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
[tex]
\rho=\sum_{n}p_n |n\rangle\langle n|
[/tex]

where [itex]p_n[/itex] is a classical probability distribution ([itex]\sum_{n} p_n=1,\; 0\leq p_n \leq 1[/itex]) and [itex]|n\rangle\langle n|[/itex] are projection operators that are not necessarily orthogonal. [itex]\rho[/itex] represents a classical statistical ensemble of quantum states where the state [itex]|n\rangle[/itex] appears with probability [itex]p_n[/itex]. The ensemble average of an operator [itex]O[/itex] is an ensemble of states described by a density matrix [itex]\rho[/itex] is
[tex]
\langle O \rangle_{\rho}=\mathbf{Tr}(O\rho )
[/tex]
Physically this is the average of a number of measurements of [itex]O[/itex] 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 [itex]s_z[/itex] in this ensemble of protons.

Homework Equations

[tex]
\mathbb{I}=\sum_{n}|n\rangle\langle n|
[/tex]

The Attempt at a Solution

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

Does that look right?

 

[vela]

Minor correction:
[tex]\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}[/tex]

 

[jfy4]

Thank you,

Then
[tex]
\begin{align}
\rho &=\frac{3}{20}\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}+\frac{7}{10}\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix} \\
&=\frac{1}{20}\begin{pmatrix} 3 & 3 \\ 3 & 17 \end{pmatrix}
\end{align}
[/tex]
Then
[tex]
\begin{align}
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} \\
&=\frac{\hbar}{40}\begin{pmatrix} 3 & 3 \\ -3 & -17 \end{pmatrix}
\end{align}
[/tex]
So
[tex]
\mathbf{Tr}(s_z\cdot\rho)=-\frac{7\hbar}{20}
[/tex]

Does this look good?

 

[vela]

Yes, looks good.

 

[jfy4]

Thanks for your help.