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

• Thread starter Zero1010
• Offset appointment Mar 2, 2020

Homework Statement:

Quick question nigh using the generalised Ehrenfest Theorem.

Relevant Equations:

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

Hi,

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

Since the generalised Ehrenfest Theorem takes the course:

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

Does this mean I too have to detect the expectation value of the Hamiltonian operator?

Thanks

Answers and Replies

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

Hello,

I have a question which asks me to apply the generalised Ehrenfest Theorem to discover 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:

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

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

Thanks

Not if you can avoid it!

Homework Argument:: 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 utilize the generalised Ehrenfest Theorem to detect expressions for
##\frac {d<Sx>} {dt}## and ##\frac {d<Sy>} {dt}## - I accept worked out <Sx> and <Sy> earlier in the question.

Since the generalised Ehrenfest Theorem takes the form:

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

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

Thanks for the replies.

Ok, does that hateful I use the values I have been given for Sx (rather than the expectation value I have calculated) and the Hamiltonian and then piece of work out the expectation of the commutator. I'm pretty sure the 2 operators commute so that should be easy.

Cheers

Thanks for the replies.

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

Thanks

Two operators commute iff the commutator is naught. Do y'all really mean that?

Yeah that'due south what I hateful. I have been given the Hamiltonian equally:

##\hat H = \omega \hat Sx##

Which I retrieve should commute with $$\lid Sx$$ as $$\omega$$ is a constant. I'm I on the correct rails?

Yep that'south what I hateful. I accept been given the Hamiltonian equally:

##\hat H = \omega \hat Sx##

Which I think should commute with $$\chapeau Sx$$ as $$\omega$$ is a abiding. I'm I on the correct runway?

Yes, information technology does commute.

Ok, one final question about this...until the adjacent one.

So the 2 operators commute so the RHS of the generalised Ehrenfest Theorem = 0. Nonetheless 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 and then far.

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

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

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

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

Thanks for your help so far.

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

Ok, one concluding question well-nigh this...until the adjacent ane.

And so the two operators commute and then the RHS of the generalised Ehrenfest Theorem = 0. However using the expectation value for Sx I calculated before

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

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

Cheers for your help so far.

Please, post your calculation so we can help you

In what way?

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

##\lid H = \omega \hat Sx##

and Sx is given every bit:

##\hat Sx = \frac{\hbar}{2}
\brainstorm{pmatrix}
0&1 \\
1&0\terminate{pmatrix}
##

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

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

Sorry, what I was maxim hither is if I work out the derivative of the expectation value of ##\lid S_{x}## information technology dosent equal zero like the RHS of the Ehrenfest equation from higher up so I am worried that the value obtained for ##\lid S_{x}## earlier in the question is wrong.

And can I see this work?

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

Thank you for your help so far.

Ok, question is well-nigh a spin ane/2 particle in a compatible magnetic field in the z-direction. At time 0, the initial spin state is given past the Spinor:

##| A \rangle = \frac{1}{five}
\brainstorm{bmatrix}
1 \\
4
\end{bmatrix} ##

In the kickoff office I need to find a spinor to represent the spin country at whatsoever 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{i}{five}
\begin{bmatrix}
3\\
4\end{bmatrix} ##

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

Using a similar method i also get

##a_{d}= \frac{5}{v}##

Using the equation to evidence a spin 1/ii particle at any time:

##| A \rangle = \frac{iii}{5}eastward^{\frac{-iEut}{\hbar}}
\brainstorm{bmatrix}
1\\
0\end{bmatrix}
+
\frac{4}{five}eastward^{\frac{-iEut}{\hbar}}
\begin{bmatrix}
0\\
1\finish{bmatrix}
##

##| A \rangle =
\brainstorm{bmatrix}
\frac{iii}{v}eastward^{\frac{-iEut}{\hbar}}\\
\frac{iv}{v}e^{\frac{-iEut}{\hbar}}\end{bmatrix} ##

Since ##e_{u}=\frac{\hbar \omega}{ii}## and ##e_{d}=-\frac{\hbar \omega}{ii}##

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

So the Spinor required is (hopefully):

##| A \rangle = \frac{i}{5}
\brainstorm{bmatrix}
3e^{\frac{-i\omega t}{2}}\\
4e^{\frac{i \omega t}{two}}\stop{bmatrix} ##

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

##\lid S_{x}## is given in the question:

##\hat S_{ten}=
\frac{\hbar}{ii}\begin{bmatrix}
0 & 1\\
1&0
\end{bmatrix}##

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

##<S_{ten}>=\frac {\hbar}{fifty}(12e^{i\omega t} + 12e^{-i\omega t})##

From this I go the expectation value:

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

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

##<\hat s_{y}>=\frac {\hbar}{ii}
\begin{bmatrix}
0 & -i\\
i & 0
\stop{bmatrix}##

I get the expectation

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

Ok, question is near a spin ane/2 particle in a uniform magnetic field in the z-direction. At fourth dimension 0, the initial spin state is given past the Spinor:

##| A \rangle = \frac{1}{5}
\begin{bmatrix}
1 \\
iv
\end{bmatrix} ##

In the showtime part I need to find a spinor to represent the spin land 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}one&0 \end{bmatrix} \frac{i}{5}
\begin{bmatrix}
iii\\
iv\end{bmatrix} ##

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

Using a similar method i too become

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

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

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

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

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

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

So the Spinor required is (hopefully):

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

Well, apart from some typos, if I empathise properly, I agree in the values of ##a_u## and ##a_d##, my question is in the step of 'Using the equation to testify a spin 1/ii particle at whatsoever time:' what is that equation?

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

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

It says the spin country tin be given at any later time past:

##| 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-management I take used ##\uparrow _{n}=\uparrow _{z}##

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

It says the magnetic field is in the z-direction in the question. I put information technology at the summit of my calculations

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

In the commencement part of the question I need to find a spinor to draw the spin state at whatever time t>0 so the expectation values of ##<S_{x}>## and ##<S_{y}>##

In the 2d role of the question (Which I was asking virtually 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 certain they are consisten with the expectation values from the first part.

The Hamiltonian is given every bit ##\chapeau H = \omega \hat S_{z}##. which I take causeless (perchance incorrectly) is the same as ##\hat H = \omega \hat S_{x}## and ##\hat H = \omega \hat S_{y}##

The Hamiltonian is given as ##\lid H = \omega \lid S_{z}##. which I take assumed (possibly incorrectly) is the same as ##\hat H = \omega \hat S_{ten}## and ##\hat H = \omega \hat S_{y}##

That makes no sense. The magnetic field is in the z-direction, which is non the same as being in the x or y directions.

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

Ok, my misunderstanding.

I'll go back to the 2d office and try again with the correct Hamiltonian.

Thanks again

Sorry, I merely want to cheque, from my working in a higher place do the expectation values look ok?

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

Ok, hither is my work for the side by side part of the question. If someone could cheque it an let me know if I am on the right runway that would be great.

I need to use the Generalised Ehrenfest's Theorem to find an expression for ##\frac{d <S_{10}>}{dt}##

The Hamiltonian given as ##\chapeau H = \omega \chapeau S_{z}##

##\chapeau H = \frac{\hbar \omega}{2}
\begin{bmatrix}
1 & 0 \\
0 & -1
\cease{bmatrix}
##

Here goes:

##\frac{d \hat S_{10}}{dt}=\frac{1}{i \hbar} <[\lid S_{x}, \hat H]>##

##[\chapeau S_{x}, \lid H] = [\frac{\hbar}{2}
\begin{bmatrix}
0&1\\
1&0
\finish{bmatrix},\frac{\hbar \omega}{2}
\begin{bmatrix}
1&0\\
0&-1
\cease{bmatrix}]
##

##[\hat S_{x}, \chapeau H]= \hat A \hat B - \lid B \chapeau A##

##\omega (\frac{\hbar}{2}
\begin{bmatrix}
0 & 1 \\
1 & 0
\finish{bmatrix}
\frac{\hbar}{two}
\begin{bmatrix}
1 & 0 \\
0 & -ane
\terminate{bmatrix} -
\frac{\hbar}{2}
\begin{bmatrix}
one & 0 \\
0 & -ane
\end{bmatrix}
\frac{\hbar}{2}
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix})
##

##\omega (\frac{\hbar}{4}
\begin{bmatrix}
0 & -1 \\
1 & 0
\finish{bmatrix} -
\frac{\hbar}{four}
\begin{bmatrix}
0 & 1 \\
-i & 0
\end{bmatrix})##

##
\frac{\hbar^{two} \omega}{2}
\begin{bmatrix}
0 & -ane \\
one & 0
\end{bmatrix}##

##\frac{d <\chapeau 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}{ii}
\begin{bmatrix}
0 & -1 \\
ane & 0
\end{bmatrix}
##

Peradventure that last step isn't quite correct!

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

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

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