One-Dimensional Bound States

When the energy of a particle is less than the potential at both positive and negative infinity, the particle is trapped in a well and it goes back and forth between the turning points in the potential with its kinetic energy; however, when the energy is larger than the potential at either infinity, the particle is said to be "unbound". For a bound state, the wavefunction must decay at least exponentially at both infinities.

Basic Properties

1. Non-degeneracy of the bound states in 1D

Let's consider a more general property that is the Wronskian for 2 solutions of the 1-D Schrodinger equation must equal to a constant.

Schrodinger equation : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi}

is a second-order differential equation. Such equation has 2 linearly independent Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(2)}} for each value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\!} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(1)}}{dx^2}+V(x)\psi_E^{(1)}=E\psi_E^{(1)}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(2)}}{dx^2}+V(x)\psi_E^{(2)}=E\psi_E^{(2)}}

By definition in mathematics, the Wronskian of these functions is: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W=\psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}}

Multiplying equation (2) by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(2)}} , equation (3) by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}} , then subtracting one equation from the other, we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2\psi_E^{(1)}}{dx^2}\psi_E^{(2)}-\frac{d^2\psi_E^{(2)}}{dx^2}\psi_E^{(1)}=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow \frac{dW}{dx}=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow W=C}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\!} is constant. So, the Wronskian for 2 solutions of the 1-D Schrodinger equation must equal to a constant.

For the bound states, the wave function vanish at infinity, i.e: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}(\infty)=\psi_E^{(2)}(\infty)=0}

From (4), (5) and (6), it follows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W=0\!} From (4) and (7), we get: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow \frac{1}{\psi_E^{(1)}}\frac{d\psi_E^{(1)}}{dx}-\frac{1}{\psi_E^{(2)}}\frac{d\psi_E^{(2)}}{dx}=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow \frac{d}{dx}\left[\ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})\right]=0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow \ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})=\mbox{const.}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow \psi_E^{(1)}=\mbox{const.}\psi_E^{(2)}}

From (8) it follows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(1)}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_E^{(2)}} describe the same state. Therefore, the bound states in 1D are non-degenerate.

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\psi(\textbf{r})=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi(\textbf{r})}{\partial x^2}+V(\textbf{r})\psi(\textbf{r})}

so: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\psi^*(\textbf{r})=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi^*(\textbf{r})}{\partial x^2}+V(\textbf{r})} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow\psi^*=C\psi} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow\psi=C^*\psi^*} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow\psi^*=|C|^2\psi^*} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow|C|^2=1}

so: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C=e^{i\theta}\!} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow\psi^*=e^{i\theta}\psi}

let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta=0\!} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow\psi^*=\psi} So, the wave function is real.

3. The nth excited state has n nodes.

See the Oscillation theorem below.

Parity operator and the symmetry of the wavefunctions

In the above problem, two basic solutions of Schrodinger equation are either odd or even. The general wavefunctions are combinations of odd and even functions. This property originates from the fact that the potential is symmetric or invariant under the inversion Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \rightarrow -x} , and so does the Hamiltonian. Therefore, the Hamiltonian commutes with the parity operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{P}} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{P} \psi(x)=\psi(-x)} , or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{P}\left |{\mathbf{r}} \right \rangle=\left |{-\mathbf{r}} \right \rangle} In this case, the wavefunctions themselves do not need to be odd or even, but they must be some combinations of odd and even functions.

One useful law about parity is that, for 1D bound state, if the potential V(x) is inversion symmetric, its wave function is either even or odd.

We will prove it step by step:

1. If potential V(x) is inversion symmetric, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (\mathbf{r}))} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (-\mathbf{r})} are wave functions with the same eigenvalue.

Prove: Just change the sign of x, and V(x)=V(-x).

2. If potential V(x) is inversion symmetric, as for every eigenvalue, we can find a complete set of eigenfunctions, which are either even or odd.

Prove: If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (\mathbf{r})} is a solution of one stationary Schrodinger function, then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (-\mathbf{r})} is another solution. Let's formalize

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{u(r)}=\psi (\mathbf{r})+\psi (-\mathbf{r})} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{v(r)=}\psi (\mathbf{r})-\psi (-\mathbf{r})}

3. So Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{u(r)}} is even and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{v(r)}} is odd. Because they only have one eigenfunction.

4. For 1D bound state, if the potential V(r) is inversion symmetric, its wave function is either even or odd.

[[1]]

Infinite square well

Let's consider the motion of a particle in an infinite and symmetric square well: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=+\infty} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \ge |L|/2} , otherwise Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=0\!}

A particle subject to this potential is free everywhere except at the two ends (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \pm L/2} ), where the infinite potential keeps the particle confined to the well. Within the well the Schrodinger equation takes the form:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\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 Schrodinger equation in this form, we see that our solution are simply

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

Now we impose that the solution must vanish at $x=\pm L/2$:

$-A\sin(kL/2)+B\cos(kL/2)=0\!$

$A\sin(kL/2)+B\cos(kL/2)=0\!$

Adding the two equations, we get:

$2B\cos(kL/2)=0\!$

It follows that either $B=0\!$ or $\cos(kL/2)=0\!$.

Case 1: $B=0\!$. In this case $A\neq 0\!$, otherwise the wavefunction vanishes every where. Furthermore, it is required that: $\sin(kL/2)=0\!$

$\rightarrow k=2n\pi /L\!$ where $n=1,2,3,...\!$

And the wave functions are odd: $\psi (x)=A\sin(2n\pi x/L)\!$

Case 2: $\cos(kL/2)=0\!$ $\rightarrow k=(2n+1)\pi /L\!$ where $n=0,1,2,...\!$

In this case $A=0\!$ and $B\neq 0\!$, and the wavefunctions are even: $\psi (x)=B\cos[(2n+1)\pi x/L]\!$

The two solutions give the eigenenergies $E_{n}={\frac {\hbar ^{2}}{2m}}\left({\frac {\pi n}{L}}\right)^{2}\!$, where $n=1,2,3,...\!$ These wave numbers are quantized as a result of the boundary conditions, thus making the energy 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. This ground state is also nodeless, the first excited state has one node, the second excited state has two nodes, and so on. The wavefunctions are also orthogonal.

[[2]]

Finite asymmetric square well

(1)Scattering State:

If we have the potential $V_{1}$for $x<0$,$0$for $0<x<a$ and $V_{2}$ for $x>a$, then we can write the wave functions for scattering states as:

${\begin{cases}x<a&\Psi _{1}=A_{1}e^{ik_{1}x}+A_{2}e^{-ik_{2}x}\\0\leqslant x<a&\Psi _{2}=B_{1}e^{ik_{2}x}+B_{2}e^{-ik_{2}x}\\x\geqslant a&\Psi _{3}=Ce^{ik_{3}x}\end{cases}}$

Further we define: $k_{1}={\frac {\sqrt {2m(E-V_{1})}}{\hbar }}$ $k_{2}={\frac {\sqrt {2mE}}{\hbar }}$ <math>k_{3}=\frac{\sqrt{2m(E-V_{2