Density matrix for a mixed neutron beam

In summary: It is a little confusing if you write the eigenvalues of ##\sigma_z## as ##1/2,-1/2##. So, please be consistent and correct in your notation.
  • #1
AwesomeTrains
116
3

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:
[tex]
\sigma_x = \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix},
\sigma_y = \begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix},
\sigma_z = \begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix}
[/tex]
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?
 
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  • #2
AwesomeTrains said:

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:
[tex]
\sigma_x = \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix},
\sigma_y = \begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix},
\sigma_z = \begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix}
[/tex]
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?
 
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  • #3
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##?
 
  • #4
AwesomeTrains said:
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##?
AwesomeTrains said:
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 ## ??
 
  • #5
nrqed said:
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} ##
 
  • #6
AwesomeTrains said:

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:
[tex]
\sigma_x = \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix},
\sigma_y = \begin{pmatrix}
0 & -i \\
i & 0 \\
\end{pmatrix},
\sigma_z = \begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix}
[/tex]
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?
AwesomeTrains said:
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##?
 
  • #7
nrqed said:
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)
 
  • #8
AwesomeTrains said:
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.
 
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  • #9
nrqed said:
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
 
  • #10
AwesomeTrains said:
##
\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 Schrodinger 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.
 
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  • #11
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 :)
 
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1. What is a density matrix for a mixed neutron beam?

A density matrix for a mixed neutron beam is a mathematical representation of the state of a neutron beam that contains a mixture of different types of neutrons, such as thermal, epithermal, and fast neutrons. It contains information about the probabilities of finding the beam in different states, as well as the correlations between the different states.

2. How is a density matrix for a mixed neutron beam calculated?

A density matrix for a mixed neutron beam is calculated using the density operator, which is a mathematical operator that represents the state of a quantum system. It takes into account the different energy levels and states of the neutrons in the beam, as well as their relative probabilities.

3. What is the significance of a density matrix for a mixed neutron beam?

The density matrix for a mixed neutron beam is significant because it allows us to describe and understand the behavior of a complex neutron beam in a more comprehensive way. It provides information about the composition of the beam and how it evolves over time, which is important for many applications in nuclear physics and engineering.

4. Can a density matrix for a mixed neutron beam change over time?

Yes, a density matrix for a mixed neutron beam can change over time as the beam interacts with its surroundings. This includes interactions with other particles, as well as interactions with external fields, such as magnetic or electric fields. These interactions can alter the probabilities and correlations of the different neutron states in the beam.

5. How is a density matrix for a mixed neutron beam used in practical applications?

A density matrix for a mixed neutron beam is used in a variety of practical applications, including neutron scattering experiments, nuclear reactor design, and neutron imaging. It provides important information for predicting and understanding the behavior of neutrons in these systems, which can help improve their efficiency and safety.

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