Plot the Expectation Value of Spin - Intro to Quantum Mechanics Homework

In summary: Yeah, that looks right. Although if you had used the exponential form you would see that is equal to:$$\frac{2\hbar}{9}\big (\cos(2\theta) - 2\sin(2\theta)\big )$$And note that ##\theta## is a function of ##t## here.
  • #1
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Homework Statement
Problem e). See picture
Relevant Equations
See picture
BF07A7E3-F097-43B8-82DF-8990F9E3BA69.jpeg

I have found an answer to all of them (a-e) but I don’t know how to plot the function.
EC020EB8-0BA7-4E19-ABC1-93849F87A94E.jpeg


Thanks!
 
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  • #2
It looks like you forgot to take the complex conjugate when writing down the adjoint.

But in any case, are you really asking how to plot a constant as a function of time?
 
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  • #3
vela said:
It looks like you forgot to take the complex conjugate when writing down the adjoint.

But in any case, are you really asking how to plot a constant as a function of time?
Aha ! Like this?
872518F3-8102-4C77-8C32-3CABE969B55A.jpeg
 
  • #4
Yes, if your original answer is correct, but like I said, I think you didn't calculate the adjoint correctly.

Also, doesn't ##2\hbar## seem awfully big for a spin-1/2 particle?
 
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  • #5
I can second that this is wrong
1660587311783.png

you need to take the complex conjugate here
Remember that the ##\dagger## (hermitian conjugate) in linear algebra means transpose ##T## and complex conjugate ##*##, i.e.
a matrix ##A## when you do the hermitian conjugate, you do this ##A^\dagger =( A^*)^T##
 
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  • #6
vela said:
Yes, if your original answer is correct, but like I said, I think you didn't calculate the adjoint correctly.

Also, doesn't ##2\hbar## seem awfully big for a spin-1/2 particle?
Aha, thanks. I get after correction ##(2/3)\hbar##.
 
  • #7
Graham87 said:
Aha, thanks. I get after correction ##(2/3)\hbar##.
🤔 Aren't you suspicious that the expectation value of spin in the x-direction is constant in a z-direction magnetic field?
 
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  • #8
Do you know the motion of a classical "rotating" charged particle in this situation? The correct solution is (perhaps surprisingly) similar.
 
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  • #9
PeroK said:
🤔 Aren't you suspicious that the expectation value of spin in the x-direction is constant in a z-direction magnetic field?
Aha, true. I think I get it. Should it look like a trigonometric function?
I will try again later.
 
  • #10
hutchphd said:
Do you know the motion of a classical "rotating" charged particle in this situation? The correct solution is (perhaps surprisingly) similar.
Not really. But I’m guessing something like sin(x)cos(y)?
I will try again later.
Thanks!
 
  • #11
Graham87 said:
Aha, true. I think I get it. Should it look like a trigonometric function?
I will try again later.
Also,the Hamiltonian drives the time evolution of the system. You should expect something to change over time in this case.
 
  • #12
Graham87 said:
Aha, thanks. I get after correction ##(2/3)\hbar##.
Still not correct.
Could you show how you did that calculation so maybe we can spot another error?
 
  • #13
malawi_glenn said:
Still not correct.
Could you show how you did that calculation so maybe we can spot another error?
I used Euler formula to convert e, but I get something messy. Might my d-answer be wrong too (see pic at beginning) ?
102E1DE6-CAFA-4B4D-9D0C-6F75DBA0F902.jpeg

EC258B14-E59A-4439-9AF5-431CFEBF597E.jpeg
 
Last edited:
  • #14
That looks right, although if you had used the exponential form you would see that is equal to:
$$\frac{2\hbar}{9}\big (\cos(2\theta) - 2\sin(2\theta)\big )$$And note that ##\theta## is a function of ##t## here.
 
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  • #15
PeroK said:
That looks right, although if you had used the exponential form you would see that is equal to:
$$\frac{2\hbar}{9}\big (\cos(2\theta) - 2\sin(2\theta)\big )$$And note that ##\theta## is a function of ##t## here.
Ah! I got the same now. Big thanks !
 

1. What is the expectation value of spin in quantum mechanics?

The expectation value of spin in quantum mechanics is a measure of the average spin of a particle in a particular state. It is calculated by taking the inner product of the state vector with the spin operator, and then taking the magnitude squared of the result.

2. How is the expectation value of spin calculated?

The expectation value of spin is calculated by taking the inner product of the state vector with the spin operator, and then taking the magnitude squared of the result. This can be represented mathematically as ⟨S⟩ = |⟨ψ|S|ψ⟩|^2.

3. What does the expectation value of spin tell us?

The expectation value of spin tells us the average spin of a particle in a particular state. It can also give us information about the orientation of the particle's spin in a given direction.

4. How does the expectation value of spin relate to the uncertainty principle?

The expectation value of spin is related to the uncertainty principle in that it is one of the properties that cannot be known with absolute certainty in quantum mechanics. The uncertainty principle states that the more precisely we know the value of one property, the less precisely we can know another related property. In the case of spin, the more precisely we know the expectation value, the less precisely we can know the orientation of the particle's spin in a given direction.

5. Can the expectation value of spin be measured experimentally?

Yes, the expectation value of spin can be measured experimentally using techniques such as Stern-Gerlach experiments. These experiments involve passing a beam of particles through a magnetic field, which causes the particles to separate based on their spin orientation. By measuring the separation of the particles, the expectation value of spin can be determined.

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