Quantum Mechanics


Quantum Oscillator

February 6th, 2010

Every Friday evening the graduate students at SFU have a problem solving session where we try to solve a problem as a group that one of the faculty proposed. This week we were asked to numerically compute the solution to the quantum oscillator. The solution to the problem is really neat and it is something that you don’t usually get to see. Most physics programs don’t bother to show you how the quantum oscillator moves.

For those of you that care, we were solving the following equation:

i \epsilon u_t + \epsilon^2 u_{xx} = (x^2) u

with periodic boundary conditions, and an off-centered Gaussian as the initial conditions. We ended up using a spectral method to solve the problem using python. You can check out the code here.

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Answer: Infinite Potential Well

February 23rd, 2008

Question: A quantum particle is inside a one-dimensional infinitely deep potential well.

Find the wavefunction that satisfies Schrödinger’s equation.

i \hbar \frac{\partial}{\partial t} \Psi (\vec{r}, t)= \frac{-\hbar^2}{2m} \nabla^2 \Psi (\vec{r}, t) + V(\vec{r}) \Psi (\vec{r}, t)

Answer:

First assume that \Psi (\vec{r}, t) can be represented as \Psi (\vec{r}, t) = \phi (\vec{r}) \Phi (t). This means that Schrödinger’s equation can be manipulated as followed:

i \hbar \frac{\partial}{\partial t} \Psi (\vec{r}, t)= \frac{-\hbar^2}{2m} \nabla^2 \Psi (\vec{r}, t) + V(\vec{r}) \Psi (\vec{r}, t)
i \hbar \phi (\vec{r}) \frac{\partial}{\partial t} \Phi (t)= \frac{-\hbar^2}{2m}\Phi (t) \nabla^2 \phi (\vec{r}) + V(\vec{r}) \phi (\vec{r}) \Phi (t)
\frac{i \hbar}{\Phi (t)} \frac{\partial}{\partial t}\Phi (t)= \frac{-\hbar^2}{2m \phi (\vec{r})} \nabla^2 \phi (\vec{r}) + V(\vec{r})

Since the this equation holds for all \vec{r} and t, both sides of the equation must equal a constant.

\frac{i \hbar}{\Phi (t)} \frac{\partial}{\partial t}\Phi (t)= \frac{-\hbar^2}{2m \phi (\vec{r})} \nabla^2 \phi (\vec{r}) + V(\vec{r}) = E

Time Component
Next, we will solve for the time component of the above equation.

\frac{i \hbar}{\Phi (t)} \frac{\partial}{\partial t}\Phi (t) = E
i \hbar \int \frac{d\Phi}{\Phi (t)} = E \int dt
i \hbar ln(\Phi (t)) = E t
\Phi (t) = e^{\frac{-iE}{\hbar}t}

Time-Independent Component
The time-independent component of the Schrödinger equation would be:

\frac{-\hbar^2}{2m} \nabla^2 \phi (\vec{r}) + \phi (\vec{r})V(\vec{r}) = E\phi (\vec{r})

The solution to this problem can be easily solved for the case x \le 0 and x \ge L. Since V(x) = \infty, the only way to solve

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

is to have \phi (x) = 0.

For the case when 0 < x < L, we know that V(x) =0. Therefore, the time-independent equation will become:

\frac{-\hbar^2}{2m} \frac{d^2}{dx^2} \phi (x) = E\phi (x)

This equation should look rather familiar because it is the differential equation describing a simple harmonic oscillator. Therefore, the general solution to this differential equation would be:

\phi (x) = A sin(\frac{\sqrt{2mE}}{\hbar}x) + B cos(\frac{\sqrt{2mE}}{\hbar}x)

Therefore, the solution would be:

\phi (x)=\begin{cases}0 & x \le 0~and~x \ge L \\A sin(\frac{\sqrt{2mE}}{\hbar}x) + B cos(\frac{\sqrt{2mE}}{\hbar}x)& 0<x<L\end{cases}

If we add a requirement that \phi (x) needs to be piecewise-smooth, we can add boundary conditions \phi (0) = 0 and \phi (L) = 0 to the problem:

\phi (x) = A sin(\frac{\sqrt{2mE}}{\hbar}x) + B cos(\frac{\sqrt{2mE}}{\hbar}x)

Therefore, when \phi (0), we can see that B=0 because

\phi (x) = A sin(0) + B cos(0) = B = 0

Now that we know B=0, we can apply the second boundary condition \phi (L)=0 to the remaining portion of the equation.

\phi (L) = A sin(\frac{\sqrt{2mE}}{\hbar}L) = 0
\therefore ~ \frac{\sqrt{2mE}}{\hbar}L = n \pi where n \in \mathbb N

This implies

E = \frac{n^2 \pi^2 \hbar^2}{2mL^2}

Therefore, we can see that the general solution can be reduced to:

\phi (x)=\begin{cases}0 & x \le 0~and~x \ge L \\A sin(\frac{n\pi}{L}x) & 0<x<L\end{cases}

The last step is to normalize this wave-function

1 = \int_{-\infty}^{\infty} |\phi (x)|^2 dx
= |A|^2 \int_{0}^{L} sin^2(\frac{n\pi}{L}x) dx
= |A|^2 \frac{L}{n\pi}(-\frac{1}{2}cos(\frac{n\pi}{L}x)sin(\frac{n\pi}{L}x) + \frac{1}{2}\frac{n\pi}{L}x)\arrowvert_{0}^{L}
= |A|^2 \frac{L}{n\pi}(\frac{1}{2}n\pi) = |A|^2 \frac{L}{2}

Therefore

A = \sqrt{\frac{2}{L}}

Therefore, the final solution would be

\Psi (x, t)=\begin{cases}0 & x \le 0~and~x \ge L \\ \sqrt{\frac{2}{L}} sin(\frac{n\pi}{L}x)e^{\frac{-iE}{\hbar}t} & 0<x<L\end{cases}
where E = \frac{n^2 \pi^2 \hbar^2}{2mL^2}

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Question: Infinite Potential Well

February 17th, 2008

Question: A quantum particle is inside a one-dimensional infinitely deep potential well.

Find the wavefunction that satisfies Schrödinger’s equation.

i \hbar \frac{\partial}{\partial t} \Psi (\vec{r}, t)= \frac{-\hbar^2}{2m} \nabla^2 \Psi (\vec{r}, t) + V(\vec{r}) \Psi (\vec{r}, t)

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Higgs Boson

November 7th, 2007

I just read an awesome overview of the Higgs Boson and the Standard Model on Cosmic Variance. Although I have an under-grad degree in physics, Profs so often brush over this stuff.

Thanks for the great article John! It was very educational.

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