How Does a Density Matrix Represent Quantum Averages?

[jfy4]

I'm really excited to get this as a homework problem. I have wanted to feel good about this formalism is quantum mechanics for a while now but my own stupidity has been getting in the way... With this homework problem hopefully I can move on to a new level.

Homework Statement

The most general observable is a density matrix. Generally it is a non-negative self-adjoint operator with trace 1. It has the general form
$$\rho=\sum_{n}p_n |n\rangle\langle n|$$

where $p_n$ is a classical probability distribution ($\sum_{n} p_n=1,\; 0\leq p_n \leq 1$) and $|n\rangle\langle n|$ are projection operators that are not necessarily orthogonal. $\rho$ represents a classical statistical ensemble of quantum states where the state $|n\rangle$ appears with probability $p_n$. The ensemble average of an operator $O$ is an ensemble of states described by a density matrix $\rho$ is
$$\langle O \rangle_{\rho}=\mathbf{Tr}(O\rho )$$
Physically this is the average of a number of measurements of $O$ in a classical probability distribution of different states. Consider a polarized beam of protons where 30% of the protons have spin up in the x-direction and 70% have spin down in the z direction. Find the density matrix for this ensemble and compute the ensemble average of $s_z$ in this ensemble of protons.

Homework Equations

$$\mathbb{I}=\sum_{n}|n\rangle\langle n|$$

The Attempt at a Solution

I set up the density matrix like this
$$\rho=\frac{3}{10}|\uparrow_{x}\rangle \langle \uparrow_{x} |+\frac{7}{10}|\downarrow_{z}\rangle \langle \downarrow_{z} |$$
and with
$$s_z=\frac{\hbar}{2}\begin{pmatrix}<br /> 1 & 0 \\<br /> 0 & -1<br /> \end{pmatrix}$$
Then
$$\langle s_z\rangle_{\rho}=\mathbf{Tr}\left[\frac{3}{10}\frac{\hbar}{2}\begin{pmatrix}<br /> 1 & 0 \\<br /> 0 & -1<br /> \end{pmatrix}|\uparrow_{x}\rangle \langle \uparrow_{x} |+\frac{7}{10}\frac{\hbar}{2}\begin{pmatrix}<br /> 1 & 0 \\<br /> 0 & -1<br /> \end{pmatrix}|\downarrow_{z}\rangle \langle \downarrow_{z} |\right]$$
Now I need help with how to compute the above...

May I have some help?

Thanks

 

[vela]

Try finding the matrix representing the density operator with respect to the S_z eigenbasis.

 

[jfy4]

Well, here what I think I know...
$$\mathbb{I}=|\uparrow_{z}\rangle\langle \uparrow_{z}|+|\downarrow_{z}\rangle \langle \downarrow_{z} |$$
so
$$\begin{align}<br /> |\uparrow_{x}\rangle &=|\uparrow_{z}\rangle\langle \uparrow_{z}|\uparrow_{x}\rangle+|\downarrow_{z} \rangle \langle \downarrow_{z}|\uparrow_{x}\rangle \\<br /> &= \begin{pmatrix} 1 \\ 0 \end{pmatrix}\frac{1}{\sqrt{2}}+\begin{pmatrix}0 \\ 1 \end{pmatrix}\frac{1}{\sqrt{2}} \\<br /> &= \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \end{pmatrix}<br /> \end{align}$$
so
$$|\uparrow_{x}\rangle\langle \uparrow_{x}|=\frac{1}{2}\begin{pmatrix}1 \\ 1\end{pmatrix}\begin{pmatrix}1 \\ 1 \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$

Does that look right?

 

[vela]

Minor correction:
$$\lvert \uparrow_{x} \rangle \langle \uparrow_{x}\rvert = \frac{1}{2} \begin{pmatrix}1 \\ 1\end{pmatrix} \begin{pmatrix}1 & 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$

 

[jfy4]

Thank you,

Then
$$\begin{align}<br /> \rho &=\frac{3}{20}\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}+\frac{7}{10}\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix} \\<br /> &=\frac{1}{20}\begin{pmatrix} 3 & 3 \\ 3 & 17 \end{pmatrix}<br /> \end{align}$$
Then
$$\begin{align}<br /> s_z \cdot \rho &=\frac{\hbar}{2}\frac{1}{20}\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix}3 & 3 \\ 3 & 17 \end{pmatrix} \\<br /> &=\frac{\hbar}{40}\begin{pmatrix} 3 & 3 \\ -3 & -17 \end{pmatrix}<br /> \end{align}$$
So
$$\mathbf{Tr}(s_z\cdot\rho)=-\frac{7\hbar}{20}$$

Does this look good?

 

[vela]

Yes, looks good.

 

[jfy4]

Thanks for your help.