- #1
Taylor_1989
- 402
- 14
Homework Statement
If possible could someone have a look at my working for this problem, I am not sure if I have carried out part b) correctly. I have done all three problem and carried through my solution to b) just to see if it did simplify out, which it didn’t which make me think I may have carried b) out incorrect. I have attached the question as a picture due to the amount of content.
Homework Equations
$$\hat{H}|n\rangle-IE_{n}|n\rangle \:[1]$$
$$P(R)=|\left\langle R|n \right\rangle|^2 \: [2]$$
The Attempt at a Solution
My working for a)
$$|n\rangle=A|n\rangle+B|n\rangle$$
$$|n\rangle=A\begin{pmatrix}1\\ 0\end{pmatrix}+B\begin{pmatrix}0\\ 1\end{pmatrix}$$
$$|n\rangle=\begin{pmatrix}A\\ 0\end{pmatrix}+\begin{pmatrix}0\\ B\end{pmatrix}$$
$$|n\rangle=\begin{pmatrix}A\\ B\end{pmatrix}$$
Finding the eingevalue
$$\hat{H}|n\rangle-IE_{n}|n\rangle=0$$
$$\hat{H}-IE_{n}=0$$
$$\begin{pmatrix}\mu &-\mu \\ -\mu &2\mu \end{pmatrix}-\begin{pmatrix}E_n&0\\ 0&E_n\end{pmatrix}=0$$
Find the Characteristic equation, using the determinate of the matrix below,
$$\begin{pmatrix}\mu -E_n&-\mu \\ -\mu &2\mu -E_n\end{pmatrix}=0$$
$$\left(\mu -E_n\right)\left(2\mu -E_n\right)-\mu ^2=0$$
Solving the quartic I get the following roots
$$E_0=\frac{3\mu -\sqrt{5}\mu }{2}\:,\:E_1=\frac{3\mu \:+\sqrt{5}\mu \:}{2}$$
b) finding the eigenvector
$$\begin{pmatrix}\mu &-\mu \\ -\mu &2\mu \end{pmatrix}\begin{pmatrix}A\\ B\end{pmatrix}-\left(\frac{3-\sqrt{5}}{2}\right)\begin{pmatrix}\mu A\\ \mu B\end{pmatrix}=0$$
Solving the top line
$$\mu A-\mu B-\left(\frac{3-\sqrt{5}}{2}\right)\mu A=0$$
I have kept ##\mu## in as it want ##b(\mu)##
if ##A=1## the I get the following
$$|0\rangle=\begin{pmatrix}\mu \\ \left(\frac{-1+\sqrt{5}}{2}\right)\mu \end{pmatrix}$$
I am going to ommit the normalising constants as they will cancel through, as they are the same, and as stated not need for this part of the problem.
$$\begin{pmatrix}\mu \\ \left(\frac{-1+\sqrt{5}}{2}\right)\mu \end{pmatrix}=\begin{pmatrix}1\\ \frac{1}{2}-b\end{pmatrix}$$
I do think this is wrong because I really can't understand why the above works, I understanding the if you scale and eignevector not matter what, the eigenvalue will still be the same, but, I can't see how the above is correct as in this case ##\mu=1## anyway carring on solving through I get the following for b.
$$b(\mu)=\frac{1}{2}+\left(\frac{\left(1-\sqrt{5}\right)\mu }{2}\right)$$
this gave me the new ground state vector
$$|0\rangle = N_0\begin{pmatrix}1\\ \frac{1}{2}-\frac{1}{2}\frac{\left(1-\sqrt{5}\right)}{2}\end{pmatrix}=N_0\begin{pmatrix}1\\ \left(\frac{\sqrt{5}-1}{2}\mu \right)\end{pmatrix}$$
c) assuming that ##b(\mu)## is correct I then did the following
$$\langle 0|0 \rangle=N_{0}^2(|L\rangle + \frac{\left(\sqrt{5}-1\right)}{2}\mu|R\rangle)(\langle L| + \frac{\left(\sqrt{5}-1\right)}{2}\mu\langle R|)$$
So multiplying out a ##\langle 0|0 \rangle=1## I make ##N_0=\sqrt{\frac{2}{2+\left(3-\sqrt{5}\right)\mu ^2}}##
So now find find the probability I did the following first I found
$$\langle R|0\rangle=\sqrt{\frac{2}{2+\left(3-\sqrt{5}\right)\mu ^2}}\left(\frac{\left(3-\sqrt{5}\right)\mu ^2}{2}\right)$$
Using equation [2] I found the probability for the electron to be in the right to be
$$P(R)=\frac{\left(14-6\sqrt{5}\right)\mu ^{4\:}}{4+2\left(3-\sqrt{5}\right)\mu ^2}$$
As I have stated I am not really sure if my b is correct, but can't really see any other way of making it a constant function