This section is about general features of the solutions to the one-dimensional Time Independent Schrödinger Equation (TISE) \[-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi(x)}{\mathrm{d}x^2} + V(x)\psi(x) = E\psi(x)\] Recall that the TISE is an eigenvalue equation $\hat{H}|\psi\rangle=E|\psi\rangle$ with the Hamiltonian, written in the position representation \[\hat{H}\rightarrow -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x} + V(x)\]

We have seen from the previous sections that the set of possible values of $E$, which is called the energy spectrum of $\hat{H}$, can be continuous (as for the free particle) or discrete (as for the infinite potential well). It can also be degenerate (as for the free particle) or nondegenerate (as for the infinite potential well). The solutions to the TISE for a free particle $Ae^{ipx/\hbar}$ are unnormalizable, and so they are not physically realizable states. In order to get a physically realizable state, we have to take superpositions of them to form a normalizable wavefunction, i.e. $\psi(x) = \frac{1}{\sqrt{2\pi\hbar}}\int_{\infty}^{+\infty}\phi(p)e^{ipx/\hbar}\,\mathrm{d}p$ for a sufficiently smoothly varying $\phi(p)$. This means that the physically realizable states of a free particle do not have a definite energy (or momentum); there is always at least a small uncertainty in their energy. On the other hand, the stationary states of the infinite potential well

\[\psi(x)=\begin{cases} 0, & x\leq 0 \\ \sqrt{\frac{2}{a}} \sin\left ( \frac{n\pi x}{a}\right ), & 0 < x < a \\ 0, & x\geq a, \end{cases}\qquad n=1,2,3,\cdots\]

are normalizable, so the definite energy states for this system are all physically realizable.

If I give you a specific Hamiltonian $\hat{H}$, and an energy $E$, can you tell, without going through the whole process of solving the TISE, which of these features the solutions will have? We will discuss the answers to these questions in this section. After doing so, we will discuss how symmetry properties of the potential $V(x)$ can be used to simplify the process of solving the TISE.

The TISE is a second order differential equation, so, for a specific $E$, its general solution is the sum of two independent solutions, i.e. it is of the form $\psi(x) = A\psi_1(x)+B\psi_2(x)$ for specified functions $\psi_1(x)$ and $\psi_2(x)$ and arbitrary constants $A$ and $B$. This means that the general solution, for a given energy $E$, spans a two-dimensional subspace. Without any additional constraints, each energy would have a two-fold degeneracy. However, there typically are additional constraints, which, depending on the form of the potential $V(x)$ and the value of $E$, may remove the degeneracy. Sections 4.iii.3 and 4.iii.4 discuss these constraints, and we will use them in section 4.iii.5 to determine the features that solutions of the TISE have for generic potentials. Section 4.ii.6 then discusses how symmetry can be used to simplify the process of solving the TISE. Before discussing all of this, we will prove an important property of the solutions to the TISE in section 4.ii.2.

The time dependent Schrödinger equation \[i\hbar\frac{\partial\Psi(x,t)}{\partial{t}}=\hat{H}\Psi(x,t)\] involves the imaginary unit $i$, so we expect its solutions to be complex-valued functions. When we solve it by separation of variables to find the stationary states, we find that they are of the form $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$, where $\psi(x)$ is a solution to the TISE with energy $E$. So, indeed $\Psi(x,t)$is a complex valued function due to the factor $e^{-iEt/\hbar}$. However, the solutions to the TISE itself, $\psi(x)$, can always be chosen to be real.

The words “chosen to” are important here because the TISE does have complex valued solutions, so let's state things a bit more carefully.

The Hamiltonian $\hat{H}$ is a Hermitian operator so, if it is nondegenerate, its eigenstates, $\psi_n(x)$, form a complete orthonormal basis. If we multiply an eigenstate by a complex constant $A$, then $A\psi_n(x)$ is still the same physical state. In this case, what we mean by “chosen to be real” is that the eigenstates are all of the form $A\psi_n(x)$ where $\psi_n(x)$ is a real-valued function and $A$ is a complex constant. Now, if we write $A = Re^{i\phi}$ then the phase $\phi$ can be chosen arbitrarily due to the global phase ambiguity, so we can always choose to only work with real solutions by setting $\phi=0$. We would then set $R$ to ensure that the appropriate normalization conditions hold.

If $\hat{H}$ is degenerate then, as we have seen, each energy value $E$ can have at most a two-fold degeneracy. The two independent solutions, $\psi_1(x)$ and $\psi_2(x)$, may be complex valued, as they are for the free particle, but any linear combination of them $\alpha\psi_1(x)+\beta\psi_2(x)$ is still an eigenstate of $\hat{H}$ with energy $E$. We can always find two orthogonal linear combinations, $\phi_1(x)$ and $\phi_2(x)$ that are real valued, so we might as well work with those eigenstates instead.

All this means that we do not ever need to consider complex-valued solutions to the TISE, although in the degenerate case it can sometimes be useful to work with them. This is nice because it puts us in the realm of solving a differential equation for a real-valued function, which is more familiar and simpler than the general complex-valued case.

To prove these results, consider a (possibly complex valued) solution to the TISE $\psi(x)$ with energy $E$ and write it as $\psi(x) = \psi_{\mathrm{r}}(x)+i\psi_{\mathrm{im}}(x)$, where $\psi_{\mathrm{r}}(x)$ and $\psi_{\mathrm{im}}(x)$ are the real and imaginary parts of $\psi(x)$ and hence are both real valued functions. If we substitute this into the TISE, we get

\[-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi_{\mathrm{r}}(x)}{\mathrm{d}x^2} + V(x)\psi_{\mathrm{r}}(x) + i \left [ -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi_{\mathrm{im}}(x)}{\mathrm{d}x^2} + V(x)\psi_{\mathrm{im}}(x) \right ]= E\psi_{\mathrm{r}}(x) + iE\psi_{\mathrm{im}}(x).\]

For this equation to hold, both the real and imaginary parts of each side must be equal, so we have

\[-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi_{\mathrm{r}}(x)}{\mathrm{d}x^2} + V(x)\psi_{\mathrm{r}}(x) = E\psi_{\mathrm{r}}(x), \qquad\quad -\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi_{\mathrm{im}}(x)}{\mathrm{d}x^2} + V(x)\psi_{\mathrm{im}}(x) = E\psi_{\mathrm{im}}(x)\]

and hence both $\psi_{\mathrm{r}}(x)$ and $\psi_{\mathrm{im}}(x)$ are also solutions to the TISE with energy $E$. Note that this works because there are no complex numbers in the TISE, so it does not couple $\psi_{\mathrm{r}}(x)$ and $\psi_{\mathrm{im}}(x)$.

Now we have two cases to deal with. If the energy $E$ is non-degenerate then there is only one linearly independent energy eigenstate for that $E$. Therefore, we must have $\psi_{\mathrm{im}}(x) = A\psi_{\mathrm{r}}(x)$ for some real constant $A$, so they are both the same solution and we can take either one of them as the energy eigenstate.

On the other hand, if $E$ is two fold degenerate and we apply the same procedure to two independent complex valued solutions, $\psi_1(x)$ and $\psi_2(x)$ then we will end up with four real-valued solutions: $\psi_{1\mathrm{r}}(x)$, $\psi_{1\mathrm{im}}(x)$, $\psi_{2\mathrm{r}}(x)$ and $\psi_{2\mathrm{im}}(x)$. Since we only have a two-fold degeneracy, only two of these can be linearly independent, so pick any two linearly independent ones, form linear combinations that are orthogonal, and then we will have an orthogonal basis for the energy eigenspace consisting of real valued functions.

Recall that $\left|\psi(x)\right|^2$ is the probability density, which means that the probability of finding the particle in the interval $a<x<b$ is

\[P(a<x<b)=\int_a^b|\psi(x)|^2 \,\mathrm{d}x.\]

In particular, we must have $\int_{-\infty}^{+\infty}|\psi(x)|^2\,\mathrm{d}x = 1$, so if we have a solution $\psi(x) = A\psi_1(x)$ then $\int_{-\infty}^{+\infty}|\psi_1(x)|^2\,\mathrm{d}x$ must be finite, so that we can choose an appropriate normalization constant $A$. We call such a solution normalizable. Since we need to have a valid probability density, all physically realizable solutions must be normalizable.

Having non-normalizable solutions to the TISE is not a deal-breaker. Although it means that the stationary states are not themselves physically realizable, superpositions of them may well be, as is the case for the free particle. If you have non-normalizable solutions then all that really tells you is that the physically realizable solutions do not have a definite energy, i.e. there must be a nonzero uncertainty $\Delta E$.

However, if, in addition to being non-normalizable, we have solutions that diverge as $x\rightarrow \infty$ or $x\rightarrow -\infty$ then we can definitely discard those solutions. For example, if $\psi(x)\approx Ae^{\alpha x}$ for $x\gg 0$ and $\alpha > 0$ then the probability density $|\psi(x)|^2\approx |A|^2e^{2\alpha x}$ also diverges as $x\rightarrow \infty$. This means that the particle is overwhelmingly likely to be at $x=\infty$, so we would never be able to observe the particle. Unlike the case of the free particle, we cannot construct physically realizable solutions by taking superpositions of divergent solutions. If we try to superpose wavefunctions that all behave like $e^{\alpha x}$ for $x\gg 0$ and positive constants $\alpha$ then the result will still diverge as $x\rightarrow \infty$. Since we cannot make physical states by superposing such divergent states, we can just reject them.

So, if the general solution to the TISE $\psi(x) = A\psi_1(x)+B\psi_2(x)$ happens to have a divergent $\psi_1(x)$ and a non-divergent $\psi_2(x)$ then we can just discard $\psi_1(x)$ as unphysical and only consider the solution $\psi(x)=B\psi_2(x)$. This removes the degeneracy.

Solutions to the TISE have to satisfy boundary conditions and, depending on the potential $V(x)$ and energy $E$, these can remove degeneracy, make the energy spectrum discrete, and make the solutions normalizable. For example, all three of these things happen for the infinite square well potential. A lot of the physics is determined by the boundary conditions.

The boundary conditions that solutions to the TISE must satisfy are:

1. At every point $x$, $\psi(x)$ must be continuous: $\lim_{\epsilon \rightarrow 0_+} \psi(x+\epsilon) = \lim_{\epsilon\rightarrow 0_+}\psi(x-\epsilon)$.
2. At every point $x'$ where $V(x')$ is not infinite, $\left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'}$ must be continuous: $\lim_{\epsilon\rightarrow 0_+}\left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'+\epsilon}=\lim_{\epsilon \rightarrow 0_+}\left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'-\epsilon}$.

The notation $\lim_{\epsilon\rightarrow 0_+}$ means that we take the limit from the positive side, i.e. we start with a small positive value of $\epsilon > 0$ and then consider what happens as it gets smaller and smaller. Similarly, the notation $\lim_{\epsilon\rightarrow 0_-}$ means taking the limit from the negative side with $\epsilon < 0$.

The free particle solutions satisfy both of these conditions everywhere, so the boundary conditions do not constrain these solutions further. For the infinite square well potential, we started with the general solution to the TISE $\psi(x)=A\sin kx + B\cos k x$ and then applied the first boundary condition. We noted that $\psi(x)$ must be zero in the regions of infinite potential to avoid our solutions having infinite energy. Then, the first boundary condition implies that the solutions must go to zero at the ends of the well, so we applied $\psi(0)=\psi(a)=0$. This did two things. First, it gave us $B=0$, which removes the degeneracy. We now only have one independent solution $A\sin k$ for each energy. The second thing it did was to constrain the possible values of $k$ to be $\frac{n\pi}{a}$ for $n=1,2,\cdots$, which gave us a discrete energy spectrum. Both the nondegeneracy and the discreteness come from the boundary conditions. The fact that the solutions have to be zero outside the potential also made the solutions normalizable.

For the infinite square well potential, the second boundary condition holds inside the well, but we did not apply it at $x=0$ or $x=a$ because the potential jumps to infinity at these points. The solutions

\[ \psi(x)=\begin{cases} 0, & x\leq 0 \\ \sqrt{\frac{2}{a}}\sin\left ( \frac{n\pi x}{a}\right ), & 0<x<a \\ 0, & x\geq a, \end{cases}\]

do not have a continuous derivative at $x=0$ or $x=a$. For example,

\[\lim_{\epsilon \rightarrow 0_+} \left . \frac{\mathrm{d}\psi}{\mathrm{d}x}\right |_{x=\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{n\pi}{a}\cos \frac{n\pi\epsilon}{a} = \frac{n\pi}{a},\]

\[\lim_{\epsilon \rightarrow 0_-} \left . \frac{\mathrm{d}\psi}{\mathrm{d}x}\right |_{x=\epsilon} = \lim_{\epsilon \rightarrow 0} 0 = 0.\]

In fact, there are no solutions with finite energy that have continuous derivatives at the two ends of the well.

Although, in this case, we just ignored the second boundary condition at points where $V(x)$ jumps to infinity, in general there are boundary conditions on the derivatives at such points. It is just that these conditions are different from continuity. How do we determine what boundary conditions to apply to the derivative in these cases?

First, note that potentials that have infinite values are an idealization, and do not really occur in nature. Even the Coulomb potential $V(r)=-\frac{e^2}{4\pi\epsilon_0r}$, which has a singularity at $r=0$, is an idealization because the proton is not really a point particle, but has its charge density spread out over a small region, and this means that the singularity is not present in the real world. So, to deal with idealized potentials that have infinite values, we can always just solve a more realistic potential that does not have infinite values and then take the appropriate limit to get the solutions to the idealized potential. We would then just apply continuity for the derivative when solving the TISE as usual. We will see this method in action when we solve the finite square well potential in section 4.vii, showing that the solutions for the infinite square well emerge in the limit, and this will justify our decision to impose no boundary condition on the derivative in that case.

However, the whole point of considering idealized potentials is that they are usually easier to solve than more realistic ones. Often, we can figure out what the boundary condition on the derivative ought to be without going through a limiting procedure. We will deal with this on a case-by-case basis, and see an example of how to do this for the delta function potential in section 4.vi.

Given that the boundary conditions determine a lot of the physics of the system, it would be nice to give a rigorous argument for why they have to hold. Doing so in a way that would make a mathematician happy is a bit beyond the scope of the course, but we can give some intuition.

If the potential $V(x)$ is a continuous function then the theory of differential equations tells us that $\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}$ also has to be continuous. In order for this to be true, $\frac{\mathrm{d}\psi}{\mathrm{d}x}$ must be continuous and differentiable, otherwise $\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}$ would not be well defined. But this means that $\psi(x)$ must also be continuous and differentiable, because otherwise $\frac{\mathrm{d}\psi}{\mathrm{d}x}$ would not be well defined. If this were the only situation we had to deal with then the two boundary conditions would always hold everywhere.

However, we want to also deal with those pesky idealized potentials, which may have discontinuities and/or infinite values. The second derivative $\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}$ can then have the same type of discontinuities or infinite values because they can cancel each other out in the expression $-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi(x)}{\mathrm{d}x^2} + V(x)\psi(x)$. The conditions we have to impose on the second derivative will then depend on just how pathological we allow the potential to be.

In this course, we will only deal with three types of discontinuity in the potential: a jump discontinuity, a delta function, and a jump to infinity. We have already dealt with the jump to infinity in the case of the infinite square well potential ($\psi(x)$ must be continuous and zero at these points, with no additional condition on the derivative), so let's deal with the other two cases in turn.

At a jump discontinuity, the potential is finite, but takes two different values depending on whether we take a limit from the left or the right. There is a jump discontinuity at $x'$ if $\lim_{\epsilon\rightarrow 0_+}V(x'+\epsilon)$ and $\lim_{\epsilon\rightarrow 0_+}V(x' - \epsilon)$ both exist, but $\lim_{\epsilon\rightarrow 0_+}V(x'+\epsilon) \neq \lim_{\epsilon\rightarrow 0_+}V(x'-\epsilon)$. This is illustrated below.

In this case, $\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}$ can also have a jump discontinuity at $x'$. Since $\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}$ is the derivative of $\frac{\mathrm{d}\psi}{\mathrm{d}x}$, $\frac{\mathrm{d}\psi}{\mathrm{d}x}$ can have a sudden change of slope at $x'$, but it will still be continuous. If it is continuous then $\psi(x)$ will be both continuous and differentiable, so both boundary conditions hold in this case.

For the delta function case, consider a potential of the form $V(x)=f(x)+a\delta(x-x')$, where $f(x)$ is a continuous function and $a$ is a constant. The second derivative will then be of the form $\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}=g(x)+b\delta(x-x')$, where $g(x)$ is a continuous function and $b$ is a constant. We will then have

\[\frac{\mathrm{d}\psi}{\mathrm{d}x}=\int g(x)\,\mathrm{d}x+b\int\delta(x-x')\,\mathrm{d}x=G(x)+b\theta(x-x')+C,\]

where $G(x)$ is the indefinite integral of $g(x)$, $C$ is an integration constant, and

\[\theta(x-x')=\begin{cases} 0, & x < x' \\ 1, & x\geq x',\end{cases}\]

is the Heaviside step-function (you should convince yourself that the delta function is the derivative of the step function). This means that $\frac{\mathrm{d}\psi}{\mathrm{d}x}$ has a jump discontinuity at $x'$. This is why we do not apply the second boundary condition at points where $V(x)$ is infinite because $\frac{\mathrm{d}\psi}{\mathrm{d}x}$ should have a discontinuity there. However, since it is only a jump discontinuity, $\psi(x)$ will have a sudden change of slope at $x'$ but is still continuous there, so we do apply the first boundary condition.

Finally, since the first boundary condition always holds in the cases we are considering, we can use it to figure out what happens to the second boundary condition at points where the potential has a delta function spike. If we take the TISE and integrate it over a small slice centered at $x=x'$, we get:

\[\int_{x'-\epsilon}^{x'+\epsilon}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}\,\mathrm{d}x=\frac{2m}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x) \,\mathrm{d}x-\frac{2mE}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon} \psi(x)\,\mathrm{d}x\]

Note that $\int_{x'-\epsilon}^{x'+\epsilon}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}\,\mathrm{d}x= \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'-\epsilon}$, so taking the limit of the right hand side as $\epsilon \rightarrow 0_+$ will give us the boundary condition on the derivative. Now, if $\psi(x)$ is continuous then the second integral on the right hand side vanishes, so we are left with

\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'-\epsilon}\right ]=\frac{2m}{\hbar^2} \lim_{\epsilon \rightarrow 0_+} \left [ \int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x)\,\mathrm{d}x \right ]\]

We will see how this can be used to determine the boundary condition for a delta function potential in section 4.vi.

In this section, we illustrate the features we expect energy eigenstates to have for general potentials. We will start with a piece-wise constant potential, as illustrated below, where we can see what happens explicitly, before moving on to something more generic.

Note that the potential has a minimum $V_0$ and values $V_+$ and $V_-$, which extend to $x = +\infty$ and $x = -\infty$ respectively. The features that the energy eigenstates have depend on how the energy $E$ is related to these three values.

It is helpful to start by rewriting the TISE in the form

\[\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} - k(x)^2\psi(x)=0,\]

where $k(x)=\sqrt{\frac{2m}{\hbar^2}\left (V(x) - E\right )}$. Let's call $x < a$ the $-$ region, $a \leq x \leq b$ region $0$, and $x>b$ the $+$ region. The function $k(x)$ is constant on these three regions with values:

\[k_- = \sqrt{\frac{2m}{\hbar^2} \left ( V_- - E\right )},\qquad k_0= \sqrt{\frac{2m}{\hbar^2} \left ( V_0-E\right )},\qquad k_+ = \sqrt{\frac{2m}{\hbar^2} \left ( V_+-E\right )}.\]

On each of these three regions we therefore have a linear, homogeneous, second order differential equation, which we can solve with the standard techniques. The characteristic equation is $m^2 - k^2=0$, where $k=k_-,k_0,k_+$ on each of the three regions, so $m = \pm k$ and the general solution is

\[\psi(x)=\begin{cases} \psi_-(x) = Ae^{k_-x} + Be^{-k_-x}, & x<a \\ \psi_0(x)=Ce^{k_0x} + De^{-k_0 x}, & a\leq x \leq b \\ \psi_+(x) = Fe^{k_+ x} + Ge^{-k_+x}, & x > b.\end{cases}\]

We next have to solve the boundary conditions at at $x=a$ and $x=b$ and check whether there are any non-divergent solutions that satisfy them. This will depend on which of the three $k_-, k_0, k_+$ are real or imaginary, which in turn depends on how $E$ is related to $V_-$, $V_0$ and $V_+$. We will deal with each possible case in turn.

If $E < V_0$ then all of $k_-, k_0, k_+$ are real and positive. This means that we can discard the terms $Be^{-k_-x}$ and $Fe^{k_+ x}$ because they diverge as $x\rightarrow -\infty$ and $x\rightarrow +\infty$ respectively, so the general solution reduces to

\[\psi(x)=\begin{cases} \psi_-(x) = Ae^{k_-x} , & x<a \\ \psi_0(x)=Ce^{k_0x} + De^{-k_0 x}, & a\leq x \leq b \\ \psi_+(x) = Ge^{-k_+x}, & x>b. \end{cases}\]

We next have to solve the boundary conditions at $x=a$ and $x=b$, but it turns out that there are no solutions. The algebra is a bit messy, so we will not go through the details, but the boundary conditions end up telling you that one of $k_-, k_0, k_+$ has to be imaginary, which is a contradiction.

This is consistent with the intuition from classical mechanics, since the particle would have to have a negative kinetic energy everywhere to exist in this potential, which is not possible.

In this case, $k_-$ and $k_+$ are real and positive, but $k_0$ is imaginary. Define $k_0' = -ik_0$ so that $k_0'= \sqrt{\frac{2m}{\hbar^2} \left ( E -V_0\right )}$ is real and positive. As before, we can discard the terms that diverge as $x\rightarrow\pm\infty$ and we are left with the solution

\[ \psi(x)=\begin{cases} \psi_-(x) = Ae^{k_-x} , & x<a \\ \psi_0(x)=Ce^{ik_0'x} + De^{-ik_0' x}, & a\leq x \leq b \\ \psi_+(x) = Ge^{-k_+x}, & x>b.\end{cases}\]

These solutions tend to zero exponentially as $x\rightarrow\pm\infty$ in the $-$ and $+$ regions, which implies that they are normalizable, and oscillate sinusoidally in region $0$. States of this form are called bound because, apart from the exponential tails, they are confined to region $0$, and hence bound to the potential well.

In this case, the boundary conditions at $x=a$ and $x=b$ do have solutions. There are four unknown constants—$A$, $C$, $D$ and $G$—but one of them will be determined by the normalization condition $\int_{-\infty}^{+\infty}|\psi(x)|^2\,\mathrm{d}x = 1$, leaving three constants to be determined. The boundary conditions give us four equations: continuity of $\psi(x)$ and of $\frac{\mathrm{d}\psi}{\mathrm{d}x}$ at both $x=a$ and $x=b$, so we have an over-determined system: four equations determining three constants. These will not have a consistent solution for a generic $E$.

However, if we also view $E$ as an undetermined constant, then we have four equations constraining four constants, so there will be consistent solutions, but only for the values of $E$ that satisfy the boundary conditions. There will be a discrete set of energies that satisfy the boundary conditions, and hence a discrete energy spectrum for bound states. Note also that the solutions will be nondegenerate, as the value of $E$ uniquely determines the other constants.

In summary, for bound states the eigenstates are normalizable and nondegenerate, and there is a discrete energy spectrum.

The cases $V_+ < E < V_-$ and $E>V_-$ are both called scattering states. There does not seem to be a standard terminology to distinguish these two cases, but I will call $V_+ < E < V_-$ semi-scattering states and $E>V_+$ full-scattering states.

For the semi-scattering case, $k_-$ is real and positive, and both $k_0$ and $k_+$ are imaginary. We can still discard the divergent term $Be^{-k_-x}$, but $Fe^{k_+ x}$ is oscillatory, so we have to keep it. Defining the real and positive constants $k_0'= \sqrt{\frac{2m}{\hbar^2} \left ( E -V_0\right )}$ and $k_+'= \sqrt{\frac{2m}{\hbar^2} \left ( E -V_+\right )}$ so that $k_0 = ik_0'$ and $k_+ = ik_+'$, we have the general solution

\[\psi(x)=\begin{cases} \psi_-(x) = Ae^{k_-x} , & x<a \\ \psi_0(x)=Ce^{ik_0'x} + De^{-ik_0' x}, & a\leq x \leq b \\ \psi_+(x) = Fe^{ik_+' x} + Ge^{-ik_+'x}, & x>b.\end{cases}\]

Since this solution oscillates as $x\rightarrow - \infty$, it will not be normalizable. We can still choose to set one of the five unknown constants arbitrarily, since multiplying a wavefunction by an overall constant does not change the physical state. As before, the boundary conditions give us four equations but, in this case, we have four unknown constants, so there is a unique solution with no further constraints on the energy $E$. Therefore, there will be a continuous energy spectrum: all possible values of $E$ satisfying $V_+ < E < V_-$ are allowed. Since the boundary conditions uniquely determine the four unknown constants, the solutions are nondegenerate.

This type of solution is called semi-scattering because the particle can escape to $x = +\infty$, but not $x=-\infty$. As we have seen, semi-scattering states are unnormalizable, nondegenerate, and have a continuous energy spectrum.

In the full scattering case, all of $k_-$, $k_0$ and $k_+$ are imaginary, so we replace them by the real positive constants $k_-'= \sqrt{\frac{2m}{\hbar^2} \left ( E -V_-\right )}$, $k_0'= \sqrt{\frac{2m}{\hbar^2} \left ( E -V_0\right )}$ and $k_+'= \sqrt{\frac{2m}{\hbar^2} \left ( E -V_+\right )}$. There are no divergent terms in the general solution, so we have

\[\psi(x) = \begin{cases} \psi_-(x) = Ae^{ik_-'x} + Be^{-ik_-'x}, & x<a \\ \psi_0(x)=Ce^{ik_0'x} + De^{-ik_0' x}, & a\leq x \leq b \\ \psi_+(x) = Fe^{ik_+' x} + Ge^{-ik_+'x}, & x>b. \end{cases}\]

This is much like the free particle solution. It oscillates in all three regions, but with a different wave number in each region. Because the solution oscillates as $x\rightarrow \pm \infty$, it is unnormalizable. As before, one of the six unknown constants can be set arbitrarily, but then we still have five unknown constants and only four equations from the boundary conditions. This means that one constant remains undetermined. We have a continuous energy spectrum: all possible values of satisfying $E>V_-$ are allowed, but because of the undetermined constant the solutions will be degenerate.

This type of solution is called full-scattering because the particle can escape to both $x=-\infty$ and $x=+\infty$. As we have seen, full-scattering states are unnormalizable, degenerate, and have a continuous energy spectrum.

Now consider more general potentials. The potential below is generic enough to understand the cases we want to consider in the following sections.

Note that the potential has a minimum $V_0$ and asymptotes to the values $V_{\pm}$ as $x\rightarrow \pm\infty$. The features that the energy eigenstates have depend on how the energy $E$ is related to these three values in the same way as for the piece-wise constant potential. In particular, wherever $E < V(x)$ the general solution will be a sum of two solutions, one of which decays to zero as $x\rightarrow +\infty$ and diverges as $x\rightarrow -\infty$, while the other diverges as $x\rightarrow +\infty$ and decays to zero as $x\rightarrow -\infty$. Wherever $E > V(x)$, we will have a sum of two oscillating solutions. The boundary conditions must be applied at the points where $E = V(x)$, since that is where transitions between the two types of solution occur. These are called the turning points because they are the points at which a classical particle would have zero kinetic energy, so it would turn around and go back the other way. The type of solution depends on the value of $E$ in the following way:

• There are no viable solutions to the TISE for values of $E$ such that $E < V(x)$ everywhere.
• The solution will be a bound state if the region in which $E > V(x)$ is bounded, i.e. does not stretch to either $x\rightarrow + \infty$ or $x \rightarrow \infty$. Bound states are:

  • Normalizable
  • Nondegenerate
  • Have a discrete energy spectrum
• The solution will be a semi-scattering state if the region in which $E > V(x)$ extends to either $x\rightarrow +\infty$ or $x\rightarrow-\infty$ but not both. Semi-scattering states are:

  • Unnormalizable
  • Nondegenerate
  • Have a continuous energy spectrum
• The solution will be a full-scattering state if the region in which $E>V(x)$ extends to both $x\rightarrow +\infty$ and $x\rightarrow -\infty$. Full-scattering states are:

  • Unnormalizable
  • Degenerate
  • Have a continuous energy spectrum

The values of $E$ that correspond these four cases are shown on the graph of the potential above.

A node of a wavefunction is a point at which it crosses the $x$-axis, from negative to positive or vice versa. In the solutions for an infinite square well potential

\[\psi(x)=\begin{cases} 0, & x\leq 0 \\ \sqrt{\frac{2}{a}}\sin\left ( \frac{n\pi x}{a}\right ), & 0<x<a \\ 0, & x\geq a, \end{cases}\]

if we exclude the end points $x=0$ and $x=a$, the ground state has no nodes, the first excited state has one node, the second excited state has two nodes, etc. This behavior is generally true for the bound states of all potentials. A proof is beyond the scope of this course, but the intuition is the same as for the infinite square well. In the regions where $E > V(x)$, the solution is oscillatory. The more times the solution oscillates in this region the higher the energy. Roughly, more oscillations indicate a smaller “wavelength” and hence a larger momentum. “Wavelength” is in scare quotes here because the solution need not be periodic, so the concept does not strictly apply, but the general idea that a larger number of nodes gives a larger mean momentum, and hence energy, is correct. Therefore, the ground state will have the smallest possible number of nodes, i.e. zero, the first excited state the next smallest, i.e. one, and so on.

Both bound and semi-scattering states have nondegenerate energy spectra. We can prove this formally based on the fact that these solutions satisfy $\psi(x) \rightarrow 0$ for at least one of $x\rightarrow +\infty$ or $x\rightarrow -\infty$ (exactly one for semi-scattering states and both for bound states).

Suppose, for the sake of a contradiction, that there are two distinct states, $\psi_1(x)$ and $\psi_2(x)$, corresponding to the same energy $E$, that both tend to zero as $x\rightarrow \infty$ (the case where it is $x\rightarrow -\infty$ is dealt with in the same way). They must both satisfy the TISE, so

\[ \frac{\mathrm{d}^2\psi_1}{\mathrm{d}x^2} - k(x)^2 \psi_1(x) = 0,\]

\[ \frac{\mathrm{d}^2\psi_2}{\mathrm{d}x^2} - k(x)^2 \psi_2(x) = 0.\]

Multiplying the top equation by $\psi_2(x)$ and the bottom equation by $\psi_1(x)$ and subtracting gives

\[\psi_2(x)\frac{\mathrm{d}^2\psi_1}{\mathrm{d}x^2} - \psi_1(x) \frac{\mathrm{d}^2\psi_2}{\mathrm{d}x^2} = 0.\]

Adding and subtracting the term $\frac{\mathrm{d}\psi_1}{\mathrm{d}x} \frac{\mathrm{d}\psi_2}{\mathrm{d}x}$ gives

\[\psi_2(x)\frac{\mathrm{d}^2\psi_1}{\mathrm{d}x^2} + \frac{\mathrm{d}\psi_1}{\mathrm{d}x} \frac{\mathrm{d}\psi_2}{\mathrm{d}x} - \frac{\mathrm{d}\psi_1}{\mathrm{d}x} \frac{\mathrm{d}\psi_2}{\mathrm{d}x} - \psi_1(x) \frac{\mathrm{d}^2\psi_2}{\mathrm{d}x^2} = 0,\]

which, by the product rule, we recognize as

\[\frac{\mathrm{d}}{\mathrm{d}x} \left [ \psi_2(x) \frac{\mathrm{d}\psi_1}{\mathrm{d}x} - \psi_1(x) \frac{\mathrm{d}\psi_2}{\mathrm{d}x} \right ] = 0.\]

Integrating both sides gives

\[\psi_2(x) \frac{\mathrm{d}\psi_1}{\mathrm{d}x} - \psi_1(x) \frac{\mathrm{d}\psi_2}{\mathrm{d}x} = C,\]

where $C$ is a constant.

Suppose $\psi_1(x)$ and $\psi_2(x)$ both tend to zero $x \rightarrow +\infty$. Then, since the derivatives are bounded, considering the limit of this expression as $x \rightarrow \infty$, the constant $C$ must be zero. The same argument holds if $x \rightarrow - \infty$ is where the solutions tend to zero. Thus, we have

\[\psi_2(x) \frac{\mathrm{d}\psi_1}{\mathrm{d}x} = \psi_1(x) \frac{\mathrm{d}\psi_2}{\mathrm{d}x} \quad \Rightarrow \quad \frac{\frac{\mathrm{d}\psi_1}{\mathrm{d}x}}{\psi_1(x)} = \frac{\frac{\mathrm{d}\psi_2}{\mathrm{d}x}}{\psi_2(x)} \quad \Rightarrow \quad \frac{\mathrm{d}}{\mathrm{d}x} \left [ \ln \psi_1(x) - \ln \psi_2(x)\right ] = 0. \]

This implies that there exists a constant $D$ such that

\[\ln \psi_1(x) = \ln \psi_2(x) + D,\]

and taking the exponential of this gives

\[\psi_1(x) = e^D \psi_2(x).\]

But since $e^D$ is just a constant, this shows that $\psi_1(x)$ and $\psi_2(x)$ are the same physical solution, so there can be no degeneracy.

At the end of section 4.ii we considered the symmetric infinite potential well

\[V(x) = \begin{cases} \infty, & x<-\frac{a}{2}, \\ 0, & -\frac{a}{2} \leq x \leq +\frac{a}{2}, \\ \infty, & x>+\frac{a}{2}, \end{cases}\]

as illustrated below.

We noted that the eigenstates

\[\psi_n(x) = \begin{cases} \sqrt{\frac{2}{a}} \cos \left ( \frac{n\pi x}{a}\right ), & n=1,3,5,7,\cdots \\ \sqrt{\frac{2}{a}} \sin \left ( \frac{n\pi x}{a} \right ), & n=2,4,6,8,\cdots,\\ \end{cases}\]

were a sequence of alternating odd and even functions. The ground state is odd, the first excited state is even, the second excited state is odd, etc.

The same thing is true of the bound states whenever $V(x)$ is an even function, i.e. if $V(-x) = V(x)$ then the bound states form an alternating sequence of odd or even functions. Knowing that all the eigenstates are either even or odd functions can simplify the process of solving the TISE, and we will see an example of that when we study the finite square well potential in section 4.vii. This section shows why this must be the case.

The parity operator $\hat{\mathcal{P}}$ is defined via its action on the position eigenstates:

\[\hat{\mathcal{P}} |x\rangle = |-x\rangle.\]

We can determine how it acts on wavefunctions by writing a general one dimensional quantum state in this basis as

\[|\psi \rangle = \int_{-\infty}^{+\infty} \mathrm{d}x\, \psi(x) |x\rangle,\]

so that

\[\hat{\mathcal{P}}|\psi\rangle = \int_{-\infty}^{+\infty} \mathrm{d}x\, \psi(x) |-x\rangle.\]

Substituting $x' = -x$ we get $\mathrm{d}x' = -\mathrm{d}x$ and

\begin{align*} x & = -\infty \quad \Rightarrow \quad x' = +\infty \\ x & = +\infty \quad \Rightarrow \quad x' = -\infty, \end{align*}

which implies

\[\hat{\mathcal{P}}|\psi\rangle = -\int_{+\infty}^{-\infty} \mathrm{d}x'\, \psi(-x') |-x'\rangle = \int_{-\infty}^{+\infty} \mathrm{d}x\, \psi(-x) |x\rangle, \]

where, in the second step we got rid of the minus sign by switching the limits and we replaced $x'$ by $x$ because $x'$ is a dummy variable. Hence, $\hat{\mathcal{P}}$ maps a wavefunction $\psi(x)$ to $\psi(-x)$.

In the in class activities, you will show that $\hat{\mathcal{P}}$ is a Hermitian operator, which implies that it has real eigenvalues. Note also that

\[\hat{\mathcal{P}}^2 \psi(x) = \hat{\mathcal{P}} \psi(-x) = \psi(x),\]

so $\hat{\mathcal{P}}^2 = \hat{I}$, the identity operator.

Now let $\psi(x)$ be an eigenstate of $\hat{\mathcal{P}}$ with eigenvalue $a$. Then,

\begin{align*} \hat{\mathcal{P}} \psi(x) & = a \psi(x) \\ \Rightarrow\, \hat{\mathcal{P}}^2 \psi(x) & = a^2 \psi(x) \\ \Rightarrow\quad\,\, \psi(x) & = a^2\psi(x), \end{align*}

so $a^2=1$ and, since $a$ is real, $a = \pm 1$.

The eigenvalues of $\hat{\mathcal{P}}$ are $\pm 1$ and the corresponding eigenstates are the odd (eigenvalue $-1$) and even (eigenvalue $+1$) wavefunctions.

An operator $\hat{A}$ is even if $\hat{\mathcal{P}}\hat{A}\hat{\mathcal{P}} = \hat{A}$, or equivalently $\hat{A}\hat{\mathcal{P}} = \hat{\mathcal{P}}\hat{A}$. It is odd if $\hat{\mathcal{P}}\hat{A}\hat{\mathcal{P}} = -\hat{A}$, or equivalently $\hat{A}\hat{\mathcal{P}} = -\hat{\mathcal{P}}\hat{A}$.

In an in class activity, you will show that $\hat{x}$ and $\hat{p}$ are both odd operators. This implies that the kinetic energy operator $\frac{\hat{p}^2}{2m}$ is an even operator because

\[\hat{\mathcal{P}} \hat{p}^2 = -\hat{p} \hat{\mathcal{P}} \hat{p} = +\hat{p}^2 \hat{\mathcal{P}}.\]

Similarly, if the potential is even, $V(-\hat{x}) = V(\hat{x})$, its Taylor expansion only includes even powers of $\hat{x}$,

\[V(\hat{x}) = a_0 + a_2 \hat{x}^2 + a_3 \hat{x}^4 + \cdots.\]

By the same reasoning we used for $\hat{p}^2$, $\hat{x}^{2n}$ is an even operator and hence so is $V(\hat{x})$.

Therefore, the Hamiltonian $\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})$ is also an even operator, so $\hat{H} \hat{\mathcal{P}} = \hat{\mathcal{P}}\hat{H}$ or equivalently $[\hat{H},\hat{\mathcal{P}}] = 0$.

Since commuting operators have joint eigenstates, if $|\psi\rangle$ is a nondegenerate eigenstate of $\hat{H}$, it must also be an eigenstate of $\hat{\mathcal{P}}$, so either $\psi(-x) = \psi(x)$ or $\psi(-x) = -\psi(x)$.

Since we know that bound states are nondegenerate, we can conclude that the bound states of symmetric potentials are either even or odd functions. However, we can also conclude that they must form an alternating sequence of even and odd functions by the following argument.

Consider the ground state. Earlier, we argued that this state has no nodes. If it were an odd function, and not equal to zero at the origin then it would have a jump discontinuity at $x=0$. Since solutions to the TISE must be continuous, it must be zero at the origin, but this would be a node, contradicting the idea that the ground state has no nodes. Therefore, the ground state must be an even function.

Next consider the first excited state. It has exactly one node. If the node were anywhere other than $x=0$, say at $x=a$, then it would also have a node at $x=-a$. Since it only has one node, the node must be at $x=0$. But since the function is supposed to go from positive values to negative values or vice versa when it crosses a node, it cannot be an even function. Therefore, it must be odd.

The same type of argument works for the rest of the excited states. If they have an even number of nodes then there cannot be a node at $x=0$ and hence the wavefunction must be even. If there are an odd number of nodes then one of them must be at $x=0$ meaning that the wavefunction must be odd. Since the number of nodes increases by one each time we increase the energy to the next allowed value in the spectrum, we get an alternating sequence of even and odd wavefunctions.

1. Prove that the parity operator $\hat{\mathcal{P}}$ is Hermitian, i.e. \[\langle \phi | \hat{\mathcal{P}} | \psi \rangle = \langle \psi | \hat{\mathcal{P}} | \phi \rangle^* \] HINT: Use the fact that $\hat{\mathcal{P}}$ maps $\psi(x)$ to $\psi(-x)$ and the definition of the inner product \[\langle \phi | \psi \rangle = \int_{-\infty}^{+\infty} \mathrm{d}x\, \phi^*(x) \psi(x).\]
2. 1. Show that $\hat{x}$ is an odd operator. It is sufficient to show that $\hat{x}\hat{\mathcal{P}}|x\rangle = - \hat{\mathcal{P}}\hat{x}|x\rangle$. Why?
   2. Show that $\hat{\mathcal{P}}|p\rangle = |-p\rangle$ using $\langle x | p \rangle = \frac{1}{\sqrt{2\pi \hbar}} e^{ipx/\hbar}$.
   3. Show that $\hat{p}$ is an odd operator.