Time dependent expectation value problems

  • #1
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Homework Statement


upload_2017-3-21_20-44-19.png


Homework Equations




The Attempt at a Solution


I tried to solve (a), but i don't know which approach is right ((1) or (2)) and how to solve (b).[/B]
upload_2017-3-21_20-57-18.png
 

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Answers and Replies

  • #2
DrClaude
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You don't start the problem correctly. The ##|S_z = + \hbar/2 \rangle## state is not an eigenstate of the Hamiltonian, so after a time ##T_2##, the state of the system is not ##e^{-i E T_2 / \hbar} |S_z = + \hbar/2 \rangle##.
 
  • #3
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You don't start the problem correctly. The ##|S_z = + \hbar/2 \rangle## state is not an eigenstate of the Hamiltonian, so after a time ##T_2##, the state of the system is not ##e^{-i E T_2 / \hbar} |S_z = + \hbar/2 \rangle##.
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H=-(e/mc)S * B , Sz and H differ just by a multiplicative constant, so they commute. The Sz eigenstates are also energy eigenstates.
(* I tried to solve it Sz not Sx by mistake)
 
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  • #4
DrClaude
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H=-(e/mc)S * B , Sz and H differ just by a multiplicative constant, so they commute. The Sz eigenstates are also energy eigenstates.
In which direction is the magnetic field?
 
  • #5
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In which direction is the magnetic field?
In the problem, megnetic field is Bx direction.
 
  • #6
DrClaude
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In the problem, megnetic field is Bx direction.
So how can you rewrite ##\mathbf{S} \cdot \mathbf{B}##?
 
  • #7
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So how can you rewrite ##\mathbf{S} \cdot \mathbf{B}##?
Sx * B
 
  • #8
BvU
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So no commuting with ##{\bf S}_z## !
 
  • #9
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Then how can i solve this problem. i would appreciate it if you can help me
 
  • #10
DrClaude
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You have to express the initial state in terms of the eigenstates of the Hamiltonian.
 
  • #11
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You have to express the initial state in terms of the eigenstates of the Hamiltonian.
upload_2017-3-22_22-9-13.png


I think it is not. Sx and H have same eigenstates, so the initial state l+> is lSx+> + lSx->
 
  • #12
DrClaude
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Sx and H have same eigenstates, so the initial state l+> is lSx+> + lSx->
Don't forget the normalization factor.

But now that the state is expressed in terms of the eigenstates of the Hamiltonian, it should be easy to calculate the time evolution.
 
  • #13
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Don't forget the normalization factor.

But now that the state is expressed in terms of the eigenstates of the Hamiltonian, it should be easy to calculate the time evolution.
bpGR.dn.gif

I think the answer is weird. These are the probability for Sz + , and 1-cos(T2/h)(E1-E2) for Sz -
So, the expectation value h/2(1+cos(T2/h)(E1-E2)) and -h/2(1-cos(T2/h)(E1-E2)) respectively
 
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  • #14
DrClaude
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I think the answer is weird
Why? This is a very standard problem in QM and illustrates nicely the concept of precession.
 
  • #15
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Why? This is a very standard problem in QM and illustrates nicely the concept of precession.
I replied the answer that i tried
 
  • #16
DrClaude
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I replied the answer that i tried
Ok, I had replied before you posted the solution.

There is a problem, as you can clearly see that the prbability you get ranges from 0 to 2. Check again how you calculate the absolute value squared.
 
  • #17
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Ok, I had replied before you posted the solution.

There is a problem, as you can clearly see that the prbability you get ranges from 0 to 2. Check again how you calculate the absolute value squared.
I got it, then how can i approach problem(b) i think the result of first measurement is same with (a)
 
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  • #18
DrClaude
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I got it, then how can i approach problem(b) i think the result of first measurement is same with (a)
The probability of measuring Sz = +ħ/2 is indeed given by the same equation, replacing T2 by T1. But you already know the measurement result, so you need to figure out what you get at T2.

By the way, you can simplify the result you get using explicit values for E1 and E2.
 
  • #19
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The probability of measuring Sz = +ħ/2 is indeed given by the same equation, replacing T2 by T1. But you already know the measurement result, so you need to figure out what you get at T2.

By the way, you can simplify the result you get using explicit values for E1 and E2.
Result of first measurement at T1 is the same with (a). I should have to get a probability of that state before i get a expectation value as i did in (a).
I think total result probability is probability(T1) * probability(T2)
bq69.dn.gif


Answer is quite complex, is it right??
 
  • #20
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I changed it again.
bq9e.dn.gif
 

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