Time dependent expectation value problems

[BREAD]

Homework Statement

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]

 

[DrClaude]

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##.

 

[BREAD]

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)

 

[DrClaude]

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?

 

[BREAD]

In which direction is the magnetic field?

In the problem, megnetic field is Bx direction.

 

[DrClaude]

In the problem, megnetic field is Bx direction.

So how can you rewrite ##\mathbf{S} \cdot \mathbf{B}##?

 

[BREAD]

So how can you rewrite ##\mathbf{S} \cdot \mathbf{B}##?

Sx * B

 

[BvU]

So no commuting with ##{\bf S}_z## !

 

[BREAD]

Then how can i solve this problem. i would appreciate it if you can help me

 

[DrClaude]

You have to express the initial state in terms of the eigenstates of the Hamiltonian.

 

[BREAD]

You have to express the initial state in terms of the eigenstates of the Hamiltonian.

I think it is not. Sx and H have same eigenstates, so the initial state l+> is lSx+> + lSx->

 

[DrClaude]

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.

 

[BREAD]

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.

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

 

[DrClaude]

I think the answer is weird

Why? This is a very standard problem in QM and illustrates nicely the concept of precession.

 

[BREAD]

Why? This is a very standard problem in QM and illustrates nicely the concept of precession.

I replied the answer that i tried

 

[DrClaude]

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.

 

[BREAD]

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)

 

[DrClaude]

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 S_z = +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.

 

[BREAD]

The probability of measuring S_z = +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)

Answer is quite complex, is it right??

 

[BREAD]

I changed it again.