited state we see that our linear superposition takes the following form now our first step is to normalize this wave function so that we can establish the constant a and to normalize the wave

function we require the following relationship to hold true if we substitute our expression for the initial wave function we find the Following yellow equation and if we then multiply out the bracket we find the following blue expression next if we rewrite the sine squared terms in terms of cosines we find the following rather overwhelming blue expression performing each of these integrals is straightforward and follows the same logic as previous examples we've looked at when the dust settles we find the Following green expression and then if we submit the limits we finally arrive at the rather

simple expression 2a squared which must be equal to 1 and therefore we see that a is equal to one over the square root of two once we've worked out the normalization constant we can sub this back into our original wave function expression and we get the following result and after expanding out the bracket we Arrive at the following green expression which summarizes our initial state wave function next we want to match this up with the most general solution of the schrodinger equation we know from our earlier analysis of the infinite square well potential that the

general solution is a superposition of eigenstates which in the specific case of the infinite square well takes the following form The next step in our strategy is to set t equals to zero and then match coefficients with our initial wave function so let's do that if we take our initial normalized wave function and compare the coefficients term by term with the most general wave function for the infinite square well we find from matching the first term that c one equals one over root two likewise by matching the second terms we find that c two also

equals one over Root two and finally we see that cn equals zero when n is greater than or equal to three and now that we've determined the coefficients we can substitute these into our general wave function and we find the following green expression where we've also substituted the values for e1 and e2 using the energy equation that we derived earlier next we recall from the einstein de bruy relation that the frequency is related To the energy by the equation equals hf and therefore it follows that the angular frequency is related to the energy by the

equation omega equals e over h-bar and therefore we see from the first exponential term that we can equate omega 1 as being equal to h bar pi squared over 2 m a squared and we see from the second exponential term that we can write omega 2 is equal to 4 times omega one It then follows that our general wave function takes the following form and we can then pull out a factor of one over root two and rewrite this explicitly in terms of the energy eigenstates of the infinite square well by noting that the term

inside the first square bracket is simply the first i can say psi one and the term inside the second square bracket is the second eigenstate psi two and therefore we can write our general Wave function explicitly in terms of the first two energy eigenstates in the following way now one of the advantages of doing this is that we can simply read off the probabilities of finding the particles in each of these energy eigenstates as this information is contained in the coefficients of the two eigenstates appearing in the general wavefunction expression in particular the probability of

Measuring the energy of the particle and finding the energy to be the eigenvalue e1 is simply equal to the square of the coefficient of the first term and we see that this gives an answer of one-half likewise the probability of finding the particle with energy e2 is also equal to one-half and so we see that the measurement of the total energy of the particle could lead to either of two results the eigenvalue e1 or the eigenvalue e2 This is telling us that there is an inherent uncertainty in the energy of the particle due to the

fact that the wave function is a linear combination of two different energy eigenstates to clarify this point let's calculate the energy uncertainty which as we saw earlier involves calculating the standard deviation of the energy let's begin by calculating the expectation value of the energy which is simply equal to the probability of Measuring e1 multiplied by the energy e1 plus the probability of measuring e2 multiplied by the energy e2 if we put the values in we find the following expression for the expectation value of the energy this is telling us that if we prepared an infinite

number of identical ensemble states and measured the energy of each of these states and then calculated an average this is the value that we would obtain Next we can calculate the expectation value of the square of the energy which after a simple calculation results in the following expression now if we sub these expressions into the standard deviation equation we see that the uncertainty in the energy is not zero and so we see that by forming a linear combination of states we have introduced uncertainty into the energy of the particle which is exactly what we expect

Based on our earlier discussion quantum states with uncertain energy are called non-stationary states because they have some observable properties that change with time for example we will now show that the probability density and position expectation value corresponding to this wave function will oscillate in time so let's do that let's begin by calculating the probability density corresponding to our General wave function as we've seen many times now the probability density is given by the modulus squared of the wave function and if we then sub in the general wave function we determined earlier we find the following

yellow expression next if we multiply out the brackets and carefully group terms we find this beautiful blue expression now the reason we have written it like this is because we can then use euler's Equation which was introduced earlier to write this term as two cos brackets 3 omega 1 times t in which case we arrive at pure perfection in the form of this purple expression now we immediately notice by staring at this expression that the probability density is a function of time this was not the case when we were looking at the probability density for

a single energy eigenstate However here the linear combination of energy eigenstates has led to a cos 3 omega t term to gain a bit more insight into how this term affects the probability density and to help visualize what's going on let's plot a graph of probability density on the y axis and x position on the x axis we see that the presence of the time cosine term has caused the probability density curve to oscillate back and Forth about the center of the box with a characteristic frequency since the probability density tells us information about where

we're most likely to find our particle and since the probability density is changing with time this is telling us that the location of where we're most likely to find the particle is also changing with time to help understand this in a bit more detail let's look at the expectation Value of the position of the particle described by this simple linear superposition now i won't go through the details of this calculation but it's safe to say that it follows in the same spirit as the other calculations that we've looked at in detail feel free to try

it yourself and if you do so you should find that when the dust settles the expectation value is given by the following compact expression So what is this telling us well this is saying that the expectation value of the position of the particle is oscillating about the center of the box to visualize this let's superimpose the expectation value on the graph of the oscillating probability density that we looked at earlier we see that the expectation value of x oscillates with a particular frequency in fact we can calculate this oscillation frequency if you recall from earlier

we found that The angular frequency corresponding to the ground state was given by the following expression and therefore three times omega one is simply three times this and we will label this angular frequency omega prime now if you look carefully you notice that we can write omega prime as the difference of two terms and if you recall the energy equation that we derived earlier for the particle in the infinite square well we see that we can Write omega prime as the difference between the energy of the first excited state and the ground state divided by

the reduced planck's constant and finally we can rewrite the angular frequency in terms of the oscillation frequency and we arrive at the following beautifully concise result now the eagle eyed amongst you might recognize this as the equation used to calculate the frequency of emitted light when an electron makes a transition Between two energy levels inside ball's model of the atom and you would be correct in thinking that there is a deep link between our analysis and this famous result and so to end this quantum mechanical journey let's discuss this exciting link in a bit more

detail if we begin by considering an electron in the ground state of a hydrogen atom then according to maxbourne's probabilistic interpretation of the wave Function the electron can be found at any location where the probability density has an appreciable value and it therefore follows that the electric charge of the electron would not be confined to a particular location within the atom but rather distributed in proportion to the electron's probability density now this might seem slightly counter-intuitive but it follows logically from the analysis so far Outlined it's simply saying that the charge should be distributed in

accordance with where it's most likely to find the electron now when an electron is in the ground state of an atom its energy is sharply defined and as we've seen from our analysis earlier this corresponds to a stationary state in which the probability density is independent of time and therefore the electrons charge distribution is also Independent of time now according to classical electromagnetism a static distribution of charge does not emit radiation and therefore we see that quantum mechanics provides a way of resolving the paradox of the old quantum theory concerning the stability of atoms if

you recall one of the great challenges facing the early pioneers of quantum theory was to understand how atoms could be stable given that any Orbital model of the atom predicted that the orbiting electrons should continuously emit radiation and therefore spiral into the nucleus according to the bohr model of the atom electrons that are excited to higher energy levels do indeed emit radiation when they make a transition from an excited state to a lower energy state if we consider an electron in the process of making a transition then as we've just seen the wave Function describing

a transition would involve a linear superposition of the first excited energy eigenstate and the ground state and therefore the probability density will oscillate as we've just seen and since the probability density is proportional to the associated electron charge distribution it follows that the charge distribution will oscillate in sync with the probability distribution with a frequency given by the following Expression where e2 is the energy of the excited state and e1 is the energy of the ground state now according to classical electromagnetism this charge distribution would be expected to emit radiation at the same frequency as

the oscillation frequency of the charge distribution but we note that this is also precisely the frequency of the photon that would be emitted according to ball's model of The atom since the energy carried by the photon is equal to the difference between the two energy levels interestingly we see that this cannot happen for an electron in the ground state of the atom because there is no state of lower energy for the ground state to mix with to produce an oscillatory probability density or charge distribution now i find it quite remarkable that even though we have

used a very crude One-dimensional particle in the box model for our analysis we've been able to capture some of the most important and interesting features of the full quantum mechanical theory of atoms in fact one of the first things that schrodinger did once he had developed his wave equation was to apply it to the hydrogen atom and he was able to show that not only does it correctly predict the energy levels of the hydrogen atom and the frequencies of the photons Emitted in atomic transitions but it also allows you to correctly predict the probabilities per

second that those transitions will take place it's interesting to note that schrodinger believed his theory provided a physical picture of the process of emission of radiation by atoms in discussing the advantages of his own theory schrodinger wrote it is hardly necessary to point out how much more gratifying it would be to conceive of a Quantum transition as an energy change from one vibrational mode to another than to regard it as a jumping of electrons there were certainly those who disagreed with schrodinger's interpretation but what is beyond or doubt is the magnitude of the contribution that

schrodinger made towards our understanding of the atom unsurprisingly schrodinger was awarded the nobel prize for his groundbreaking Work and the equation that now bears his name lies at the heart of our modern understanding of particle physics and i think this is as good a place as any to bring this quantum mechanical journey to an end for now at least the good news is that if you've made it this far then you have covered some of the most important ideas in any introductory course in quantum mechanics and perhaps what's even better news is that in my

next video i'm going to put Schrodinger's equation to the test when i explore quantum tunneling and how it enables the stars to shine it feels only appropriate that i end by quoting the words of schrodinger himself who said giving expression to thought by the observable medium of words is like the work of a silk worm in being made into silk the material achieves its value but in the light of day it stiffens it Becomes something alien no longer malleable true we can then more easily and freely recall the same thought but perhaps we can never

experience it again in its original freshness thank you for watching