2 A Quantum Particle in One Dimension

As an extreme example, we don’t know if our universe is infinite or if it is finite, with a radius $R\gtrsim 20$ light years. If the latter is true, then you wouldn’t be able to travel with arbitrary momentum, but only very specific values finely graded by $\sim 1/R$. Needless to say, we have no way of measuring momentum to an accuracy which can distinguish these two options.

Conversely, suppose that there are spatial dimensions in our universe beyond the obvious three that we see. If such dimensions exist, we need a reason why we’ve not observed them. One obvious possibility is that they wrap around to form a circle or some other compact space. In this case, the lowest energy wavefunctions simply spread uniformly around the extra dimension. If you want to move in the extra dimension, then you need a minimum amount of momentum $p=\mathrm{\hslash }/R$ and, correspondingly, a minimum amount of energy ${\mathrm{\hslash }}^2/2m{R}^2$. If we can’t muster this minimum energy then we are unaware of these extra dimensions. But, ironically, we’re unaware of them because we inhabit them uniformly, rather than because we’re stuck at one point.

There are no pressing reasons to believe that extra dimension exist in our universe. Our best particle accelerators reach energies of around ${10}^{12}$ eV, also known as a TeV, and there is no sign of them knocking particles into the next dimension. This puts a limit on the size of these putative small extra dimensions of m or so. (To give you the fuller story, you really need relativistic dynamics by the time we get to these energies. The result is that the minimum momentum in the extra dimension requires energy $E=\mathrm{\hslash }c/R$, but the general conclusion remains unchanged.)

The discreteness in quantities like momentum and energy is one of the characteristic features of quantum mechanics. However, as the example above illustrates, there is no discreteness in the fundamental equations. Instead, the discreteness is an emergent phenomenon. The integers $n$ arise only when you solve these equations. In the current example, there is a nice intuitive explanation for how this arises: the only states that are allowed are those for which an integer number $n$ of de Broglie wavelengths fits nicely around the circle. Indeed, the quantisation condition (2.16) becomes

$\lambda ={\displaystyle \frac{2\pi R}{|n|}}\mathit{\hspace{1em}\hspace{1em} }\mathrm{with}\mathit{\hspace{1em}}n\in 𝐙$

Finally, we can return to the normalisability issue that plagued our earlier example of a particle on a line. The wavefunction (2.12) is not normalised but, when viewed on a circle, it is at least normalisable. We have

${\displaystyle {\int }_0^{2\pi R}}𝑑x{|\psi |}^2=2\pi R$

which tells us that the correctly normalised wavefunction is

$\psi (x)={\displaystyle \frac{e^{ikx}}{\sqrt{2\pi R}}}$

You can see that this doesn’t fare well if we try to return to a particle on the real line by taking $R\to \mathrm{\infty }$. In this limit, the wavefunction becomes smaller and smaller in its attempt to remain normalised, until it vanishes completely.

To see the restriction in more detail, first recall that in our previous example the wavevector $k$ can take either sign. Moreover, $e^{+ikx}$ and $e^{-ikx}$ describe two different states; from (2.14) they correspond to a particle with positive or negative momentum. For the particle in the interval, we will consider the more general ansatz

$\psi (x)=A{e}^{ikx}+B{e}^{-ikx}$

now with $k>0$. The requirement that $\psi (x=0)=0$ tells us that $B=-A$, so the wavefunction must be of the form

$\psi (x)=A\left(e^{ikx}-e^{-ikx}\right)=2iA\sin kx$

The factor of $2iA$ in front simply changes the normalisation of the wavefunction and doesn’t affect the physics. Indeed, if we want the wavefunction to be normalised correctly, should take

$\psi (x)=\sqrt{{\displaystyle \frac{2}{L}}}\sin kx$

(2.17)

But now we have to impose the requirement that the wavefunction vanishes at the other end of the integral, $\psi (x=L)=0$, or

$\sin kL=0\mathit{\hspace{1em}\hspace{1em} }\Rightarrow \mathit{\hspace{1em}\hspace{1em} }k={\displaystyle \frac{\pi n}{L}}\mathit{\hspace{1em}\hspace{1em} }\mathrm{with}n\in {𝐙}^{+}$

The wavefunctions for the $n=1,2,3$ and 4 states are shown in Figure 8.

There is one slightly subtle difference from our previous example: in current case $n$ should be a positive integer, while for the particle on the circle it could have either sign. This is because reversing the sign of $k$ in (2.17) merely flips the overall sign of the wavefunction which, as we have seen, does not give a different state. This means that the $+n$ and $-n$ states are the same. Meanwhile, if you set $n=0$, the wavefunction simply vanishes and and so the upshot is that we restrict $n$ to be a positive integer.

In this example, we again see that energy is quantised, taking values

$E={\displaystyle \frac{{\mathrm{\hslash }}^2{\pi }^2{n}^2}{2m{L}^2}}$

(2.18)

This is very similar to the result of a particle on a circle.

What about the momentum? Recall our previous discussion: the state $e^{ikx}$ has momentum $p=\mathrm{\hslash }k$. But for the current example, we have (ignoring the normalisation) $\psi =e^{ikx}-e^{-ikx}$. This is the superposition of two states, one with momentum $+p$ and the other with momentum $-p$. This means that these states do not have a well defined momentum! You can think of them as standing waves, bouncing backwards and forwards between the two walls but not going anywhere. We’ll become more precise about how to identify the momentum of a state as we go on.

The discreteness of energy levels in the infinite well has an important application, similar in spirit to the “extra dimension” story that we told above, but significantly less science fiction. Consider particles moving, as particles do, in three spatial dimensions. Suppose that you trap them in a well in one dimension, but still allow them to wander in the other two. Then, provided that you can restrict their energies to be small enough, the particles will act, to all intents and purposes, as if they’re really two-dimensional objects.

This is in sharp distinction to what happens in classical mechanics where the particles would move approximately in two dimensions but there would always be small oscillations in the well that can’t be ignored. However, the discreteness of quantum mechanics turns an approximate statement into an exact one: if the particle doesn’t have enough energy to jump to the $n=2$ state then it really should be thought of as a 2d particle. Of course, we can also restrict its motion once more and make it a 1d particle.

This may not seem like a big deal at this stage but, as we will learn in later courses, interesting things can happen in low dimensions that don’t happen in our 3d world. But these things aren’t mere mathematical curiosities but can be constructed in the lab using the method above. (If you want an example, in 2d — but not in 1d or in 3d — it’s possible for an electron to split into $N$ objects each carrying fractional electric charge $1/N$. And this can be seen in experiments!)

For our purposes, let’s consider a wavefunction built by taking

$A(k)=\exp \left(-{\displaystyle \frac{{(k-k_0)}^2}{2\sigma }}+{\displaystyle \frac{k_0^2}{2\sigma }}\right)$

This is a Gaussian distribution over momentum states ${\psi }_k$, centred around the wavenumber $k=k_0$, with width $\sigma >0$. A sketch is shown to the right. The additional factor of $e^{k_0^2/2\sigma }$ is just a constant and only affects the normalisation of the wavefunction; we’ve included it so that things look simpler below.

When used to construct the linear superposition (2.19), the Gaussian distribution means, roughly speaking, that we include significant contributions only from wavenumbers that lie more or less within the window

${(k-k_0)}^2\lesssim \text{a few times}\sigma$

(2.20)

Outside of this window, the coefficients $A(k)$ drop off very quickly. What does the resulting wavefunction look like? We have

	$\psi (x,t)$	$=$	${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑k\exp \left(-{\displaystyle \frac{k^2-2k{k}_0}{2\sigma }}-i{\displaystyle \frac{\mathrm{\hslash }t{k}^2}{2m}}+ikx\right)$
		$=$	${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑k\exp \left(-{\displaystyle \frac{\alpha }{2}}{\left(k-{\displaystyle \frac{\beta }{\alpha }}\right)}^2+{\displaystyle \frac{{\beta }^2}{2\alpha }}\right)$

where, in the second line, we have simply completed the square and the various parameters are given by

$\alpha ={\displaystyle \frac{1}{\sigma }}+{\displaystyle \frac{i\mathrm{\hslash }t}{m}},\beta ={\displaystyle \frac{k_0}{\sigma }}+ix$

(2.21)

At this stage, we have an integral to do. This is well-known Gaussian integral, with

${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑k{e}^{-\alpha {(k-\gamma )}^2/2}=\sqrt{{\displaystyle \frac{2\pi }{\alpha }}}$

This result holds for any $\gamma$ and for any $\alpha$ with $\mathrm{Re}(\alpha )>0$. A quick look at the $\alpha$ defined above confirms that we’re in business, and the final result for the wavefunction is

$\psi (x,t)=\sqrt{{\displaystyle \frac{2\pi }{\alpha (t)}}}\exp \left(-{\displaystyle \frac{1}{2\alpha (t)}}{\left(x-{\displaystyle \frac{i{k}_0}{\sigma }}\right)}^2\right)$

(2.22)

where the function $\alpha (t)$ is defined in (2.21). We see that the wavefunction also takes the form of a Gaussian, now in space, and with a time dependent width $\alpha (t)$.

There’s a lot to unpack in this formula. First, note that the wavefunction is normalisable. (Although, as we shall see shortly, not actually normalised.) This follows from the fact that $\psi (x,t)$ decays exponentially quickly as $x\to \pm \mathrm{\infty }$. So although we built the wavefunction from the non-normalisable, and hence illegal, waves (2.12) there’s nothing wrong with the end product. This is the first sign that we shouldn’t be so hasty in simply discarding the $e^{ikx}$ wavefunctions on the line. They may not be genuine states of the system, but they’re still useful.

The state described by (2.22) has neither definite momentum nor energy nor, indeed, position. This is actually the lot of most states. Moreover, the state does not take the simple separable form (2.8) of the stationary states that we’ve discussed until now. Indeed, the time dependence is fairly complicated. It turns out that one can construct all solutions of the Schrödinger equation through similar linear superpositions of stationary states. When, as in the present case, the stationary states are simply $e^{ikx}$, the linear superposition is just the Fourier transform of a function. But it also holds in more complicated cases. This is why its sensible to solve the time dependent Schrödinger equation by first looking for solutions to the simpler time independent Schrödinger equation. We’ll address this further in Section 3.

Next, we can try to extract some physics from the state (2.22) which goes by the name of the Gaussian wavepacket. Clearly it describes a state that is fairly well localised in space. But, as time goes on, it becomes more and more spread out¹¹ 1 The Mathematica website has a nice demonstration of an evolving Gaussian wavepacket.. This is what happens to quantum probability if it is not trapped by some potential: it disperses. It is only the stationary states that have a definite energy and, correspondingly, the simple $e^{-iEt/\mathrm{\hslash }}$ time dependence that stick in one place.

There is also something interesting in the size of the dispersion. Recall that we built the state by integrating over momentum modes in the window (2.20). We can think of this in terms of the variance, or uncertainty, of the wavenumber which we write as

$\mathrm{\Delta }{k}^2\sim \sigma$

where $\sim$ is an important mathematical symbol that means the thing of the left is roughly like the thing on the right. It’s useful in physics when trying to tease out some relation without bothering about all the details like annoying numerical coefficients. Meanwhile, the spread in position is determined by the function $\alpha (t)$ given in (2.21). It is at its minimum when $t=0$, so we have

$\mathrm{\Delta }{x}^2\sim \alpha (t)\ge \alpha (t=0)={\displaystyle \frac{1}{\sigma }}$

We see that the spread of the wavefunction in position space is inversely proportional to its spread in wavenumber or, equivalently, its spread in momentum $p=\mathrm{\hslash }k$. Multiplying these together, we get

$\mathrm{\Delta }{x}^2\mathrm{\Delta }{p}^2\gtrsim {\mathrm{\hslash }}^2$

This is our first glimpse at the famous Heisenberg uncertainty relation. You can localise a particle in space only at the expense of having a broad range of momenta, and vice versa. We’ll derive the proper mathematical expression of the Heisenberg uncertainty relation in Section 3.4.1.

This also gives us a new perspective on our original attempt at writing states as $\psi =e^{ikx}$. These wavefunctions have a definite momentum, $p=\mathrm{\hslash }k$. But they are also spread out everywhere in space and this is what resulted in their non-normalisable downfall.

To get a better sense of what’s going on, we should compute the probability density $P\propto {|\psi |}^2$. At this point we should be more careful about the overall normalisation, so we introduce a constant $C$ that we’ll figure out later and write $P=C{|\psi |}^2$. We then have

	$P(x,t)$	$=$	${\displaystyle \frac{2\pi C}{|\alpha |}}\exp \left(-{\displaystyle \frac{1}{2\alpha }}{(x-i{k}_0/\sigma )}^2-{\displaystyle \frac{1}{2{\alpha }^{\star }}}{(x+i{k}_0/\sigma )}^2\right)$		(2.23)
		$=$	${\displaystyle \frac{2\pi C{e}^{k_0^2/\sigma }}{|\alpha |}}\exp \left(-{\displaystyle \frac{1}{\sigma {|\alpha |}^2}}{\left(x-{\displaystyle \frac{\mathrm{\hslash }{k}_{0}t}{m}}\right)}^2\right)$

where you need a little bit of algebra to get to the second line. You can check that the probability is correctly normalised if we take $C=e^{-k_0^2/\sigma }/\sqrt{4\sigma {\pi }^3}$. Importantly, the normalisation factor itself doesn’t depend on time. This had to be the case because we’ve constructed this wavefunction to obey the time dependent Schrödinger equation and, as we saw in Section 1.2.2, if you normalise the wavefunction at one time then it stays normalised for all times.

From the form of the probability (2.23), you can clearly see the growing width of the distribution, with

${|\alpha |}^2={\displaystyle \frac{1}{{\sigma }^2}}+{\displaystyle \frac{{\mathrm{\hslash }}^2{t}^2}{m^2}}$

However, it is now also clear that the probability distribution doesn’t just sit at the origin and spread out, but instead moves. This follows because the $x$ dependence takes the form $x-vt$ where the velocity is

$v={\displaystyle \frac{\mathrm{\hslash }{k}_0}{m}}$

If we think in classical terms, and define the momentum to be $p=mv$, then we have $p=\mathrm{\hslash }{k}_0$ as expected.

There should be a more systematic way to extract this information about the momentum of a state, one that doesn’t involve staring at complicated formulae to figure out what’s going on. And there is. Our equation (2.23) is a probability distribution for the position of the particle. As with any probability distribution, we can use it to compute averages and variances and so on.

Given a probability distribution $P(x)$, with $x\in 𝐑$, the average of any function $f(x)$ is given by

$⟨f(x)⟩={\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑xf(x)P(x)$

In quantum mechanics the average is usually called the expectation value and, as above, is denoted by angular brackets $⟨\cdot ⟩$. We can then ask: what is the average position of the particle given the probability distribution (2.23)? We simply need to compute the integral

	$⟨x⟩$	$=$	${\displaystyle \frac{1}{\sqrt{\pi \sigma {|\alpha |}^2}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑xx\exp \left(-{\displaystyle \frac{1}{\sigma {|\alpha |}^2}}{\left(x-{\displaystyle \frac{\mathrm{\hslash }{k}_{0}t}{m}}\right)}^2\right)$
		$=$	${\displaystyle \frac{1}{\sqrt{\pi \sigma {|\alpha |}^2}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑\tilde{x}\left(\tilde{x}+{\displaystyle \frac{\mathrm{\hslash }{k}_{0}t}{m}}\right)\exp \left(-{\displaystyle \frac{{\tilde{x}}^2}{\sigma {|\alpha |}^2}}\right)$

where, in the second line, we’ve shifted the integration variable to $\tilde{x}=x-\mathrm{\hslash }{k}_{0}t/m$. We can now use the general result for Gaussian integrals,

${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑x{e}^{-a{x}^2}=\sqrt{{\displaystyle \frac{\pi }{a}}}\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑xx{e}^{-a{x}^2}=0$

(2.24)

where the second equality follows because the integrand is odd (and suitably well behaved at infinity). Using these, we get

$⟨x⟩={\displaystyle \frac{\mathrm{\hslash }{k}_{0}t}{m}}$

showing, once again, that the wavepacket travels with velocity $v=\mathrm{\hslash }{k}_0/m$.

We could do further calculations to compute the average of any other function $f(x)$. But instead we’re going to pause and ask a different question. How do we compute the average momentum of the wavepacket?

At first glance, this might seem like an odd question. After all, we’ve just computed the velocity $v$ and so the average momentum is surely $p=mv$. However, in later examples things won’t be so straightforward and this is a useful place to pause and see what’s going on.

One reason this is an interesting question is because it gets to the heart of the difference between classical and quantum mechanics. In classical mechanics, the state of the system is determined by both $x$ and $p$. But in quantum mechanics we have only the wavefunction, $\psi (x)$ and this has to contain information about both position and momentum. So how is this encoded? This is one of the steps in this section that involves a leap of faith and no small amount of creativity.

In fact, we saw a hint of how to proceed back in the introduction. Recall that our Hamiltonian for a one-dimensional particle is

$\widehat{H}=-{\displaystyle \frac{{\mathrm{\hslash }}^2}{2m}}{\displaystyle \frac{d^2}{d{x}^2}}+V(x)$

while, in classical mechanics, the energy of a particle is

$E_{\mathrm{classical}}={\displaystyle \frac{p^2}{2m}}+V(x)$

This suggests a relationship between momentum $p$ and the act of taking a derivative

$p⟷-i\mathrm{\hslash }{\displaystyle \frac{d}{dx}}$

where I’ve picked the minus sign, as opposed to plus sign, because I’ve studied this course before. This is the clue that we need. Given a wavefunction $\psi (x)$, the momentum is encoded in how fast it varies in space. We can see this in the simple non-normalisable states $\psi =e^{ikx}$ where ${\psi }^{\prime }=ik{e}^{ikx}$ and is clearly bigger for higher momentum. In general, the correct relationship between momentum and the derivative manifests itself in the following formula

$⟨p⟩=-i\mathrm{\hslash }{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑x{\psi }^{\star }{\displaystyle \frac{d\psi }{dx}}$

(2.25)

We’ll have a lot more to say about this in Section 3. For now, let’s just see what it gives for our Gaussian wavepacket. We have

	$⟨p⟩$	$=$	${\displaystyle \frac{1}{\sqrt{\pi \sigma {|\alpha |}^2}}}{e}^{-k_0^2/\sigma }{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}dx{e}^{-{(x+i{k}_0/\sigma )}^2/2{\alpha }^{\star }}\times -i\mathrm{\hslash }{\displaystyle \frac{d}{dx}}{e}^{-{(x-i{k}_0/\sigma )}^2/2\alpha }$
		$=$	${\displaystyle \frac{1}{\sqrt{\pi \sigma {|\alpha |}^2}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑x{\displaystyle \frac{i\mathrm{\hslash }}{\alpha }}\left(x-{\displaystyle \frac{i{k}_0}{\sigma }}\right)\exp \left(-{\displaystyle \frac{1}{\sigma {|\alpha |}^2}}{\left(x-{\displaystyle \frac{\mathrm{\hslash }{k}_{0}t}{m}}\right)}^2\right)$

where, after taking the derivative, the algebra needed to go from the first to the second line is identical to that in (2.23). Again shifting the integration variable to $\tilde{x}=x-\mathrm{\hslash }{k}_{0}t/m$, we have

$⟨p⟩={\displaystyle \frac{1}{\sqrt{\pi \sigma {|\alpha |}^2}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}𝑑\tilde{x}{\displaystyle \frac{i\mathrm{\hslash }{k}_0}{\alpha }}\left({\displaystyle \frac{\tilde{x}}{k_0}}+{\displaystyle \frac{\mathrm{\hslash }t}{m}}-{\displaystyle \frac{i}{\sigma }}\right)\exp \left(-{\displaystyle \frac{{\tilde{x}}^2}{\sigma {|\alpha |}^2}}\right)=\mathrm{\hslash }{k}_0$

where, to get the final equality, we did the Gaussian integrals (2.24) and then used the expression for $\alpha$ given in (2.21).

That was a fair bit of work just to get the answer that we expected all along: $⟨p⟩=\mathrm{\hslash }{k}_0$. As we go on, we’ll see that the expression (2.25) is the right way to think about the average momentum of any state.

There are a bunch of constants sitting in (2.28) and life is simpler if we can just get rid of them. To this end, define

$y=\sqrt{{\displaystyle \frac{m\omega }{\mathrm{\hslash }}}}x\mathit{\hspace{1em}\hspace{1em} }\mathrm{and}\mathit{\hspace{1em}\hspace{1em} }\tilde{E}={\displaystyle \frac{2E}{\mathrm{\hslash }\omega }}$

(2.29)

Then the Schrödinger equation takes the cleaner form

${\displaystyle \frac{d^2\psi }{d{y}^2}}-y^2\psi =-\tilde{E}\psi$

(2.30)

Before we get going, we can find one solution just by staring. This is the Gaussian wavefunction

$\psi (y)=e^{-y^2/2}$

(2.31)

The derivatives are ${\psi }^{\prime }=-y\psi$ and ${\psi }^{\prime \prime }=y^2\psi -\psi$, so we see that this obeys the Schrödinger equation with (rescaled) energy $\tilde{E}=1$.

Furthermore, it’s simple to see that all normalisable solutions should fall off in the same exponential fashion, with $\psi \sim e^{-y^2/2}$ as $y\to \pm \mathrm{\infty }$. This follows from looking at the large $y$ behaviour of (2.30), where the $\tilde{E}\psi$ term is necessarily negligible compared to the $y^2\psi$. This motivates the general ansatz

$\psi (y)=h(y)e^{-y^2/2}$

(2.32)

where we will shortly take $h(y)$ to be a polynomial. Taking derivatives of this wavefunction gives ${\psi }^{\prime }=(h^{\prime }-hy)e^{-y^2/2}$ and ${\psi }^{\prime \prime }=(h^{\prime \prime }-2h^{\prime }y+h{y}^2-h)e^{-y^2/2}$ so the Schrödinger equation (2.30) becomes

${\displaystyle \frac{d^{2}h}{d{y}^2}}-2y{\displaystyle \frac{dh}{dy}}+(\tilde{E}-1)h=0$

(2.33)

You can check that this is indeed satisfied by our earlier solution (2.31) with $h=1$. To find the general solution, we take the polynomial ansatz

$h(y)={\displaystyle \sum _{p=0}^{\mathrm{\infty }}}{a}_p{y}^p$

Here $p$ is just a dummy summation index; do not confuse it with momentum! Plugging this ansatz into (2.33) gives a relation between the coefficients,

	$(p+2)(p+1)a_{p+2}-2p{a}_p-(\tilde{E}-1)a_p=0$
	$\mathrm{ }\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\Rightarrow \mathit{\hspace{1em}\hspace{1em} }{a}_{p+2}={\displaystyle \frac{2p-\tilde{E}+1}{(p+2)(p+1)}}{a}_p$		(2.34)

All we have to do is solve this simple recursion relation.

First note, that the recursion relation involves two independent sets of coefficients: $a_p$ with $p$ even, and $a_p$ with $p$ odd. These two sets don’t talk to each other and so we have two classes of solutions. We’ll see the interpretation of this shortly.

Next, let’s look at what happens for large $p$. There are two options: either the recursion relation terminates, so that $a_p=0$ for all $p>n$ for some $n$. Or the recursion relation doesn’t terminate and $a_p\ne 0$ for all $p$. We’re going to argue that only the first option is allowed. If the recursion relation keeps going forever, the resulting wavefunction will be non-normalisable.

Why is this? If the recursion relation doesn’t terminate then, for very large $p$, we have $a_{p+2}\approx 2a_p/p$. But this is the kind of expansion that we get from an exponentially growing function. To see this, look at

The appropriate recursion relation for this exponential function is then

$b_{p+2}={\displaystyle \frac{(p/2)!}{(p/2+1)!}}{b}_n={\displaystyle \frac{1}{p/2+1}}{b}_p⟶{\displaystyle \frac{2}{p}}{b}_p\mathit{\hspace{1em}}\mathrm{as}p\to \mathrm{\infty }$

The upshot of this argument is that, if the recursion relation (2.34) fails to terminate then, for large $|y|$, the wavefunction actually looks like $\psi (y)=h(y)e^{-y^2/2}\to e^{+y^2}{e}^{-y^2/2}=e^{+y^2/2}$

The emergence of such solutions isn’t entirely unexpected. We know that, at large $y$, both $e^{-y^2/2}$ and $e^{+y^2/2}$ are equally valid solutions. Our preference for the former over the latter is for physical reasons, but the power-law ansatz that we’re playing with has no knowledge of this preference. Therefore it’s no surprise that it offers up the $e^{+y^2/2}$ solution as an option. Nonetheless, this doesn’t change the fact that it’s not an option we want to make use of. The $e^{+y^2/2}$ wavefunctions are non-normalisable and therefore illegal. Moreover, unlike the more straightforward $\psi =e^{ikx}$ wavefunctions that we met in the last section, there’s no redemption for wavefunctions that grow exponentially quickly. They are not of interest and should be discarded.

All of which is to say that we should be looking for solutions to (2.34) for which the sequence $a_p$ terminates. This means that there must be some positive integer $n$ for which

$2n-\tilde{E}+1=0$

(2.35)

But this is the spectrum of the theory that we were looking for! The allowed energy states of the harmonic oscillator take the form

${\tilde{E}}_n=1+2n\mathit{\hspace{1em}\hspace{1em}}\mathrm{with}n=0,1,2,\mathrm{\dots }$

Recalling the scaling (2.29), the energies are

$E_n=\mathrm{\hslash }\omega \left({\displaystyle \frac{1}{2}}+n\right)\mathit{\hspace{1em}\hspace{1em} }\mathrm{with}n=0,1,2,\mathrm{\dots }$

(2.36)

All energies are proportional to $\mathrm{\hslash }\omega$, with $\omega$ the frequency of the harmonic oscillator. Furthermore the states are equally spaced, with

$E_{n+1}-E_n=\mathrm{\hslash }\omega$

The next few (unnormalised) wavefunctions are

	$\psi =2y{e}^{-y^2/2}$	$\Rightarrow$	$\mathrm{ }\mathit{\hspace{1em}}{E}_1={\displaystyle \frac{3}{2}}\mathrm{\hslash }\omega$
	$\psi =(-2+4y^2)e^{-y^2/2}$	$\Rightarrow$	$\mathrm{ }\mathit{\hspace{1em}}{E}_2={\displaystyle \frac{5}{2}}\mathrm{\hslash }\omega$
	$\psi =(-12y+8y^3)e^{-y^2/2}$	$\Rightarrow$	$\mathrm{ }\mathit{\hspace{1em}}{E}_3={\displaystyle \frac{7}{2}}\mathrm{\hslash }\omega$

These, together with the Gaussian, are shown in Figure 9. In general, the functions $h(y)$ are known as Hermite polynomials and have a number of nice properties.

We can now return to one issue that we left hanging. The recursion relation (2.34) does not relate $a_p$ with $p$ even to to those with $p$ odd. This manifests itself in the solutions above where the polynomials contain only even powers of $y$ or only odd powers of $y$. Correspondingly, the two classes of solutions that we anticipated earlier are simply even or odd functions, with

			$\psi (y)=\psi (-y)\mathit{\hspace{1em}\hspace{1em} }\mathrm{when}{a}_p\ne 0\text{for }p\text{ even}$
	$\mathrm{and}$		$\psi (y)=-\psi (-y)\mathit{\hspace{1em}\hspace{1em} }\mathrm{when}{a}_p\ne 0\text{for }p\text{ odd}$

There is a general lesson here. Whenever the potential is an even function, meaning $V(x)=V(-x)$, then the energy eigenstates will arrange themselves into even and odd functions. Underlying this is the concept of parity symmetry, which is the statement that the physics is unchanged under $x\to -x$. We’ll make use of this idea of parity symmetry later in these lectures.

At first glance, the wavefunctions that we’ve found don’t seem to capture much of the familiar classical physics of a particle bouncing back and forth in a potential. Because they’re all stationary states, the time dependence is simply an overall phase $e^{-iEt/\mathrm{\hslash }}$ in front of the wavefunction. You can compute the average position and momentum in any of the states above and you will find $⟨x⟩=⟨p⟩=0$. In some sense, this is what you expect because it is also the average behaviour of the classical solution! Still, it would reassuring if we could see some remnant of our classical intuition in these wavefunctions.

A general property of quantum systems is that they tend to look more classical as you go to higher energies. For example, the discretisation effects may be less noticeable if they’re small compared to the overall energy of the system. For the harmonic oscillator, the wavefunctions for the ${20}^{\mathrm{th}}$ and ${60}^{\mathrm{th}}$ excited states are shown in Figure 10. Although it may not be obvious, some key elements of the classical trajectories can be seen hiding in these wavefunctions.

First, look at the way the wavefunction oscillates. We know that a free particle with definite momentum $p=\mathrm{\hslash }k$ is associated to $e^{ikx}$. The bigger $k$, the smaller the wavelength, and the higher the momentum. In the wavefunctions shown in Figure 10, you can see that the wavelength of oscillations is much smaller near the bottom of the potential and then gets stretched towards the edges. This coincides with the classical expectation, where the particle is travelling much faster at the bottom of the potential and slows as it rises.

The way to quantify this idea uses a technique known as the WKB approximation. We’ll discuss this in more detail in the lectures on Topics in Quantum Mechanics, but here we just describe the basic idea which is enough to extract the physics that we care about. We write the Schrödinger equation equation as

$-{\displaystyle \frac{{\mathrm{\hslash }}^2}{2m}}{\displaystyle \frac{d^2\psi }{d{x}^2}}=\left(E-{\displaystyle \frac{1}{2}}m{\omega }^2{x}^2\right)\psi$

The idea is that, for large $E$, we might be able to think of the $E-\frac{1}{2}m{\omega }^2{x}^2$ as roughly over some small region of $x$. We then get two different kinds of behaviour: if then the wavefunction oscillates, approximately as $e^{ikx}$ for some $k$. Alternatively, when $x^2>2E/m{\omega }^2$, the wavefunction drops off as $e^{-k|x|}$ for some $k$. The WKB approximation builds on this intuition by looking for solutions where $k$ itself is varies with $x$, so $k=k(x)$. This, of course, is what’s seen in the wavefunctions plotted in Figure 10.

One consequence of this is that we expect that, as $E$ increases, the wavefunctions extend further out. Indeed, the $n=60$ wavefunction extends out further than the $n=20$ wavefunction. Mathematically this follows simply because there are higher powers of $y$ in the polynomial $h(y)$. The kind of ideas sketched above can be used to show that, for large $E$, the final turning point of the wavefunction occurs at

$x_{\max }^2\approx {\displaystyle \frac{2E}{m{\omega }^2}}$

This is precisely the turning point of the classical particle, where the kinetic energy in (2.26) vanishes.

Finally, look at the overall envelope of the wavefunction, a curve that peaks at the edges and dips in the middle. This is telling us that if you do a measurement of a particle in this high energy state, you’re more likely to find it near the edges than near the origin. But this, too, is the same as the classical picture. If you take a photograph of a ball oscillating in a harmonic potential, you’re most likely to catch it sitting towards the end of its trajectory, simply because it’s going slower and so spends more time in those regions. Conversely, the ball is less likely to be caught as it whizzes past the origin. This is, again, captured by the form of the wavefunction. We see that the quantum world is not completely disconnected from the classical. You just have to know where to look.

Clearly the potential is discontinuous at $x=\pm a$ and this raises the question of what kind of wavefunction $\psi (x)$ we should be looking for. The answer is that $\psi (x)$ itself should be continuous, as too should ${\psi }^{\prime }(x)$. But ${\psi }^{\prime \prime }(x)$ inherits the discontinuity of the potential.

To see the statements above, first integrate the Schrödinger equation (2.38) over a small interval at $x=a$,

			$-{\displaystyle \frac{{\mathrm{\hslash }}^2}{2m}}{\displaystyle {\int }_{a-ϵ}^{a+ϵ}}𝑑x{\displaystyle \frac{d^2\psi }{d{x}^2}}={\displaystyle {\int }_{a-ϵ}^{a+ϵ}}𝑑x(E-V(x))\psi$
	$\Rightarrow$		${{\displaystyle \frac{d\psi }{dx}}|}_{a+ϵ}-{{\displaystyle \frac{d\psi }{dx}}|}_{a-ϵ}=-{\displaystyle \frac{2m}{{\mathrm{\hslash }}^2}}{\displaystyle {\int }_{a-ϵ}^{a+ϵ}}𝑑x(E-V(x))\psi$

Although $V(x)$ is discontinuous at $x=a$, it is finite. This means that the integral on the right-hand side vanishes as we take limit $ϵ\to 0$, telling us that ${\psi }^{\prime }$ is continuous at $x=a$. But if ${\psi }^{\prime }(x)$ is continuous then so too is $\psi (x)$ itself. Returning to the Schrödinger equation (2.38), we then see that the discontinuity in $V(x)$ can only show up in the second derivative ${\psi }^{\prime \prime }(a)$.

Now our strategy for solving the Schrödinger equation is clear: we find solutions inside and outside the well and then patch them together, making sure that both $\psi$ and ${\psi }^{\prime }$ are continuous at the join.

Before we proceed, there is one last idea that will make our life easier. This is the idea of parity, which comes from the observation that the potential is an even function with $V(x)=V(-x)$. This means that all solutions to the Schrödinger equation will be either even functions or odd functions. (We saw this in the example of the harmonic oscillator.) To see why this is necessarily the case, first note that if $\psi (x)$ solves the Schrödinger equation for some value of $E$, then so too does $\psi (-x)$. Under the assumption that there aren’t two different wavefunction with the same energy (a so-called non-degenerate spectrum), we must have

$\psi (x)=\alpha \psi (-x)$

for some $\alpha \in 𝐂$. But we have

$\psi (x)=\psi (-(-x))=\alpha \psi (-x)={\alpha }^2\psi (x)$

which tells us that $\alpha =\pm 1$, corresponding to either odd or even wavefunctions. (It’s possible to extend this proof even in the case of degenerate spectra but this won’t be needed for the examples that we’ll consider here.)

As we’ll now see, knowing in advance that we’re looking for even or odd solutions to the Schrödinger equation greatly simplifies our task of finding solutions. We’ll look for each in turn.

Inside the potential well, the Schrödinger equation reads

This is, of course, once again the Schrödinger equation for a free particle, just with shifted energy. As before, there are two kinds of solutions:

• •

  Solutions with $E>-V_0$ have wavefunctions $e^{\pm ikx}$.

• •

  Solutions with have wavefunctions $e^{\pm {\eta }^{\prime }x}$.

It turns out that the former solutions are the ones of interest and all our bound states will have energies . This is perhaps not surprising give that $-V_0$ is the lowest value of the potential.

Because we’re looking for parity even solutions, we should consider $\psi =e^{+ikx}+e^{-ikx}$ or,

$\psi (x)=B\cos kx$

(2.41)

where $B$ is again a normalisation constant and $k$ is, like $\eta$, related to the energy, now with

$E=-{\displaystyle \frac{{\mathrm{\hslash }}^2{\eta }^2}{2m}}={\displaystyle \frac{{\mathrm{\hslash }}^2{k}^2}{2m}}-V_0$

(2.42)

Our next step is to patch the exponentially decaying solutions (2.40) outside the well with the oscillatory solution (2.41) inside the well. Because both solutions are even functions, we only have to do this patching once at $x=+a$; the solution will then be automatically patched at $x=-a$ as well. We have

	$\psi (x)\text{cts at }x=a$	$\Rightarrow$	$\mathrm{ }\mathit{\hspace{1em}\hspace{1em}}B\cos ka=A{e}^{-\eta a}$
	${\psi }^{\prime }(x)\text{cts at }x=a$	$\Rightarrow$	$\mathrm{ }-kB\sin ka=-\eta A{e}^{-\eta a}$

There is no simple solution to this transcendental equation, but it’s not difficult to understand the property of the solutions using graphical methods. In Figure 11 we first plot the graph $\eta =k\tan ka$ in blue, and then superpose this with the circle

$k^2+{\eta }^2={\displaystyle \frac{2m{V}_0}{{\mathrm{\hslash }}^2}}$

shown in red. We restrict to the range $\eta \ge 0$ as befits our normalisable wavefunction. We get a solution whenever the red curve intersects the blue one. As expected, there are a only discrete solutions, happening for specific values of $\eta$. Moreover, there are also only a finite number of them.

We see that the first solution is guaranteed: no matter how small the radius of the circle, it will always intersect the first blue line. However, the number of subsequent solutions depends on the parameters in the game. We can see that the number of solutions will grow as increase $V_0$ the depth of the well, since this increases the radius of the red circle. The number of parameters will also grow as we increase the width of the well; this is because the separation between the blue lines is determined by the divergence of the tan function, so occurs when $k\approx \pi /2a$. As we increase $a$, keeping $V_0$ fixed, the blue lines get closer together while the red circle stays the same size.

We can be more precise about this. From the graph, the $n^{\mathrm{th}}$ crossing occurs somewhere in the region

giving an estimate of energy of the $n^{\mathrm{th}}$

$E_n={\displaystyle \frac{{\mathrm{\hslash }}^2{k}_n^2}{2m}}-V_0$

In the limit of an infinite well, we have $V_0\to \mathrm{\infty }$ so the red circle becomes infinitely large, intersecting the blue lines only asymptotically where $k_{n}a=(n-1/2)\pi$. Clearly all energies descend to $-\mathrm{\infty }$ in this limit, but we can instead measure energies with respect to the floor of the potential. We then get

$E+V_0\to {\displaystyle \frac{{\mathrm{\hslash }}^2{\pi }^2{(2n-1)}^2}{8m{a}^2}}$

But we’ve met this result before: it coincides with the energy spectrum (2.18) of a particle in an infinite well. (To see the agreement, we have to note that $L=2a$ and, furthermore, remember that we have restricted to even parity states only which is why we’ve got only odd integers $(2n-1)$ in the numerator.)

There is one last mild surprise in our analysis. All our bound states have energy

This is what we would expect for a classical particle, trapped inside the well. The quantum novelty is that the wavefunction itself is not restricted only to the well: it leaks out into the surrounding region $|x|>a$, albeit with an exponentially suppressed wavefunction (2.40). This means that there is some finite probability to find the particle outside the well, in a region that would be classically inaccessible.