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
<br /> \rho=\sum_{n}p_n |n\rangle\langle n|<br />

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
<br /> \langle O \rangle_{\rho}=\mathbf{Tr}(O\rho )<br />
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

<br /> \mathbb{I}=\sum_{n}|n\rangle\langle n|<br />

The Attempt at a Solution

I set up the density matrix like this
<br /> \rho=\frac{3}{10}|\uparrow_{x}\rangle \langle \uparrow_{x} |+\frac{7}{10}|\downarrow_{z}\rangle \langle \downarrow_{z} |<br />
and with
<br /> s_z=\frac{\hbar}{2}\begin{pmatrix}<br /> 1 &amp; 0 \\<br /> 0 &amp; -1<br /> \end{pmatrix}<br />
Then
<br /> \langle s_z\rangle_{\rho}=\mathbf{Tr}\left[\frac{3}{10}\frac{\hbar}{2}\begin{pmatrix}<br /> 1 &amp; 0 \\<br /> 0 &amp; -1<br /> \end{pmatrix}|\uparrow_{x}\rangle \langle \uparrow_{x} |+\frac{7}{10}\frac{\hbar}{2}\begin{pmatrix}<br /> 1 &amp; 0 \\<br /> 0 &amp; -1<br /> \end{pmatrix}|\downarrow_{z}\rangle \langle \downarrow_{z} |\right]<br />
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...
<br /> \mathbb{I}=|\uparrow_{z}\rangle\langle \uparrow_{z}|+|\downarrow_{z}\rangle \langle \downarrow_{z} |<br />
so
<br /> \begin{align}<br /> |\uparrow_{x}\rangle &amp;=|\uparrow_{z}\rangle\langle \uparrow_{z}|\uparrow_{x}\rangle+|\downarrow_{z} \rangle \langle \downarrow_{z}|\uparrow_{x}\rangle \\<br /> &amp;= \begin{pmatrix} 1 \\ 0 \end{pmatrix}\frac{1}{\sqrt{2}}+\begin{pmatrix}0 \\ 1 \end{pmatrix}\frac{1}{\sqrt{2}} \\<br /> &amp;= \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \end{pmatrix}<br /> \end{align}<br />
so
<br /> |\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 &amp; 1 \\ 1 &amp; 1 \end{pmatrix}<br />

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 &amp; 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 &amp; 1 \\ 1 &amp; 1 \end{pmatrix}

 

[jfy4]

Thank you,

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

Does this look good?

 

[vela]

Yes, looks good.

 

[jfy4]

Thanks for your help.