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# Posts Tagged ‘infinite potential well’

## PHY456H1F: Quantum Mechanics II. Recitation 3 (Taught by Mr. Federico Duque Gomez). WKB method and Stark shift.

Posted by peeterjoot on October 28, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# WKB method.

Consider the potential

\begin{aligned}V(x) = \left\{\begin{array}{l l}v(x) & \quad \mbox{iflatex x \in [0,a]} \\ \infty & \quad \mbox{otherwise} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.1)

as illustrated in figure (\ref{fig:qmTwoR3:qmTwoR3fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoR3fig1}
\caption{Arbitrary potential in an infinite well.}
\end{figure}

Inside the well, we have

\begin{aligned}\psi(x) = \frac{1}{{\sqrt{k(x)}}} \left( C_{+} e^{i \int_0^x k(x') dx'}+C_{-} e^{-i \int_0^x k(x') dx'}\right)\end{aligned} \hspace{\stretch{1}}(2.2)

where

\begin{aligned}k(x) = \frac{1}{{\hbar}} \sqrt{ 2m( E - v(x) }\end{aligned} \hspace{\stretch{1}}(2.3)

With

\begin{aligned}\phi(x) = e^{\int_0^x k(x') dx'}\end{aligned} \hspace{\stretch{1}}(2.4)

We have

\begin{aligned}\psi(x) &= \frac{1}{{\sqrt{k(x)}}} \left( C_{+}(\cos \phi + i\sin\phi) + C_{-}(\cos\phi - i \sin\phi)\right) \\ &= \frac{1}{{\sqrt{k(x)}}} \left( (C_{+} + C_{-})\cos \phi + i(C_{+} - C_{-}) \sin\phi\right) \\ &= \frac{1}{{\sqrt{k(x)}}} \left( (C_{+} + C_{-})\cos \phi + i(C_{+} - C_{-}) \sin\phi\right) \\ &\equiv \frac{1}{{\sqrt{k(x)}}} \left( C_2 \cos \phi + C_1 \sin\phi\right),\end{aligned}

Where

\begin{aligned}C_2 &= C_{+} + C_{-} \\ C_1 &= i( C_{+} - C_{-})\end{aligned} \hspace{\stretch{1}}(2.5)

Setting boundary conditions we have

\begin{aligned}\phi(0) = 0\end{aligned} \hspace{\stretch{1}}(2.7)

Noting that we have $\phi(0) = 0$, we have

\begin{aligned}\frac{1}{{\sqrt{k(0)}}} C_2 = 0\end{aligned} \hspace{\stretch{1}}(2.8)

So

\begin{aligned}\psi(x) \sim\frac{1}{{\sqrt{k(x)}}} \sin\phi\end{aligned} \hspace{\stretch{1}}(2.9)

At the other boundary

\begin{aligned}\psi(a) = 0\end{aligned} \hspace{\stretch{1}}(2.10)

So we require

\begin{aligned}\sin \phi(a) = \sin(n \pi)\end{aligned} \hspace{\stretch{1}}(2.11)

or

\begin{aligned}\frac{1}{{\hbar}} \int_0^a \sqrt{2 m (E - v(x')} dx' = n \pi\end{aligned} \hspace{\stretch{1}}(2.12)

This is called the Bohr-Sommerfeld condition.

Check with $v(x) = 0$.

We have

\begin{aligned}\frac{1}{{\hbar}} \sqrt{2m E} a = n \pi\end{aligned} \hspace{\stretch{1}}(2.13)

or

\begin{aligned}E = \frac{1}{{2m}} \left(\frac{n \pi \hbar}{a}\right)^2\end{aligned} \hspace{\stretch{1}}(2.14)

# Stark Shift

Time independent perturbation theory

\begin{aligned}H = H_0 + \lambda H'\end{aligned} \hspace{\stretch{1}}(3.15)

\begin{aligned}H' = e \mathcal{E}_z \hat{Z}\end{aligned} \hspace{\stretch{1}}(3.16)

where $\mathcal{E}_z$ is the electric field.

To first order we have

\begin{aligned}{\left\lvert {\psi_\alpha^{(1)}} \right\rangle} = {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} + \sum_{\beta \ne \alpha} \frac{ {\left\lvert {\psi_\beta^{(0)}} \right\rangle} {\left\langle {\psi_\beta^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} }{E_\alpha^{(0)} -E_\beta^{(0)} }\end{aligned} \hspace{\stretch{1}}(3.17)

and

\begin{aligned}E_\alpha^{(1)} = {\left\langle {\psi_\alpha^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} \end{aligned} \hspace{\stretch{1}}(3.18)

With the default basis $\{{\left\lvert {\psi_\beta^{(0)}} \right\rangle}\}$, and $n=2$ we have a 4 fold degeneracy

\begin{aligned}l,m &= 0,0 \\ l,m &= 1,-1 \\ l,m &= 1,0 \\ l,m &= 1,+1\end{aligned}

but can diagonalize as follows

\begin{aligned}\begin{bmatrix}\text{nlm} & 200 & 210 & 211 & 21\,-1 \\ 200 & 0 & \Delta & 0 & 0 \\ 210 & \Delta & 0 & 0 & 0 \\ 211 & 0 & 0 & 0 & 0 \\ 21\,-1 & 0 & 0 & 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.19)

FIXME: show.

where

\begin{aligned}\Delta = -3 e \mathcal{E}_z a_0\end{aligned} \hspace{\stretch{1}}(3.20)

We have a split of energy levels as illustrated in figure (\ref{fig:qmTwoR3:qmTwoR3fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoR3fig2}
\caption{Energy level splitting}
\end{figure}

Observe the embedded Pauli matrix (FIXME: missed the point of this?)

\begin{aligned}\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.21)

Proper basis for perturbation (FIXME:check) is then

\begin{aligned}\left\{\frac{1}{{\sqrt{2}}}( {\left\lvert {2,0,0} \right\rangle} \pm {\left\lvert {2,1,0} \right\rangle} ), {\left\lvert {2, 1, \pm 1} \right\rangle}\right\}\end{aligned} \hspace{\stretch{1}}(3.22)

and our result is

\begin{aligned}{\left\lvert {\psi_{\alpha, n=2}^{(1)}} \right\rangle} = {\left\lvert {\psi_{\alpha}^{(0)}} \right\rangle} +\sum_{\beta \notin \text{degenerate subspace}} \frac{ {\left\lvert {\psi_\beta^{(0)}} \right\rangle} {\left\langle {\psi_\beta^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} }{E_\alpha^{(0)} -E_\beta^{(0)} }\end{aligned} \hspace{\stretch{1}}(3.23)

# Adiabatic perturbation theory

Utilizing instantaneous eigenstates

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = \sum_{\alpha} b_\alpha(t) {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.24)

where

\begin{aligned}H(t) {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}= E_\alpha(t) {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.25)

We found

\begin{aligned}b_\alpha(t) = \bar{b}_\alpha(t) e^{-\frac{i}{\hbar} \int_0^t (E_\alpha(t') - \hbar \Gamma_\alpha(t')) dt'}\end{aligned} \hspace{\stretch{1}}(4.26)

where

\begin{aligned}\Gamma_\alpha = i{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.27)

and

\begin{aligned}\frac{d{{}}}{dt}\bar{b}_\alpha(t)=-\sum_{\beta \ne \alpha} \bar{b}_\beta(t)e^{-\frac{i}{\hbar} \int_0^t (E_{\beta\alpha}(t') - \hbar \Gamma_{\beta\alpha}(t')) dt'}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.36)

Suppose we start in a subspace

\begin{aligned}\text{span} \left\{\frac{1}{{\sqrt{2}}}( {\left\lvert {2,0,0} \right\rangle} \pm {\left\lvert {2,1,0} \right\rangle} ), {\left\lvert {2, 1, \pm 1} \right\rangle}\right\}\end{aligned} \hspace{\stretch{1}}(4.29)

Now expand the bra derivative kets

\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}=\left({\left\langle {\psi_{\alpha}^{(0)}} \right\rvert} +\sum_{\gamma} \frac{ {\left\langle {\psi_\gamma^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} {\left\langle {\psi_\gamma^{(0)}} \right\rvert} }{E_\alpha^{(0)} -E_\gamma^{(0)} }\right)\frac{d{{}}}{dt}\left({\left\lvert {\psi_{\beta}^{(0)}} \right\rangle} +\sum_{\gamma'} \frac{ {\left\lvert {\psi_{\gamma'}^{(0)}} \right\rangle} {\left\langle {\psi_{\gamma'}^{(0)}} \right\rvert} H' {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\gamma'}^{(0)} }\right)\end{aligned} \hspace{\stretch{1}}(4.30)

To first order we can drop the quadratic terms in $\gamma,\gamma'$ leaving

\begin{aligned}\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}&\sim\sum_{\gamma'} \left\langle{{\psi_{\alpha}^{(0)}}} \vert {{\psi_{\gamma'}^{(0)}}}\right\rangle \frac{ {\left\langle {\psi_{\gamma'}^{(0)}} \right\rvert} \frac{d{{H'(t)}}}{dt} {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\gamma'}^{(0)} }&=\frac{ {\left\langle {\psi_{\alpha}^{(0)}} \right\rvert} \frac{d{{H'(t)}}}{dt} {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\alpha}^{(0)} }\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.31)

so

\begin{aligned}\frac{d{{}}}{dt}\bar{b}_\alpha(t)=-\sum_{\beta \ne \alpha} \bar{b}_\beta(t)e^{-\frac{i}{\hbar} \int_0^t (E_{\beta\alpha}(t') - \hbar \Gamma_{\beta\alpha}(t')) dt'}\frac{ {\left\langle {\psi_{\alpha}^{(0)}} \right\rvert} \frac{d{{H'(t)}}}{dt} {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\alpha}^{(0)} }\end{aligned} \hspace{\stretch{1}}(4.32)

## A different way to this end result.

A result of this form is also derived in [1] section 20.1, but with a different approach. There he takes derivatives of

\begin{aligned}H(t) {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} = E_\beta(t) {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.33)

\begin{aligned}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} + H(t) \frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} = \frac{d{{E_\beta(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}+ E_\beta(t) \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.34)

Bra’ing ${\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}$ into this we have, for $\alpha \ne \beta$

\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} + {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}H(t) \frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} &= \not{{{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{E_\beta(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}}}+ {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}E_\beta(t) \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} \\ {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} + E_\alpha(t) {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} &=\end{aligned}

or

\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} =\frac{{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} }{E_\beta(t) - E_\alpha(t)}\end{aligned} \hspace{\stretch{1}}(4.35)

so without the implied $\lambda$ perturbation of ${\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}$ we can from 4.36 write the exact generalization of 4.32 as

\begin{aligned}\frac{d{{}}}{dt}\bar{b}_\alpha(t)=-\sum_{\beta \ne \alpha} \bar{b}_\beta(t)e^{-\frac{i}{\hbar} \int_0^t (E_{\beta\alpha}(t') - \hbar \Gamma_{\beta\alpha}(t')) dt'}\frac{{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} }{E_\beta(t) - E_\alpha(t)}\end{aligned} \hspace{\stretch{1}}(4.36)

# References

[1] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

## PHY356F: Quantum Mechanics I. Lecture 9 — Bound states.

Posted by peeterjoot on November 16, 2010

My notes from Lecture 9, November 16, 2010. Taught by Prof. Vatche Deyirmenjian.

[Click here for a PDF of this post with nicer formatting]

Motivation. Motivation for today’s physics is Solar Cell technology and quantum dots.

## Problem:

What ware the eigenvalues and eigenvectors for an electron trapped in a 1D potential well?

### MODEL.

Quantum state $\lvert {\Psi} \rangle$ describes the particle. What $V(X)$ should we choose? Try a quantum well with infinite barriers first.

These spherical quantum dots are like quantum wells. When you trap electrons in this scale you’ll get energy quantization.

### VISUALIZE.

Draw a picture for $V(X)$ with infinite spikes at $\pm a$. (ie: figure 8.1 in the text).

### SOLVE.

First task is to solve the time independent Schr\”{o}dinger equation.

\begin{aligned}H {\lvert {\Psi} \rangle} = E {\lvert {\Psi} \rangle}\end{aligned} \hspace{\stretch{1}}(5.98)

derivable from

\begin{aligned}H {\lvert {\Psi} \rangle} = i \hbar \frac{\partial {}}{\partial {t}} {\lvert {\Psi} \rangle}\end{aligned} \hspace{\stretch{1}}(5.99)

In the position representation, we project ${\langle {x} \rvert}$ onto $H {\lvert {\Psi} \rangle}$ and solve for $\left\langle{{x}} \vert {{\Psi}}\right\rangle = \Psi(x)$. For the problems in Chapter 8,

\begin{aligned}H = \frac{\mathbf{P}^2}{2m} + V(X,Y,Z),\end{aligned} \hspace{\stretch{1}}(5.100)

where

\begin{aligned}P &= \text{momentum operator} \\ X &= \text{position operator} \\ m &= \text{electron mass}\end{aligned}

We should be careful to be strict about the notation, and not interchange the operators and their specific representations (ie: not interchanging “little-x” and “big-x”) as we see in the text in this chapter.

Here the potential energy operator $V(X,Y,Z)$ is time independent.

If $i \hbar \frac{d{\lvert {\Psi} \rangle}}{dt} = H {\lvert {\Psi} \rangle}$ and $H$ is time independent then ${\lvert {\Psi} \rangle} = {\lvert {u} \rangle} e^{-i E t/\hbar}$ implies

\begin{aligned}i \hbar \frac{ -i E }{\hbar} {\lvert {u} \rangle} e^{-i E t/\hbar} = H {\lvert {u} \rangle} e^{-i E t/\hbar}\end{aligned}

or

\begin{aligned}E {\lvert {u} \rangle} = H {\lvert {u} \rangle}\end{aligned} \hspace{\stretch{1}}(5.101)

Here $E$ is the energy eigenvalue, and ${\lvert {u} \rangle}$ is the energy eigenstate. Our differential equation now becomes

\begin{aligned}-\frac{\hbar^2 }{2m} \frac{d^2 u(x)}{dx^2} + V(x) u(x) = E u(x)\end{aligned} \hspace{\stretch{1}}(5.102)

where $V(x) = 0$ for ${\left\lvert{x}\right\rvert} < a$. We won't find anything like this for real, but this is our first approximation to the quantum dot.

Our differential equation in the well is now

\begin{aligned}-\frac{\hbar^2 }{2m} \frac{d^2 u(x)}{dx^2} = E u(x)\end{aligned} \hspace{\stretch{1}}(5.103)

or with $\alpha = \sqrt{2m E/\hbar^2}$

\begin{aligned}\frac{d^2 u(x)}{dx^2} u(x) = -\frac{2 m E}{\hbar^2} u(x) = - \alpha^2 u(x)\end{aligned} \hspace{\stretch{1}}(5.104)

Our solution for ${\left\lvert{x}\right\rvert} < a$ is then

\begin{aligned}u(x) = A \cos \alpha x + B \sin\alpha x\end{aligned} \hspace{\stretch{1}}(5.105)

and for ${\left\lvert{x}\right\rvert} > a$ we have $u(x) = 0$ since $V(x) = \infty$.

Setting $u(a) = u(-a) = 0$ we have

\begin{aligned}A \cos \alpha a + B \sin\alpha a &= 0 \\ A \cos \alpha a - B \sin\alpha a &= 0\end{aligned}

### Type I.

$B=0$, $A \cos\alpha a = 0$. For $A \ne 0$ we must have

\begin{aligned}\cos \alpha a = 0\end{aligned}

or $\alpha a = n \frac{\pi}{2}$, where $n = 1, 3, 5, ...$, so our solution is

\begin{aligned}u(x) = A \cos \left( \frac{n \pi}{2 a} x \right) \end{aligned} \hspace{\stretch{1}}(5.106)

### Type II.

$A=0$, $B \sin\alpha a = 0$. For $B \ne 0$ we must have

\begin{aligned}\sin \alpha a = 0\end{aligned}

or $\alpha a = n \frac{\pi}{2}$, where $n = 1, 2, 4, ...$, so our solution is

\begin{aligned}u(x) = B \sin \left( \frac{n \pi}{2 a} x \right) \end{aligned} \hspace{\stretch{1}}(5.107)

### Via determinant

We could also write

\begin{aligned}\begin{bmatrix}\cos \alpha a & \sin\alpha a \\ \cos \alpha a & - \sin\alpha a \end{bmatrix}\begin{bmatrix}A \\ B\end{bmatrix}= 0\end{aligned}

and then must have zero determinant, or

\begin{aligned}-2 \sin\alpha a \cos\alpha a = -\sin 2 \alpha a\end{aligned} \hspace{\stretch{1}}(5.108)

so we must have

\begin{aligned}2 \alpha a = n \pi\end{aligned}

or

\begin{aligned}\alpha = \frac{n \pi}{2a}\end{aligned}

regardless of $A$ and $B$. We can then determine the solutions 5.106, and 5.107 simply by noting that this value for $\alpha$ kills off either the sine or cosine terms of 5.105 depending on whether $n$ is even or odd.

## CHECK.

\begin{aligned}u_n(x) &= A \cos \left( \frac{n \pi}{2 a} x \right) \\ u_n(x) &= B \sin \left( \frac{n \pi}{2 a} x \right) \end{aligned}

satisfy the time independent Schr\”{o}dinger equation, and the corresponding eigenvalues from from

\begin{aligned}\alpha = \sqrt{\frac{2 m E}{\hbar^2}},\end{aligned}

or

\begin{aligned}E = \frac{\hbar^2 \alpha^2}{2m} = \frac{\hbar^2}{2m} \left( \frac{n \pi}{2a} \right)^2 \end{aligned}

for $n = 1, 2, 3, \cdots$.

## On the derivative of $u$ at the boundaries

Integrating

\begin{aligned}-\frac{\hbar^2 }{2m} \frac{d^2 u(x)}{dx^2} u(x) + V(x) u(x) = E u(x),\end{aligned} \hspace{\stretch{1}}(5.109)

over $[a-\epsilon,a+\epsilon]$ we have

\begin{aligned}-\frac{\hbar^2 }{2m} &\int_{a-\epsilon}^{a-\epsilon}\frac{d^2 u(x)}{dx^2} dx+ \int_{a-\epsilon}^{a-\epsilon}V(x) u(x) dx = \int_{a-\epsilon}^{a-\epsilon}E u(x) dx \\ -\frac{\hbar^2 }{2m} &\left( \left.\frac{du}{dx}\right\vert_{a-\epsilon}^{a+\epsilon} + 0 = 0\right)\end{aligned} \hspace{\stretch{1}}(5.110)

which gives us

\begin{aligned}\left.\frac{du}{dx}\right\vert_{a + \epsilon}-\left.\frac{du}{dx}\right\vert_{a - \epsilon} = 0\end{aligned} \hspace{\stretch{1}}(5.112)

or

\begin{aligned}\left.\frac{du}{dx}\right\vert_{a + \epsilon}&=\left.\frac{du}{dx}\right\vert_{a - \epsilon} \end{aligned} \hspace{\stretch{1}}(5.113)

We can infer how the derivative behaves over the potential discontinuity, so in the limit where $\epsilon \rightarrow 0$ we must have wave function continuity at despite the potential discontinuity.

This sort of analysis, which is potential dependent, we see that for this infinite well potential, our derivative must be continuous at the boundary.

## Problem:

non-infinite step well potential.

Given a zero potential in the well ${\left\lvert{x}\right\rvert} < a$

\begin{aligned}-\frac{\hbar^2 }{2m} \frac{d^2 u(x)}{dx^2} u(x) + 0 = E u(x),\end{aligned} \hspace{\stretch{1}}(5.114)

and outside of the well ${\left\lvert{x}\right\rvert} > a$

\begin{aligned}-\frac{\hbar^2 }{2m} \frac{d^2 u(x)}{dx^2} u(x) + V_0 u(x) = E u(x)\end{aligned} \hspace{\stretch{1}}(5.115)

Inside of the well, we have the solution worked previously, with $\alpha = \sqrt{2m E/\hbar^2}$

\begin{aligned}u(x) &= A \cos\alpha x + B \sin\alpha x \end{aligned} \hspace{\stretch{1}}(5.116)

Then we have outside of the well the same form

\begin{aligned}-\frac{\hbar^2 }{2m} \frac{d^2 u(x)}{dx^2} u(x) = (E - V_0 )u(x) \end{aligned} \hspace{\stretch{1}}(5.117)

With $\beta = \sqrt{ 2m (V_0 - E)/\hbar^2}$, this is

\begin{aligned}\frac{d^2 u(x)}{dx^2} u(x) = \beta^2 u(x) \end{aligned} \hspace{\stretch{1}}(5.118)

If $V_0 - E > 0$, we have $V_0 > E$, and the states are “bound” or “localized” in the well.

Our solutions for this $V_0 > E$ case are then

\begin{aligned}u(x) &= D e^{\beta x} \\ u(x) &= C e^{-\beta x}\end{aligned} \hspace{\stretch{1}}(5.119)

for $x \le a$, and $x \ge a$ respectively.

Question: Why can we not have

\begin{aligned}u(x) = D e^{\beta x} + C e^{-\beta x}\end{aligned} \hspace{\stretch{1}}(5.121)

for $x \le -a$?

Answer: As $x \rightarrow -\infty$ we would then have

\begin{aligned}u(x) \rightarrow C e^{\beta \infty} \rightarrow \infty\end{aligned}

This is a non-physical solution, and we discard it based on our normalization requirement.

Our total solution, in regions $x a$ respectively

\begin{aligned}u_1(x) &= D e^{\beta x} \\ u_2(x) &= A \cos\alpha x + B \sin\alpha x \\ u_3(x) &= C e^{-\beta x}\end{aligned}

To find the coefficients, set $u_1(-a) = u_2(-a)$, $u_2(a) = u_3(a)$ $u_1'(-a) = u_2'(-a)$, $u_2'(a) = u_3'(a)$, and NORMALIZE $u(x)$.

Now, how about in region 2 ($x < -a$), $V_0 < E$ implies that our equation is

\begin{aligned}\frac{d^2 u(x)}{dx^2} u(x) = - \frac{2m}{\hbar^2} (E - V_0) u(x) = - k^2 u(x)\end{aligned} \hspace{\stretch{1}}(5.122)

We no longer have quantized energy for such a solution. These correspond to the “unbound” or “continuum” states. Even though we do not have quantized energy we still have quantum effects. Our solution becomes

\begin{aligned}u_1(x) &= C_2 e^{i k x} +D_2 e^{-i k x} \\ u_2(x) &= A e^{i \alpha x} +B e^{-i \alpha x} \\ u_3(x) &= C_3 e^{i k x} \end{aligned}

Question. Why no $D_2 e^{-i k x}$, in the $u_3(x)$ term?

Answer. We can, but this is not physically relevant. Why is because we associate $e^{ikx}$ with an incoming wave, with reflection in the $x < -a$ interval, and both $e^{\pm i \alpha x}$ in the $latex {\left\lvert{x}\right\rvert} a$ region.

FIXME: scan picture: 9.1 in my notebook.

Observe that this is not normalizable as is. We require “delta-function” normalization. What we can do is ask about current densities. How much passes through the barrier, and so forth.

Note to self. We probably really we want to consider a wave packet of states, something like:

\begin{aligned}\Psi_1(x) &= \int dk f_1(k) e^{i k x} \\ \Psi_2(x) &= \int d\alpha f_2(\alpha) e^{i \alpha x} \\ \Psi_3(x) &= \int dk f_3(k) e^{i k x}\end{aligned}

Then we’d have something that we can normalize. Play with this later.

# Setup for next week’s hydrogen atom lecture.

We’ll want to solve this using the formalism we’ve discussed. The general problem is a proton, positively charged, with a nearby negative charge (the electron).

Our equation to solve is

\begin{aligned}\left(-\frac{\hbar^2}{2 m_1} \boldsymbol{\nabla}_1^2-\frac{\hbar^2}{2 m_2} \boldsymbol{\nabla}_2^2\right)u(\mathbf{r}_1, \mathbf{r}_2) + V(\mathbf{r}_1, \mathbf{r}_2)u(\mathbf{r}_1, \mathbf{r}_2)=E u(\mathbf{r}_1, \mathbf{r}_2).\end{aligned} \hspace{\stretch{1}}(6.123)

Here $\left( -\frac{\hbar^2}{2 m_1} \boldsymbol{\nabla}_1^2 -\frac{\hbar^2}{2 m_2} \boldsymbol{\nabla}_2^2 \right)$ is the total kinetic energy term. For hydrogen we can consider the potential to be the Coulomb potential energy function that depends only on $\mathbf{r}_1 - \mathbf{r}_2$. We can transform this using a center of mass transformation. Introduce the centre of mass coordinate and relative coordinate vectors

\begin{aligned}\mathbf{R} &= \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{ m_1 + m_2 } \\ \mathbf{r} &= \mathbf{r}_1 - \mathbf{r}_2\end{aligned} \hspace{\stretch{1}}(6.124)

With this transformation we can reduce the problem to a single coordinate PDE.