Electron Spin State and Values

[Rahmuss]

[SOLVED] Electron Spin State and Values

Homework Statement

An electron is in the spin state:

$$X = A\begin{pmatrix} 1-2i \\ 2 \end{pmatrix}$$

(a) Determine the constant A by normalizing $$X$$

(b) If you measured $$S_{z}$$ on this electron, what values could you get, and what is the probability of each? What is the expectation value of $$S_{z}$$?

Homework Equations

$$S_{z} = \frac{\hbar}{2}\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$$

$$\left\langle S_{z}\right\rangle = \left\langle X | S_{z}X\right\rangle$$

The Attempt at a Solution

Part a):
$$A^{2}\left[ |1-2i|^{2} + |2|^{2}\right] = 1 \Rightarrow$$
$$A^{2}\left[ 1-4i+4+4\right] = 1 \Rightarrow$$
$$A^{2}\left[ 9-4i\right] = 1 \Rightarrow$$
$$A^{2} = \frac{1}{9-4i} \Rightarrow$$
$$A = \sqrt{\frac{1}{9-4i}} \Rightarrow$$

Part b):
For this part I'm a bit confused (thus the posting). I'm not sure what they mean when they talk about measuring $$S_{a}$$ on the electron. Are they just saying, if you measured the z-component of the spin of the electron? And if so would I have something like:

$$\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix}1-2i \\ 2 \end{pmatrix} \Rightarrow$$
$$\frac{\hbar}{2}\begin{pmatrix} 1-2i \\ -2 \end{pmatrix}$$

And where do I go from there? And as far as the probabilities I may be able to get that if I know the normalization constant. And I think I can get the expectation value from the normalization constant as well.

 

[Rahmuss]

Any thoughts on normalizing this? I tried another way and got $$\frac{1}{3}$$ and another way and I got $$\frac{16\hbar - 2\hbar i}{18}$$. And neither of those seem to work. I thought that 1/3 would work; but when I try to find the expectation value I get $$\frac{\hbar}{2} \frac{13}{3}$$.

 

[Rahmuss]

Ok, I have from my teacher that $$A = \frac{1}{3}$$, and that $$\left\langle S_{x}\right\rangle = \frac{2\hbar}{9}$$.

I can normalize it and get $$A = \frac{1}{3}$$; but I can't get the correct expectation value. Here is what I am doing:

$$\left\langle X | S_{x}X\right\rangle = A^{2}\frac{h}{2}[(1+2i), 2]\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} (1-2i) \\ 2 \end{pmatrix} \Rightarrow$$

$$A^{2}\frac{h}{2}[(1+2i), 2][2 + (1-2i)] \Rightarrow$$

$$A^{2}\frac{h}{2}(2 + 4i + 4 + 1 + 4 + 2 -4i) = \frac{13\hbar }{18}$$

So what am I doing wrong? I try including the the constants during the calculations and it comes out the same:

$$\left\langle X | S_{x}X\right\rangle = \begin{pmatrix}\frac{(1+2i)}{3} & \frac{2}{3} \end{pmatrix} \begin{pmatrix} 0 & \frac{h}{2} \\ \frac{h}{2} & 0 \end{pmatrix} \begin{pmatrix} (1-2i)/3 \\ 2/3 \end{pmatrix} \Rightarrow$$

$$\frac{h}{2}\begin{pmatrix} \frac{(1+2i)}{3} & \frac{2}{3} \end{pmatrix} \begin{pmatrix} \frac{2}{3} & \frac{(1-2i)}{3} \end{pmatrix} \Rightarrow$$

$$\frac{h}{2}\frac{(2 + 4i + 4 + 1 + 4 + 2 -4i)}{18} = \frac{13\hbar }{18}$$

So, it still doesn't work. Any thoughts?

 

[2Tesla]

Here is what I am doing:

$$\left\langle X | S_{x}X\right\rangle = A^{2}\frac{h}{2}[(1+2i), 2]\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} (1-2i) \\ 2 \end{pmatrix} \Rightarrow$$

$$A^{2}\frac{h}{2}[(1+2i), 2][2 + (1-2i)] \Rightarrow$$

$$A^{2}\frac{h}{2}(2 + 4i + 4 + 1 + 4 + 2 -4i) = \frac{13\hbar }{18}$$

A matrix times a vector is a vector, so you want:

$$\left\langle X | S_{x}X\right\rangle = A^{2}\frac{\hbar}{2}[(1+2i), 2]\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} (1-2i) \\ 2 \end{pmatrix} \Rightarrow$$

$$A^{2}\frac{\hbar}{2}[(1+2i), 2][2, (1-2i)] \Rightarrow$$

Now take the dot product of the two vectors:

$$A^{2}\frac{\hbar}{2}(2 + 4i + 2 - 4i) = \frac{2\hbar }{9}$$

 

[Rahmuss]

Ah... Ok. Great! Thanks 2Tesla. That helps a lot. That makes sense now. Ok, I can see why I wasn't getting it. Thanks again.