Density matrix for a mixed neutron beam

[AwesomeTrains]

Homework Statement

A beam of neutrons (moving along the z-direction) consists of an incoherent superposition of two beams that were initially all polarized along the x- and y-direction, respectively.
Using the Pauli spin matrices:
<br /> \sigma_x = \begin{pmatrix}<br /> 0 &amp; 1 \\<br /> 1 &amp; 0 \\<br /> \end{pmatrix},<br /> \sigma_y = \begin{pmatrix}<br /> 0 &amp; -i \\<br /> i &amp; 0 \\<br /> \end{pmatrix},<br /> \sigma_z = \begin{pmatrix}<br /> 1 &amp; 0 \\<br /> 0 &amp; -1 \\<br /> \end{pmatrix}<br />
give the density matrix $\hat{\rho}$ for this beam and determine the mean polarization $\langle\vec{\sigma}\rangle = tr \hat{\rho}\vec{\sigma}$

2. Homework Equations
$\hat{\rho} = \sum_n c_n |n\rangle \langle n |$ Where $c_n$ is the relative probability to be in state n. Spin states $|+\rangle = |1/2,1/2\rangle, |-\rangle = |1/2,-1/2\rangle$

The Attempt at a Solution

I'm not sure what it means that the beams were initially polarized along the x- and y-axis and I don't know where to start with this problem.
I'm guessing with polarization they mean the direction of the spin?
Does the polarization change with time since they explicitly mention that it's the initial state?

 

[nrqed]

Homework Statement

A beam of neutrons (moving along the z-direction) consists of an incoherent superposition of two beams that were initially all polarized along the x- and y-direction, respectively.
Using the Pauli spin matrices:
<br /> \sigma_x = \begin{pmatrix}<br /> 0 &amp; 1 \\<br /> 1 &amp; 0 \\<br /> \end{pmatrix},<br /> \sigma_y = \begin{pmatrix}<br /> 0 &amp; -i \\<br /> i &amp; 0 \\<br /> \end{pmatrix},<br /> \sigma_z = \begin{pmatrix}<br /> 1 &amp; 0 \\<br /> 0 &amp; -1 \\<br /> \end{pmatrix}<br />
give the density matrix $\hat{\rho}$ for this beam and determine the mean polarization $\langle\vec{\sigma}\rangle = tr \hat{\rho}\vec{\sigma}$

2. Homework Equations
$\hat{\rho} = \sum_n c_n |n\rangle \langle n |$ Where $c_n$ is the relative probability to be in state n. Spin states $|+\rangle = |1/2,1/2\rangle, |-\rangle = |1/2,-1/2\rangle$

The Attempt at a Solution

I'm not sure what it means that the beams were initially polarized along the x- and y-axis and I don't know where to start with this problem.
I'm guessing with polarization they mean the direction of the spin?
Does the polarization change with time since they explicitly mention that it's the initial state?

Hint: when you wrote $|+\rangle = |1/2,1/2\rangle,$, this is a spin up state with respect to what axis?

 

[AwesomeTrains]

I think $|+\rangle$ is with respect to the z-axis $\hat{S_z} |+\rangle = 1/2 |+\rangle$.
On wikipedia it says that spin oriented along the x-axis is represented by: $|1/2,1/2\rangle _x = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and
$|1/2,1/2\rangle _y = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ i \end{bmatrix}$ for the y-axis.
Does this mean the beam is made up of a sum of these two states? $(\hat{\rho} =c_1|1/2,1/2\rangle _x \langle1/2,1/2| _x + c_2|1/2,1/2\rangle _y \langle1/2,1/2| _y )$ with $c_1+c_2=1$?

 

[nrqed]

I think $|+\rangle$ is with respect to the z-axis $\hat{S_z} |+\rangle = 1/2 |+\rangle$.
On wikipedia it says that spin oriented along the x-axis is represented by: $|1/2,1/2\rangle _x = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and
$|1/2,1/2\rangle _y = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ i \end{bmatrix}$ for the y-axis.
Does this mean the beam is made up of a sum of these two states? $(\hat{\rho} =c_1|1/2,1/2\rangle _x \langle1/2,1/2| _x + c_2|1/2,1/2\rangle _y \langle1/2,1/2| _y )$ with $c_1+c_2=1$?

I think $|+\rangle$ is with respect to the z-axis $\hat{S_z} |+\rangle = 1/2 |+\rangle$.

Let's make sure that the basic concepts are clear. How do you see that $\hat{S_z} |+\rangle = 1/2 |+\rangle$ ??

 

[AwesomeTrains]

Let's make sure that the basic concepts are clear. How do you see that $\hat{S_z} |+\rangle = 1/2 |+\rangle$ ??

It's because it's the eigenstate of $\hat{S}_z = \frac{1}{2} \sigma_z =\frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$ if $|+\rangle$ is represented as $\begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}$

 

[nrqed]

Homework Statement

A beam of neutrons (moving along the z-direction) consists of an incoherent superposition of two beams that were initially all polarized along the x- and y-direction, respectively.
Using the Pauli spin matrices:
<br /> \sigma_x = \begin{pmatrix}<br /> 0 &amp; 1 \\<br /> 1 &amp; 0 \\<br /> \end{pmatrix},<br /> \sigma_y = \begin{pmatrix}<br /> 0 &amp; -i \\<br /> i &amp; 0 \\<br /> \end{pmatrix},<br /> \sigma_z = \begin{pmatrix}<br /> 1 &amp; 0 \\<br /> 0 &amp; -1 \\<br /> \end{pmatrix}<br />
give the density matrix $\hat{\rho}$ for this beam and determine the mean polarization $\langle\vec{\sigma}\rangle = tr \hat{\rho}\vec{\sigma}$

2. Homework Equations
$\hat{\rho} = \sum_n c_n |n\rangle \langle n |$ Where $c_n$ is the relative probability to be in state n. Spin states $|+\rangle = |1/2,1/2\rangle, |-\rangle = |1/2,-1/2\rangle$

The Attempt at a Solution

I'm not sure what it means that the beams were initially polarized along the x- and y-axis and I don't know where to start with this problem.
I'm guessing with polarization they mean the direction of the spin?
Does the polarization change with time since they explicitly mention that it's the initial state?

It's because it's the eigenstate of $\hat{S}_z = \frac{1}{2} \sigma_z =\frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$ if $|+\rangle$ is represented as $\begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}$

But you had written that $|+\rangle = |1/2,1/2\rangle$, so I thought that the notation you had in mind was that $|1/2,0\rangle$ was the eigenstate of $S_z$ with eigenvalue $\hbar/2$. But ok, I can accept your notation for $|+ \rangle$.

Ok, my next question is then: what are the eigenstates of $S_x$? What about those of $S_y$?

 

[AwesomeTrains]

But you had written that $|+\rangle = |1/2,1/2\rangle$, so I thought that the notation you had in mind was that $|1/2,0\rangle$ was the eigenstate of $S_z$ with eigenvalue $\hbar/2$. But ok, I can accept your notation for $|+ \rangle$.

Ok, my next question is then: what are the eigenstates of $S_x$? What about those of $S_y$?

Sorry about the misconception I got a bit confused myself since I've seen quite a few different notations in my lecture and on the internet.
$|+\rangle_y=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \\ \end{pmatrix}$ and
$|+\rangle_x=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \\ \end{pmatrix}$ (The $\hbar$ is in front of the operators in my lecture)

 

[nrqed]

Sorry about the misconception I got a bit confused myself since I've seen quite a few different notations in my lecture and on the internet.
$|+\rangle_y=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \\ \end{pmatrix}$ and
$|+\rangle_x=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \\ \end{pmatrix}$ (The $\hbar$ is in front of the operators in my lecture)

If we apply $S_x$ to your $|+\rangle_x$, we don't get that state back, right?

Now, once you have the correct eigenstates, we just need to use the definition $\rho = \sum_n c_n | n \rangle \langle n |$. The question is incomplete since they don't specify what fraction of the beam in each state so I assume they want a 50/50 mix. In that case $c_1 = c_2 = 0.5$. Have you seen how to calculate an outer product $| n \rangle \langle n |$? This will give a matrix.

 

[AwesomeTrains]

If we apply $S_x$ to your $|+\rangle_x$, we don't get that state back, right?

Now, once you have the correct eigenstates, we just need to use the definition $\rho = \sum_n c_n | n \rangle \langle n |$. The question is incomplete since they don't specify what fraction of the beam in each state so I assume they want a 50/50 mix. In that case $c_1 = c_2 = 0.5$. Have you seen how to calculate an outer product $| n \rangle \langle n |$? This will give a matrix.

$\rho = \sum_n c_n | n \rangle \langle n |= c_1|+\rangle_x \langle + |_x+c_2|+\rangle_y \langle + |_y\\=\frac{c_1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1^{*} & 1^{*}\\ \end{pmatrix}+ \frac{c_2}{\sqrt{2}} \begin{pmatrix} 1 \\ i \\ \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1^{*} & i^{*}\\ \end{pmatrix} \\=\frac{c_1}{2} \begin{pmatrix} 1*1 & 1*1 \\ 1*1 & 1*1 \\ \end{pmatrix}+ \frac{c_2}{2} \begin{pmatrix} 1*1 & 1*(-i) \\ i*1 & i*(-i) \\ \end{pmatrix}\\= \frac{1}{4}\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}+ \frac{1}{4} \begin{pmatrix} 1 & -i \\ i & 1 \\ \end{pmatrix}\\=\frac{1}{4}(\sigma_x+I)+\frac{1}{4}(\sigma_y+I)\\= \frac{1}{4}(\sigma_x+\sigma_y+2I)$
Just one last question then I think I got it
Why is it that we can use $|+\rangle_{x/y}$ here?
I thought the $| n \rangle$ were the eigenstates of the hamilton operator. I'm probably missing the interpretation of it

 

[nrqed]

$\rho = \sum_n c_n | n \rangle \langle n |= c_1|+\rangle_x \langle + |_x+c_2|+\rangle_y \langle + |_y\\=\frac{c_1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1^{*} & 1^{*}\\ \end{pmatrix}+ \frac{c_2}{\sqrt{2}} \begin{pmatrix} 1 \\ i \\ \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1^{*} & i^{*}\\ \end{pmatrix} \\=\frac{c_1}{2} \begin{pmatrix} 1*1 & 1*1 \\ 1*1 & 1*1 \\ \end{pmatrix}+ \frac{c_2}{2} \begin{pmatrix} 1*1 & 1*(-i) \\ i*1 & i*(-i) \\ \end{pmatrix}\\= \frac{1}{4}\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}+ \frac{1}{4} \begin{pmatrix} 1 & -i \\ i & 1 \\ \end{pmatrix}\\=\frac{1}{4}(\sigma_x+I)+\frac{1}{4}(\sigma_y+I)\\= \frac{1}{4}(\sigma_x+\sigma_y+2I)$
Just one last question then I think I got it
Why is it that we can use $|+\rangle_{x/y}$ here?
I thought the $| n \rangle$ were the eigenstates of the hamilton operator. I'm probably missing the interpretation of it

Good work.

In general a quantum state does not have to be an eigenstate of the Hamiltonian. A general solution to the time dependent Schrödinger equation for example is not an eigenstate of the Hamiltonian. And here we do not even have a Hamiltonian - it could be a free particle Hamiltonian for example - so that's not even something we can check.

 

[AwesomeTrains]

Thanks a lot for the help and patience! The last part of the question I think I can do on my own now. Have a nice weekend :)