Electron in a Quantum State: Finding Eigenvectors and Eigenvalues

[Taylor_1989]

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

 

[TSny]

OK for part (a). For part (b), I think you've been thrown off course by the statement of the problem which implies that the parameter b will be a function of μ. Note the form of the Hamiltonian. μ is just an overall multiplicative constant. So, should the eigenvectors of H depend on μ?

 

[Taylor_1989]

No, I don’t think so because the of the fact alls you effectively is extending the vector in the direction it already going in. I originally thought this but then beacuse the question stated it as a function I was not sure.

 

[TSny]

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
$$\mu A-\mu B-\left(\frac{3-\sqrt{5}}{2}\right)\mu A=0$$

Since $\mu$ cancels out in this equation, $A$ and $B$ do not depend on $\mu$. So, $b$ will not depend on $\mu$.

 

[Taylor_1989]

Thank for the response, I have realized what going on, I did cancel through originally and just had the value. But for some reason I seem to think when it comes to quantum physics "maybe this could be true" and forget the basics of maths, in part iv only just been introduced this year so I am still trying to wrap my head around more the concepts that anything, I have started going through Griffith which seem to be a very good book for the concepts, so hopefully I won't make silly mistakes like this again. Once again much appreciated.