Uncertainty Relation for Eigenstate of Spin-1/2 Particle

[Robben]

Homework Statement

Calculate $\triangle S_x$ and $\triangle S_y$ for an eigenstate of $\hat{S}_z$ for a spin$-\frac12$ particle. Check to see if the uncertainty relation $\triangle S_x\triangle S_y\ge \hbar|\langle S_z\rangle|/2$ is satisfied.

Homework Equations

$S_x =\frac12(S_+ +S_-)$
$S_y = \frac{1}{2i}(S_+-S_-)$

The Attempt at a Solution

I am not sure what I have to do in this problem. For the matrix representation I have that:

$S_x = \frac{\hbar}{2}\left[\begin{array}{ c c }0 & 1 \\1 & 0\end{array} \right]$
$S_y = \frac{\hbar}{2i}\left[\begin{array}{ c c }0 & -1 \\1 & 0\end{array} \right].$

 

[Orodruin]

What is the definition of uncertainty? I suggest you start from that and see what you can find based on it.

 

[Robben]

What is the definition of uncertainty? I suggest you start from that and see what you can find based on it.

I was more concerned on the first part of the question.

 

[Orodruin]

Yes? The uncertainty is $\Delta S_x$, how is it defined? Later you can go on to the uncertainty relation, which as the name suggests is a relation of uncertainties.

 

[Robben]

Yes? The uncertainty is $\Delta S_x$, how is it defined? Later you can go on to the uncertainty relation, which as the name suggests is a relation of uncertainties.

Ops, I thought you were talking about the uncertainty relation. But regarding $\triangle S_x$ the uncertainty is defined as $\sqrt{\langle S_x^2\rangle -\langle S_x\rangle^2}$, where $S_x = \frac{\hbar}{2}P_1 -\frac{\hbar}{2}P_2$.

So all I do is do matrix multiplication with $S_z$, i.e.

$(S_x) (S_z)=\frac{\hbar}{2}\left[\begin{array}{ c c }0 & 1 \\1 & 0\end{array} \right] \left[\begin{array}{ c c }1 & 0 \\0 & -1\end{array} \right].$

 

[Orodruin]

So what do you get when you evaluate the expectation values for an eigenstate of $S_z$? How do you compute the expectation value of any operator in a given state?

 

[Robben]

To compute the expectation value of any operator $\hat{\mathbb{O}}$ for a particle in the state $|\phi\rangle$ is defined as $\langle \hat{\mathbb{O}}\rangle = \langle \phi|\hat{\mathbb{O}}|\phi\rangle.$ But what will the eigenstate for $S_z$ be?

 

[Orodruin]

But what will the eigenstate for SzS_z be?

I suggest you take one of the ones required by the problem statement, i.e., any of the eigenstates of $S_z$.

 

[Robben]

I suggest you take one of the ones required by the problem statement, i.e., any of the eigenstates of $S_z$.

That doesn't help me understand.

 

[Orodruin]

So let us try it this way: What are the eigenstates of $S_z$?

Edit: Also, you should perhaps look for a different way of computing your expectation values. I suspect the one you quoted will not be very helpful. How would you compute $P_1$ and $P_2$ and how would you compute $\left< S_x^2 \right>$?

 

[Robben]

So let us try it this way: What are the eigenstates of $S_z$?

Well, in the book it states that $\hat{\mathbb{J}}_+|\lambda,m\rangle$ is an eigenstate of $\mathbb{J}_z$ (where I was told that $\mathbb{J}_z$ and $\mathbb{S}_z$ are interchangeable)##.

Edit: Also, you should perhaps look for a different way of computing your expectation values. I suspect the one you quoted will not be very helpful. How would you compute $P_1$ and $P_2$ and how would you compute $\left< S_x^2 \right>$?

So I can't use $S_x = \frac{\hbar}{2}P_1 -\frac{\hbar}{2}P_2$? I do not know any other way of computing the expectation values other than that definition.

 

[Orodruin]

Well, in the book it states that $\hat{\mathbb{J}}_+|\lambda,m\rangle$ is an eigenstate of $\mathbb{J}_z$ (where I was told that $\mathbb{J}_z$ and $\mathbb{S}_z$ are interchangeable)##.

Yes, but it requires that you already are familiar with another eigenstate. Are you familiar with how to find the eigenvectors of a matrix?

So I can't use $S_x = \frac{\hbar}{2}P_1 -\frac{\hbar}{2}P_2$? I do not know any other way of computing the expectation values other than that definition.

Not unless you figure out how to comput $P_1$ and $P_2$. What about the way you quoted in post #7?

 

[Robben]

Yes, but it requires that you already are familiar with another eigenstate. Are you familiar with how to find the eigenvectors of a matrix?

Yes, I am familiar with how to find eigenvectors. In quantum mechanics, I have difficulty in setting up the problem correctly. Computing it, I can do but setting it up I need a lot more practice with.

Not unless you figure out how to compute $P_1$ and $P_2$. What about the way you quoted in post #7?

That I know how to compute, its just that I am having trouble finding $|\phi\rangle$. Since we have $S_x$, i.e., $S_x = \frac{\hbar}{2}\left[\begin{array}{ c c }0 & 1 \\1 & 0\end{array} \right].$ Thus, I need to compute $\langle\phi|S_x|\phi\rangle$.

 

[Orodruin]

So, how do you represent $S_z$ in matrix form? What are the eigenvectors of that matrix? (The eigenvectors of the matrix represent the eigenstates.)

 

[Robben]

So, how do you represent $S_z$ in matrix form? What are the eigenvectors of that matrix? (The eigenvectors of the matrix represent the eigenstates.)

In matrix form $(S_z)=\frac{\hbar}{2} \left[\begin{array}{ c c }1 & 0 \\0 & -1\end{array} \right].$

 

[Orodruin]

... and so the eigenvectors are ...

 

[Robben]

... and so the eigenvectors are ...

Will the eigenvectors just be $|+z\rangle = {1\choose 0}$ and $|-z\rangle ={0 \choose 1}?$

 

[Orodruin]

You tell me. Is $S_z |+z\rangle = \lambda_+ |+z\rangle$ for some constant $\lambda_+$?

 

[Robben]

You tell me. Is $S_z |+z\rangle = \lambda_+ |+z\rangle$ for some constant $\lambda_+$?

Yup, it does satisfy that. I made this more difficult than it is. -__-. Thank you very much!

 

[Orodruin]

Just to resolve the problem itself: What do you get for $\langle S_x\rangle$ and $\langle S_x^2 \rangle$, respectively? What is the resulting uncertainty relation?

 

[Robben]

Just to resolve the problem itself: What do you get for $\langle S_x\rangle$ and $\langle S_x^2 \rangle$, respectively? What is the resulting uncertainty relation?

Using $\langle \hat{\mathbb{O}}\rangle = \langle \phi|\hat{\mathbb{O}}|\phi\rangle \implies \langle +z| S_x|+z\rangle.$ Therefore, $\langle S_x\rangle = 0$

 

[Orodruin]

Yes, this is correct. And $S_z^2$?

 

[Robben]

Yes, this is correct. And $S_z^2$?

$S_x^2$ will also equal $0$ since , $\langle S_x\rangle^2=0$.

 

[DelcrossA]

$S_x^2$ will also equal $0$ since , $\langle S_x\rangle^2=0$.

No, $S_x^2$ is a different operator than $S_x$. So you need to actually compute its uncertainty by itself.

 

[Robben]

No, $S_x^2$ is a different operator than $S_x$. So you need to actually compute its uncertainty by itself.

Opps, thank you very much for catching that! It will equal $\frac{\hbar^2}{4}$ instead of $0$.