Do i need to calculate the expectation value of the Hamiltonian?

[Zero1010]

Homework Statement

Quick question about using the generalised Ehrenfest Theorem.

Relevant Equations

$$\frac {d<A>} {dt} = \frac{1}{i\hbar}<[\hat A,\hat H]>$$

Hi,

I have a question which asks me to use the generalised Ehrenfest Theorem to find expressions for
$\frac {d<Sx>} {dt}$ and $\frac {d<Sy>} {dt}$ - I have worked out <Sx> and <Sy> earlier in the question.

Since the generalised Ehrenfest Theorem takes the form:

$$\frac {d<A>} {dt} = \frac{1}{i\hbar}<[\hat A,\hat H]>$$

Does this mean I also have to find the expectation value of the Hamiltonian operator?

Thanks

 

[PeroK]

Homework Statement:: Quick question about using the generalised Ehrenfest Theorem.
Relevant Equations:: $$\frac {d<A>} {dt} = \frac{1}{i\hbar}<[\hat A,\hat H]>$$

Hi,

I have a question which asks me to use the generalised Ehrenfest Theorem to find expressions for
$\frac {d<Sx>} {dt}$ and $\frac {d<Sy>} {dt}$ - I have worked out <Sx> and <Sy> earlier in the question.

Since the generalised Ehrenfest Theorem takes the form:

$$\frac {d<A>} {dt} = \frac{1}{i\hbar}<[\hat A,\hat H]>$$

Does this mean I also have to find the expectation value of the Hamiltonian operator?

Thanks

Not if you can avoid it!

 

[Gaussian97]

Homework Statement:: Quick question about using the generalised Ehrenfest Theorem.
Relevant Equations:: $$\frac {d<A>} {dt} = \frac{1}{i\hbar}<[\hat A,\hat H]>$$

Hi,

I have a question which asks me to use the generalised Ehrenfest Theorem to find expressions for
$\frac {d<Sx>} {dt}$ and $\frac {d<Sy>} {dt}$ - I have worked out <Sx> and <Sy> earlier in the question.

Since the generalised Ehrenfest Theorem takes the form:

$$\frac {d<A>} {dt} = \frac{1}{i\hbar}<[\hat A,\hat H]>$$

Does this mean I also have to find the expectation value of the Hamiltonian operator?

Thanks

You have to compute the expectation value of the commutator

 

[Zero1010]

Thanks for the replies.

Ok, does that mean I use the values I have been given for Sx (rather than the expectation value I have calculated) and the Hamiltonian and then work out the expectation of the commutator. I'm pretty sure the two operators commute so that should be easy.

Thanks

 

[PeroK]

Thanks for the replies.

Ok, does that mean I use the values I have been given for Sx (rather than the expectation value I have calculated) and the Hamiltonian and then work out the expectation of the commutator. I'm pretty sure the two operators commute so that should be easy.

Thanks

Two operators commute iff the commutator is zero. Do you really mean that?

 

[Zero1010]

Yes that's what I mean. I have been given the Hamiltonian as:

$\hat H = \omega \hat Sx$

Which I think should commute with $$\hat Sx$$ as $$\omega$$ is a constant. I'm I on the right track?

 

[nrqed]

Yes that's what I mean. I have been given the Hamiltonian as:

$\hat H = \omega \hat Sx$

Which I think should commute with $$\hat Sx$$ as $$\omega$$ is a constant. I'm I on the right track?

Yes, it does commute.

 

[Zero1010]

Ok, one last question about this...until the next one.

So the two operators commute so the RHS of the generalised Ehrenfest Theorem = 0. However using the expectation value for Sx I calculated earlier

$\frac {d<Sx>} {dt} \neq 0$

Does this mean my value for <Sx> is wrong or am I missing the point of the Ehrenfest Theorem?

Thanks for your help so far.

 

[PeroK]

Ok, one last question about this...until the next one.

So the two operators commute so the RHS of the generalised Ehrenfest Theorem = 0. However using the expectation value for Sx I calculated earlier

$\frac {d<Sx>} {dt} \neq 0$

Does this mean my value for <Sx> is wrong or am I missing the point of the Ehrenfest Theorem?

Thanks for your help so far.

Something is wrong somewhere. That Hamiltonian looks suspicious to me.

 

[Gaussian97]

Ok, one last question about this...until the next one.

So the two operators commute so the RHS of the generalised Ehrenfest Theorem = 0. However using the expectation value for Sx I calculated earlier

$\frac {d<Sx>} {dt} \neq 0$

Does this mean my value for <Sx> is wrong or am I missing the point of the Ehrenfest Theorem?

Thanks for your help so far.

Please, post your calculation so we can help you

 

[Zero1010]

In what way?

The value of the Hamiltonian and Sx were given in the question. As I said above, the Hamiltonian is given as:

$\hat H = \omega \hat Sx$

and Sx is given as:

$\hat Sx = \frac{\hbar}{2} \begin{pmatrix} 0&1 \\ 1&0\end{pmatrix}$

 

[Gaussian97]

And from here how do you conclude that $$\frac{d \left<\hat{S}_x\right>}{dt}\neq 0$$?

 

[Zero1010]

Sorry, what I was saying here is if I work out the derivative of the expectation value of $\hat S_{x}$ it dosent equal zero like the RHS of the Ehrenfest equation from above so I am worried that the value obtained for $\hat S_{x}$ earlier in the question is wrong.

 

[Gaussian97]

Sorry, what I was saying here is if I work out the derivative of the expectation value of $\hat S_{x}$ it dosent equal zero like the RHS of the Ehrenfest equation from above so I am worried that the value obtained for $\hat S_{x}$ earlier in the question is wrong.

And can I see this work?

 

[Zero1010]

I've run out of time now so I will put up my working tomorrow when I get a chance.

Thanks for your help so far.

 

[Zero1010]

Ok, question is about a spin 1/2 particle in a uniform magnetic field in the z-direction. At time 0, the initial spin state is given by the Spinor:

$| A \rangle = \frac{1}{5} \begin{bmatrix} 1 \\ 4 \end{bmatrix}$

In the first part I need to find a spinor to represent the spin state at any time and then the expectation values <Sx> and <Sy>

My calculations:

$| A \rangle = a_{u} | \uparrow_{z} \rangle +a_{s} | \downarrow_{z} \rangle$

$a_{u} = \langle \uparrow_{z} |A \rangle$

$a_{u} = \begin{bmatrix}1&0 \end{bmatrix} \frac{1}{5} \begin{bmatrix} 3\\ 4\end{bmatrix}$

$a_{u}= \frac{3}{5}$

Using a similar method i also get

$a_{d}= \frac{5}{5}$

Using the equation to show a spin 1/2 particle at any time:

$| A \rangle = \frac{3}{5}e^{\frac{-iEut}{\hbar}} \begin{bmatrix} 1\\ 0\end{bmatrix} + \frac{4}{5}e^{\frac{-iEut}{\hbar}} \begin{bmatrix} 0\\ 1\end{bmatrix}$

$| A \rangle = \begin{bmatrix} \frac{3}{5}e^{\frac{-iEut}{\hbar}}\\ \frac{4}{5}e^{\frac{-iEut}{\hbar}}\end{bmatrix}$

Since $e_{u}=\frac{\hbar \omega}{2}$ and $e_{d}=-\frac{\hbar \omega}{2}$

$| A \rangle = \begin{bmatrix} \frac{3}{5}e^{\frac{-i\omega t}{2}}\\ \frac{4}{5}e^{\frac{i \omega t}{2}}\end{bmatrix}$

So the Spinor required is (hopefully):

$| A \rangle = \frac{1}{5} \begin{bmatrix} 3e^{\frac{-i\omega t}{2}}\\ 4e^{\frac{i \omega t}{2}}\end{bmatrix}$

To work out the expectation value of $<S_{x}>= \langle \ A|\hat s_{x} |A \rangle$

$\hat S_{x}$ is given in the question:

$\hat S_{x}= \frac{\hbar}{2}\begin{bmatrix} 0 & 1\\ 1&0 \end{bmatrix}$Therefore:
$<S_{x}>=\frac{1}{5}\begin{bmatrix} 3e^{\frac{i\omega t}{2}} & 4e^{\frac{-i \omega t}{2}}\\ \end{bmatrix} \frac{\hbar}{2}\begin{bmatrix} 0 & 1\\ 1&0 \end{bmatrix} \frac{1}{5} \begin{bmatrix} 3e^{\frac{-i\omega t}{2}}\\ 4e^{\frac{i \omega t}{2}}\end{bmatrix}$

$<S_{x}>=\frac {\hbar}{50}(12e^{i\omega t} + 12e^{-i\omega t})$

From this I get the expectation value:

$<S_{x}>=\frac {6\hbar}{25}(e^{i\omega t} + e^{-i\omega t})$

Using the same method except using the matrix given in the question:

$<\hat s_{y}>=\frac {\hbar}{2} \begin{bmatrix} 0 & -i\\ i & 0 \end{bmatrix}$

I get the expectation

$<S_{y}> = \frac {6i\hbar}{25}(e^{-i\omega t} - e^{i\omega t})$

 

[Gaussian97]

Ok, question is about a spin 1/2 particle in a uniform magnetic field in the z-direction. At time 0, the initial spin state is given by the Spinor:

$| A \rangle = \frac{1}{5} \begin{bmatrix} 1 \\ 4 \end{bmatrix}$

In the first part I need to find a spinor to represent the spin state at any time and then the expectation values <Sx> and <Sy>

My calculations:

$| A \rangle = a_{u} | \uparrow_{z} \rangle +a_{s} | \downarrow_{z} \rangle$

$a_{u} = \langle \uparrow_{z} |A \rangle$

$a_{u} = \begin{bmatrix}1&0 \end{bmatrix} \frac{1}{5} \begin{bmatrix} 3\\ 4\end{bmatrix}$

$a_{u}= \frac{3}{5}$

Using a similar method i also get

$a_{d}= \frac{5}{5}$

Using the equation to show a spin 1/2 particle at any time:

$| A \rangle = \frac{3}{5}e^{\frac{-iEut}{\hbar}} \begin{bmatrix} 1\\ 0\end{bmatrix} + \frac{4}{5}e^{\frac{-iEut}{\hbar}} \begin{bmatrix} 0\\ 1\end{bmatrix}$

$| A \rangle = \begin{bmatrix} \frac{3}{5}e^{\frac{-iEut}{\hbar}}\\ \frac{4}{5}e^{\frac{-iEut}{\hbar}}\end{bmatrix}$

Since $e_{u}=\frac{\hbar \omega}{2}$ and $e_{d}=-\frac{\hbar \omega}{2}$

$| A \rangle = \begin{bmatrix} \frac{3}{5}e^{\frac{-i\omega t}{2}}\\ \frac{4}{5}e^{\frac{i \omega t}{2}}\end{bmatrix}$

So the Spinor required is (hopefully):

$| A \rangle = \frac{1}{5} \begin{bmatrix} 3e^{\frac{-i\omega t}{2}}\\ 4e^{\frac{i \omega t}{2}}\end{bmatrix}$

Well, apart from some typos, if I understand properly, I agree in the values of $a_u$ and $a_d$, my question is in the step of 'Using the equation to show a spin 1/2 particle at any time:' what is that equation?

 

[Zero1010]

Sorry about the typos - I'm pretty new to LaTex.

The equation is one I've taken from my text.

It says the spin state can be given at any later time by:

$| A \rangle = a_{u} e^{\frac{-iE_{u}t}{\hbar}} | \uparrow _{n} \rangle + a_{d} e^{\frac{-iE_{d}t}{\hbar}} | \downarrow _{n} \rangle$

Since the magnetic field points in the z-direction I have used $\uparrow _{n}=\uparrow _{z}$

 

[Gaussian97]

From where do you get that the magnetic field is pointing in the z-direction?

 

[Zero1010]

It says the magnetic field is in the z-direction in the question. I put it at the top of my calculations

 

[Gaussian97]

Oh I see, my mistake sorry, and then they give to you that the Hamiltonian is $\hat{H}=\hat{S}_x$? Can you please post exactly what is the question asking?

 

[Zero1010]

In the first part of the question I need to find a spinor to describe the spin state at any time t>0 and then the expectation values of $<S_{x}>$ and $<S_{y}>$

In the second part of the question (Which I was asking about at the start of the post, I need to use the generalized Erhenfest Theorem to find expressions for $\frac{d<S_{x}>}{dt}$ and $\frac{d<S_{y}>}{dt}$ and make sure they are consisten with the expectation values from the first part.

The Hamiltonian is given as $\hat H = \omega \hat S_{z}$. which I have assumed (possibly incorrectly) is the same as $\hat H = \omega \hat S_{x}$ and $\hat H = \omega \hat S_{y}$

 

[PeroK]

The Hamiltonian is given as $\hat H = \omega \hat S_{z}$. which I have assumed (possibly incorrectly) is the same as $\hat H = \omega \hat S_{x}$ and $\hat H = \omega \hat S_{y}$

That makes no sense. The magnetic field is in the z-direction, which is not the same as being in the x or y directions.

 

[Gaussian97]

Ok, so now all make more sense! No, of course, these three Hamiltonians are not equal at all. Now we can return to Erhenfest Theorem, if you try to do it with $\hat{H}=\omega \hat{S}_z$ we should get the correct answer.

 

[Zero1010]

Ok, my misunderstanding.

I'll go back to the second part and try again with the correct Hamiltonian.

Thanks again

 

[Zero1010]

Sorry, I just want to check, from my working above do the expectation values look ok?

 

[Gaussian97]

Yes, look good, although you can write them more compactly using sine and cosine (also because it may seem that these expectation values are complex numbers, so it is always better to write them in a way that shows explicitly that they are real, this will help you to check your computations).

 

[Zero1010]

Ok, here is my work for the next part of the question. If someone could check it an let me know if I am on the right track that would be great.

I need to use the Generalised Ehrenfest's Theorem to find an expression for $\frac{d <S_{x}>}{dt}$

The Hamiltonian given as $\hat H = \omega \hat S_{z}$

$\hat H = \frac{\hbar \omega}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$

Here goes:

$\frac{d \hat S_{x}}{dt}=\frac{1}{i \hbar} <[\hat S_{x}, \hat H]>$

$[\hat S_{x}, \hat H] = [\frac{\hbar}{2} \begin{bmatrix} 0&1\\ 1&0 \end{bmatrix},\frac{\hbar \omega}{2} \begin{bmatrix} 1&0\\ 0&-1 \end{bmatrix}]$

$[\hat S_{x}, \hat H]= \hat A \hat B - \hat B \hat A$

$\omega (\frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} - \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix})$

$\omega (\frac{\hbar}{4} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} - \frac{\hbar}{4} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix})$

$\frac{\hbar^{2} \omega}{2} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

$\frac{d <\hat S_{x}>}{dt}=\frac{1}{i \hbar} <[ \frac{\hbar^{2} \omega}{2} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} ]>$

$\frac{d <\hat S_{x}>}{dt}= \frac{-i\hbar \omega}{2} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

 

[PeroK]

Maybe that last step isn't quite right!

How is the expectation value of an observable/operator defined?

 

[Zero1010]

I wasn't to sure about that. Any suggestions?

 

[PeroK]

I wasn't to sure about that. Any suggestions?

Look up the definition of the expected value $\langle \hat X \rangle$.

 

[Zero1010]

Ok thanks