# One-Dimensional Bound States

When the energy of a particle moving in one dimension is less than the potential at $x\to\pm\infty$, the particle is trapped in a potential well and is said to be in a bound state. However, when the energy is larger than the potential at either infinity, the particle is said to be in a scattering state. The wave function for a bound state must decay at least exponentially at both infinities.

## Basic Properties

1. The bound states of a one-dimensional potential are non-degenerate.

To prove this, let us suppose that there are two different wave functions, $\psi_E^{(1)}$ and $\psi_E^{(2)},$ for a given bound state with energy $E\!$. They must then both satisfy the same Schrödinger equation:

$-\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(1)}}{dx^2}+V(x)\psi_E^{(1)}=E\psi_E^{(1)}$

$-\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(2)}}{dx^2}+V(x)\psi_E^{(2)}=E\psi_E^{(2)}$

Multiplying the first equation by $\psi_E^{(2)}$ and the second by $\psi_E^{(1)}$, then subtracting one equation from the other, we get

$\frac{d^2\psi_E^{(1)}}{dx^2}\psi_E^{(2)}-\frac{d^2\psi_E^{(2)}}{dx^2}\psi_E^{(1)}=0,$

or

$\rightarrow \frac{d}{dx}\left (\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}\right )=0.$

We recognize the quatity $W(x)=\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}$ as the Wronskian of the two wave functions. We therefore see that W(x) = C, where C is constant. For bound states, the wave functions must vanish at infinity, and therefore W = 0, or

$\psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}=0.$

Rearranging, we obtain

$\frac{1}{\psi_E^{(1)}}\frac{d\psi_E^{(1)}}{dx}-\frac{1}{\psi_E^{(2)}}\frac{d\psi_E^{(2)}}{dx}=0.$

We can now integrate this equation to obtain

$\ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})=A,$

or

$\psi_E^{(1)}(x)=a\psi_E^{(2)}(x).$

We see that the two wave functions must be proportional to each other; after normalization, they will become completely identical. Therefore, we see that the two wave functions correspond to the same state, and thus all bound states in one dimension are non-degenerate.

2. The wave function for a real potential V(x) can be chosen to be real.

If we take the complex conjugate of the Schrödinger equation for the system,

$E\psi=-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi,$

then we obtain $E\psi^{\ast}=-\frac{\hbar^2}{2m}\frac{d^2\psi^{\ast}}{dx^2}+V(x)\psi^{\ast}.$

Since both $\psi\!$ and $\psi^{\ast}$ satisfy the same Schrödinger equation, their sum $\psi+\psi^{\ast}$ must also be a solution. This sum is real, and thus we see that the wave function, as asserted, can be chosen to be real.

3. The nth excited state has n nodes.

See the Oscillation Theorem.

## Parity and the Symmetry of the Wave Functions

One useful theorem about one-dimensional potentials is that, if the potential $V(x)\!$ is even (i.e. $V(x)=V(-x)\!$), then the eigenstates can be taken to be even or odd.

To prove this, we first note that, if ψ(x) satisfies the Schrödinger equation for the potential,

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi,$

then so will $\psi(-x)\!$. We may see this by simply taking $x\to -x\!$ in the above equation. Since both $\psi(x)\!$ and $\psi(-x)\!$ are solutions of the Schrödinger equation, any linear combination of the two must also be a solution. In particular, $\psi_e(x)=\psi(x)+\psi(-x)\!$ and $\psi_o(x)=\psi(x)-\psi(-x)\!$ are solutions. We may easily see that $\psi_e(x)\!$ is even, while $\psi_o(x)\!$ is odd.

In particular, this implies that all bound states in such a potential must be either even or odd. Since $\psi(x)\!$ and $\psi(-x)\!$ both solve the Schrödinger equation, and because the eigenstates must be non-degenerate, we see that $\psi(x)\!$ and $\psi(-x)\!$ must be proportional to one another. We may easily show that the proportionality constant must be $\pm 1\!$, corresponding to either an even function or an odd function.

The above discussion is a simple illustration of how the symmetry of the Hamiltonian of a system dictates the transformation properties of the eigenstates, as described earlier.

## Infinite Square Well

Let's consider the motion of a particle in an infinite and symmetric square well: $V(x)=+\infty$ for $|x| \ge L/2,$ and $V(x)=0\!$ otherwise.

A particle subject to this potential is free everywhere except at the two ends $(x = \pm L/2),$ where the infinite potential keeps the particle confined to the well. Within the well the Schrödinger equation takes the form,

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi,$

or equivalently,

$\frac{d^2\psi}{dx^2}=-k^2\psi,$

where

$k=\frac{\sqrt{2mE}}{\hbar}.$

Writing the Schrödinger equation in this form, we see that our solution are simply

$\psi(x) = A \sin (kx) + B \cos (kx).\!$

As we showed earlier, the eigenstates of this potential, all of which are bound states, must be either even or odd. This means that we must either set A = 0 or B = 0. Let us consider each case in turn.

Case 1: $B=0\!$.

In this case,

$\psi(x)=A\sin{kx}.\!$

We require that the wave function vanish at the ends, so that

$\sin\left (\frac{kL}{2}\right )=0.\!$

The solutions of this equation are

$k=\frac{2n\pi}{L},\!$

where $n=1,2,3,\ldots\!$

Case 2: $A=0.\!$

In this case,

$\psi(x)=B\cos{kx}.\!$

Again requiring that the wave function vanish at the ends, i.e.

$\cos\left (\frac{kL}{2}\right )=0,\!$

we find that

$k=\frac{(2n+1)\pi}{L}.\!$

Combining these two results, we find that the energy eigenvalues are $E_n = \frac{\hbar^2}{2m} \left(\frac{\pi n}{L} \right )^2 \!$, where $n = 1,2,3,\ldots\!$ The wave numbers are quantized as a result of the boundary conditions, and thus the energies are quantized as well. The lowest energy state is the ground state, and it has a non-zero energy, which is due to quantum zero point motion. The ground state is nodeless, the first excited state has one node, the second excited state has two nodes, and so on. The wavefunctions are also orthogonal.

## Finite Asymmetric Square Well

We now consider an asymmetric square well potential V(x), defined as

$V(x)=\begin{cases} V_1,\,x<0 \\ 0,\,0a \end{cases},$

where both $V_1\!$ and $V_2\!$ are positive. We may then split our one-dimensional space into three regions; region I is given by $x<0\!$, region II by $0, and region III by $x>a\!$. The Schrödinger equation may be written in each region as

$-\frac{\hbar^2}{2m}\frac{d^2\psi_{I}(x)}{dx^2}+V_1\psi_{I}(x)=E\psi_{I}(x),$

$-\frac{\hbar^2}{2m}\frac{d^2\psi_{II}(x)}{dx^2}=E\psi_{II}(x),$

and

$-\frac{\hbar^2}{2m}\frac{d^2\psi_{III}(x)}{dx^2}+V_2\psi_{III}(x)=E\psi_{III}(x).$

We will be considering bound states here, so that $E<\min(V_1,V_2).\!$ In this case, the solutions to the above Schrödinger equations are

$\psi_{I}(x)=Ae^{k_{1}x},$

$\psi_{II}(x)=Be^{ik_{2}x}+Ce^{-ik_{2}x},$

and

$\psi_{III}(x)=De^{-k_{3}x},$

where, $k_{1}=\frac{\sqrt{2m(V_{1}-E)}}{\hbar},$ $k_{2}=\frac{\sqrt{2mE}}{\hbar},$ and $k_{3}=\frac{\sqrt{2m(V_{2}-E)}}{\hbar}.$

The wave function and it's first derivative must both be continuous at $x=0\!$ and $x=a.\!$ The condition at $x=0\!$ yields

$A=B+C\!$

and

$k_{1}A=ik_{2}(B-C),\!$

while that at $x=a\!$ gives us

$Be^{ik_{2}a}+Ce^{-ik_{2}a}=De^{-k_{3}a}$

and

$Be^{ik_{2}a}-Ce^{-ik_{2}a}=\frac{ik_{3}}{k_{2}}De^{-k_{3}a}.$

If we reduce the above set of equations to equations only involving $B\!$ and $C\!$, we obtain

$(k_1-ik_2)B+(k_1+ik_2)C=0\!$

and

$(k_3+ik_2)Be^{ik_2a}+(k_3-ik_2)Ce^{-ik_2a}=0.$

In order for this system of homogeneous equations to possess a non-trivial solution in $B\!$ and $C,\!$ the following condition must be met:

$(k_1-ik_2)(k_3-ik_2)e^{-ik_2a}-(k_1+ik_2)(k_3+ik_2)e^{ik_2a}=0,$

or

$\tan{k_{2}a}=\frac{(k_1+k_3)k_2}{k_2^2-k_1k_3}.$

If we now substitute in our expressions for $k_1,\,k_2,\!$ and $k_3,\!$ we get

$\tan\left (\frac{\sqrt{2mE}}{\hbar}a\right )=\frac{\sqrt{E}(\sqrt{V_{1}-E}+\sqrt{V_{2}-E})}{E-\sqrt{(V_{1}-E)(V_{2}-E)}}.$

The solutions to this equation then give us the bound-state energy spectrum of the system.

## Problems

(1) An electron is moving freely inside of a one-dimensional box with walls at $x=0\!$ and $x=a.\!$ If the electron is initially in the ground state of the box and we suddenly increase the size of the box by moving the right-hand wall instantaneously from $x=a\!$ to $x=4a,\!$ then calculate the probability of finding the electron in

(a) the ground state of the new box, and

(b) the first excited state of the new box.

(2) Consider a particle of mass $m\!$ bouncing vertically from a smooth floor in Earth's gravitational field. The potential is given by

$V(z)=\begin{cases} mgz, \; z>0 \\ \infty,\; z\leq 0 \end{cases},$

where $g\!$ is the acceleration due to gravity at Earth's surface. Find the energy levels and corresponding wave functions of the particle.