Quantum Mechanics

Schrödinger Equation

We start off with a standard equation for a wave.

A(x,t) means that what this wave looks like depends on position(x) and time(t). The description is set out in complex number form and can be displayed with an Argand diagram (For more info see here). This wave is a solution of the Wave Equation, and what we want to see is if the wave equation can be used to describe matter waves. The wave equation is

What this equation is saying is that, if you partially differentiate your wave, A, with respect to x twice, it will equal the partial differential of your wave with respect to t twice, multiplied by a constant, which in this case is 1/c².

So now we need to see if it will work, so first we take our wave [1] and differentiate it twice with respect to x (If you are unsure how to do this see here for help).

You may be wondering why i have changed the original equation whilst doing the differentiations. Originally we had A₀e^i(kx-ωt) and now we have A₀e^ikxe^-iωt. This is just a maths "trick" you can do to exponential powers and I personally think it makes the differentiation easier.

We now differentiate the wave twice with respect to time to get

We now substitute the two values, -k²A and -ω²A, back into the wave equation [2] in place of the differentials. This gives

Conveniently the minus signs, the A's and the squares cancel to give us

Now, if we take the 2 base quantum formulas

and try and substitute [3] into them we get a problem

E≠pc for matter, for non-relativistic matter energy and momentum obey the following law

So it looks like we have a problem. The Wave Equation [2] doesn't work for matter. One way to try and get it to work is to say that instead of ω²∝k², what if we tried to get it so it was ω∝k²? To do this we would need a wave equation that was differentiated twice with x and only once with t. Also if we replace the constant we can make life easier for ourselves.

So now we have

So here is our new wave equation. The reason I have changed the A to a ψ(Psi, pronounced "psigh" or sometimes just "sigh") is because this method is the one that works so I might as well use the proper symbols.

So now we do the same things we did before. Differentiate twice with respect to x

and only once this time with respect to t

This gives the relationship

Which is what we were after.

However if we want to make sure we get the energy correct (equation [5]) we have to make α equal to specific things, like so

If we now substitute this value of α into [6] and do a bit of rearranging we get

If we then substitute this into [4] we get

Which is correct!! So far so good, our new wave equation for matter gives us the correct energy. So let's put the constant back in

And now once again rearrange to get it in a nicer looking form.

We're nearly there now. The equation is almost complete. However when we solve it for the energy of a particle we get

but sometimes a particle can get energy from its surroundings, for example if it was in a potential, so we have to make one slight adjustment to account for all of the particles possible energies

Which means our wave equation becomes

This is called the One Dimensional Time Dependent Schrödinger Equation

Time Independence

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Most cases you can have when it comes to matter waves, the potentials (V) that do not depend on time, they don't suddenly change shape after so many seconds. If this is the case (and most of the time it is) then we can use the (Separation of Variables) method on the Schrödinger Equation.

First thing we do is assume that the Ψ can be split into two functions, one that only acts on x and one that only acts on t like so

You then get your Schrödinger Equation

and wherever there is a Ψ you just replace it with uT, so you get

Now, you divide by u, you get rid of the one on the left as that differential doesn't depend on u, and if you divide through by T you get rid of the T on the right a the right differential doesn't depend on T. So you get

Now lets say t changed, that would mean that the left hand of the equation would now have a different value, however as u is independent of t the right hand side of the equation wouldn't change. This would cause an error. The two sides of the equation were equal before, now one side has changed and they still have to be equal. To get around this problem you set both sides equal to a constant, in this case we shall call it E.

So now we have two seperate equations,

Which can be solved to get

and

Infinite Square Well

We have just considered the simple situation where our particle is not in a potential, and the solution it gave us is that the particle behaves like a standard wave, now we consider the situation where our particle is in a potential well with infinitely high sides. This may seem like an impossible situation but if a potential is deep enough it can be modelled as infinite and the equations and solutions still work. So the well we have will have these constraints

And it will look like this

If we assume that the well doesn't change shape with time then we can use the equation [8], the time independent equation

We only have to consider the area inside the well as the particle can't exist in the region of infinite potential, so inside the well V(u) is 0 so the equation becomes

To solve this equation we try the solution

So lets see if this will work. First we differentiate with respect to x, then with respect to x again like so

Which is simply

We have already defined k above, so k² is

This means that the second differential is

Which when substituted back into the wave equation gives us

This shows our general solution and clever choice of k in [9] is the solution for the electron in an infinite square well.

For the solution to work however, the wavefunction must go to zero at the edges of the box i.e. at x=±a, this means the solutions to the wave equation can be either of these 2

In order for these two solutions to hold they have to equal 0. The ways of holding this condition are shown below

As stated in [10], k has already been defined in terms of energy, so we can now equate energy to the width of the box and an integer, n

Which is rearranged to give

This means that the energy you can get is not continuous; you get only get certain values for the energy that depends upon n². We can now bring all of this together and write our equations of the solution of the time independent schrodinger equation, now written as u_n

These equations are shown below for n=1, n=2 and n=3.

Depending on the value if n the wave function will either be symmetrical or anti-symmetrical. This property is known as Parity, with a symmetric waves having even parity and anti-symmetric waves having odd parity. The probability of a particle to be at a given position is |u²|, which for the 3 wavefunctions above is

Finite Square Well

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Not all physical situations can be modelled as an infinite square well. In a lot of situations the well is of finite height as shown below

For the situation where -a≤x≤a, the solution is the same as the infinite case i.e. sinusoidal shapes that follow the form

Outside the region of -a to a however the potential is no longer infinite so we cannot get rid of the V(u) term, so we set the potential to the potential outside the well i.e. V₀

Now if we rearrange to get all of the U terms on one side we have a relitively easy second order differential to solve

[13]

If we try a solution of the following form

Then we can differentiate it twice with respect to x to obtain the d/dx² part of the first term

This gives us that d²u/dx² is just κ²u, so that means that equation [13] becomes

In order for the equation to hold then the value of κ must make the first term equal to the second term. If you divide everything by u and then rearrange for κ you get a value for κ of

IMPORTANT NOTE: The κ used in the finite square well is not the same as the k used in the solution to the infinite square well. The one in the solution to the infinite square well is just an ordinary k, the one just used in the solution for the finite square well is a greek kappa. This can lead to confusion having multiple symbols that look almost identical, if anyone gets really confused fell free to email me and I'll try to make it clearer.

So the solution to the wave equation outside the limits of -a and a is

Now this is odd. We have a barrier at -a and a stopping the particle from escaping, if the particle is inside the box then it doesn't have enough energy to get over the barrier. And yet we now have an equation that describes the position of the particle inside the forbidden area. We would have expected the solution outside of -a and a to be zero, but quantum mechanics says that there is a probability that the particle can get inside the potential barrier (provided it isn't infinite).

There are a few constraints however. If x is bigger than 0 then the term with C will tend to infinity as x tends to infinity which isn't allowed. And similarly if x is less than 0 then the term with D in would tent to infinity as x tends to minus infinity, so we get

One of the rules is that the wave function is continuous at the boundaries of the well, and so is du/dx. So at the edges of the well the wave function inside and the wave function outside must be equal, and so must du/dx, so we get the following conditions at the 2 walls, 2 equations for equating the wave equation inside to outside and 2 for equating the differentials inside and outside

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This page was written by The Rev and edited by Tom Glossop.

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