Transition between two states probability

[unscientific]

Homework Statement

Part (a): Show probability to transit from state i to j is given by:
Part (b)i: Use answer in part (a) to find probability
Part (b)ii: Use time evolution to find probability

Homework Equations

The Attempt at a Solution

Part (a) was alright, bookwork question on time dependent perturbation theory.

What's interesting was, I got different answers for part (b)i and (b)ii.

Part (b)i

Amplitude of finding particle in state j:

$$a_j = \frac{1}{i\hbar}\int V_{ji} exp(\frac{i\Delta E}{\hbar}t) dt$$

In this case, $\Delta E = 2\mu B$, and since the additional hamiltonian is $-\mu b \sigma_x$, hence $V_{ji} = \mu b$.

$$a_j = \frac{1}{i\hbar} V_{ij}^{*} \int_0^T exp(\frac{i2\mu B}{\hbar}t) dt$$

$$|a_j|^2 = (\frac{b}{B})^2 sin^2(\frac{\mu B l}{\hbar v})$$

Part (b)ii

The time evolution operator is given by $U_{(t)} = e^{-\frac{iH}{\hbar}t}$ such that $|\psi_t> = U_t |\psi_0>$.

To find the time evolution operator:
$$e^{-\frac{iH}{\hbar}t} = exp(\frac{i\mu B \sigma_z}{\hbar}t) exp(\frac{i\mu b \sigma_x}{\hbar}t)$$

For each of the exponentials, I found them in matrix form:

Now to overlap the desired |-,B> = (0 1) state with evolved state $|\psi_t>$:

$$<-,B|\psi_t> = i sin (\frac{\mu b}{\hbar}t) exp(\frac{-i\mu B}{\hbar}t)$$

Thus, probability is $sin^2 (\frac{\mu b l}{\hbar v})$

 

[TSny]

To find the time evolution operator:
$$e^{-\frac{iH}{\hbar}t} = exp(\frac{i\mu B \sigma_z}{\hbar}t) exp(\frac{i\mu b \sigma_x}{\hbar}t)$$

Keeping in mind that $\sigma_z$ and $\sigma_x$ do not commute, is this correct?

 

[unscientific]

Keeping in mind that $\sigma_z$ and $\sigma_x$ do not commute, is this correct?

I was thinking about that, as the original operator was: $U = e^{-\frac{iH}{\hbar}t} = exp(\frac{i\mu B \sigma_z}{\hbar}t + \frac{i\mu b \sigma_x}{\hbar}t)$

Do I use $\sigma_x \sigma_z$ or $\sigma_z \sigma_x$?

And is the answer to (b)i correct?

 

[TSny]

And is the answer to (b)i correct?

Edit: I think your answer to (b)i is correct.

For (b)ii, can you find the exact eigenvalues and eigenvectors for $H=-\mu( B \sigma_z+ b \sigma_x)$?

 

[unscientific]

Edit: I think your answer to (b)i is correct.

For (b)ii, can you find the exact eigenvalues and eigenvectors for $H=-\mu( B \sigma_z+ b \sigma_x)$?

The eigenvalues are $\lambda = \pm \mu \sqrt{b^2 + B^2}$. I'm having trouble finding the eigenstates. What's the point of finding them anyway?

 

[TSny]

The eigenvalues are $\lambda = \pm \mu \sqrt{b^2 + B^2}$.

OK.

I'm having trouble finding the eigenstates. What's the point of finding them anyway?

How have you gone about it?

The eigenstates of H have a simple time evolution and they can be used as basis states for expressing the spin state of the particle.

 

[unscientific]

OK.
How have you gone about it?

The eigenstates of H have a simple time evolution and they can be used as basis states for expressing the spin state of the particle.

For example, consider $\lambda = \mu \sqrt{b^2 + B^2}$, then I get these two simultaneous equations:

$$Be_1 + be_2 = \sqrt{b^2 + B^2}(e_1)$$
$$be_1 - Be_2 = \sqrt{b^2 + B^2}(e_2)$$

By the way, what's wrong with my method with the time evolution operator $U_t$ above?

 

[TSny]

For example, consider $\lambda = \mu \sqrt{b^2 + B^2}$, then I get these two simultaneous equations:

$$Be_1 + be_2 = \sqrt{b^2 + B^2}(e_1)$$
$$be_1 - Be_2 = \sqrt{b^2 + B^2}(e_2)$$

Looks like you might have forgotten the minus sign in the expression for the Hamiltonian, so some signs are incorrect here. The two equations are not independent, but you can use normalization to get a second condition.

There is another way to get the eigenstates if you note that the particle will be in a net magnetic field that lies in the xz plane at an angle of θ = tan^-1(b/B) to the z-axis. So, the eigenstates will be spin "up" and "down" along the direction of the net magnetic field. So, if you know how to rotate spin 1/2 states, you could rotate a spin up along the z-axis to produce a spin up or spin down along the net magnetic field direction.

By the way, what's wrong with my method with the time evolution operator $U_t$ above?

I don't think the method is wrong, but you can't factor the time evolution operator as you did.

 

[unscientific]

Looks like you might have forgotten the minus sign in the expression for the Hamiltonian, so some signs are incorrect here. The two equations are not independent, but you can use normalization to get a second condition.

ok I ended up with this pair of simultaneous equations for the positive eigenvalue:

$$-Be_1 - be_2 = \sqrt{b^2+B^2}e_1$$
$$-be_1+Be_2 = \sqrt{b^2+B^2}e_2$$

Still can't solve it simply, I'm trying to get something simple like $e_1=\pm e_2$.

There is another way to get the eigenstates if you note that the particle will be in a net magnetic field that lies in the xz plane at an angle of θ = tan^-1(b/B) to the z-axis. So, the eigenstates will be spin "up" and "down" along the direction of the net magnetic field. So, if you know how to rotate spin 1/2 states, you could rotate a spin up along the z-axis to produce a spin up or spin down along the net magnetic field direction.



I don't think the method is wrong, but you can't factor the time evolution operator as you did.

For a spin orientation (θ,ø) where θ = tan^-1(b/B) to z-axis, the spin of the particle can be characterized as:

$$|+,t\rangle = e^{\frac{-i\phi}{2}} cos \left(\frac{\theta}{2}\right) |+,z\rangle + e^{\frac{i\theta}{2}} sin \left(\frac{\theta}{2}\right)|-,\rangle$$

Since $\phi = 0$:

$$|+,t=0\rangle = cos \left(\frac{\theta}{2}\right) |+,z\rangle + sin \left(\frac{\theta}{2}\right)|-,z\rangle$$

This is the particle's spin state at t=0.

So at time t, the particle evolves as:

$$|+,t\rangle = cos (\frac{\theta}{2})*exp\left(\frac{-i\mu\sqrt{b^2+B^2}}{\hbar}t\right) |+,z\rangle + sin (\frac{\theta}{2})*exp\left(\frac{i\mu\sqrt{b^2+B^2}}{\hbar}t\right) |-,z\rangle$$

Thus the probability of it being in the $|-,z\rangle$ state at time t is

$$sin^2\left(\frac{\theta}{2}\right) = \frac{1}{2}\left[1 - sin\theta\right]$$

Using $tan\theta = \frac{b}{B}$, $sin\theta = \frac{b}{\sqrt{b^2+B^2}}$:


Probability is $\frac{1}{2}\left[1 - \frac{b}{\sqrt{b^2+B^2}}\right]$

This is clearly a different answer as it is a constant and there should be a factor of $sin^2(\theta)$ according to perturbation theory..

 

[TSny]

ok I ended up with this pair of simultaneous equations for the positive eigenvalue:

$$-Be_1 - be_2 = \sqrt{b^2+B^2}e_1$$
$$-be_1+Be_2 = \sqrt{b^2+B^2}e_2$$

Still can't solve it simply, I'm trying to get something simple like $e_1=\pm e_2$.

Looks good.
I found it helpful to divide both sides of the first equation by $\small \sqrt{b^2+B^2}$ and then express the coefficients in terms of cos θ and sin θ where θ = tan^-1(b/B).

$$|+,t=0\rangle = cos \left(\frac{\theta}{2}\right) |+,z\rangle + sin \left(\frac{\theta}{2}\right)|-,z\rangle$$
This is the particle's spin state at t=0.

The right side of your equation represents a state with spin that is parallel to the net B-field.

But this is not the spin state at t = 0. At t = 0 the particle has spin in the z-direction.

So at time t, the particle evolves as:

$$|+,t\rangle = cos (\frac{\theta}{2})*exp\left(\frac{-i\mu\sqrt{b^2+B^2}}{\hbar}t\right) |+,z\rangle + sin (\frac{\theta}{2})*exp\left(\frac{i\mu\sqrt{b^2+B^2}}{\hbar}t\right) |-,z\rangle$$

No. The states $|+,z\rangle$ and $|-,z\rangle$ do not evolve with time this way. It's the spin states that are parallel or antiparallel to B_net that have simple time evolution.

 

[unscientific]

Looks good.
I found it helpful to divide both sides of the first equation by $\small \sqrt{b^2+B^2}$ and then express the coefficients in terms of cos θ and sin θ where θ = tan^-1(b/B).

Yeah I got these two eigenstates:

At t = 0, the state of the particle is along z-axis, but we wish to express it in basis along net B field, B'.

$$|\psi_0\rangle = |+,z\rangle = a|+,B'\rangle + c|-,B'\rangle$$

This implies that $a = \langle+,B'|+,z\rangle = \frac{1}{\sqrt{2sec\theta(sec \theta + 1)}}(1+sec\theta) = \sqrt{\frac{1+cos\theta}{2}}$ and $c = \langle-,B'|+,z\rangle = \frac{1}{\sqrt{2sec\theta(sec\theta-1)}}(1-sec\theta) = -\sqrt{\frac{1-cos\theta}{2}}$

The right side of your equation represents a state with spin that is parallel to the net B-field.

But this is not the spin state at t = 0. At t = 0 the particle has spin in the z-direction.
No. The states $|+,z\rangle$ and $|-,z\rangle$ do not evolve with time this way. It's the spin states that are parallel or antiparallel to B_net that have simple time evolution.

That's true, because these are basis kets in the z-direction, and not in direction of tan θ. I suppose the above method can only be used.

Following from above, the ket at time t is:
$$|\psi_t\rangle = a*exp(\frac{-i\lambda_{+}}{\hbar}t)|+,B'\rangle + c*exp(\frac{-i\lambda_{-}}{\hbar}t)|-,B'\rangle$$

Thus probability amplitude is:

$$\langle -,z|\psi_t\rangle = a*exp(\frac{-i\lambda_{+}}{\hbar}t)\langle-,z|+,B'\rangle + c*exp(\frac{-i\lambda_{-}}{\hbar}t)\langle -,z|-,B'\rangle$$

where $\langle-,z|+,B'\rangle = \langle -,z|-,B'\rangle = \frac{1}{\sqrt{2sec\theta(sec \theta + 1)}} tan \theta$

Separating real and imaginary parts:

$$\langle -,z|\psi_t\rangle = cos \left(\frac{\lambda_+}{\hbar}t\right)\langle-,z|+,B'\rangle\left(a+c\right) + i sin \left( \frac{\lambda_-}{\hbar}t \right)\langle-,z|+,B'\rangle (a-c)$$
$$= cos \left(\frac{\lambda_+}{\hbar}t\right)\langle-,z|+,B'\rangle\left(a+c\right) - i sin \left( \frac{\lambda_+}{\hbar}t \right)\langle-,z|+,B'\rangle (a-c)$$

Finally, the probability is:
$$|\langle -,z|\psi_t\rangle|^2 = \left(\langle-,z|+,B'\rangle\right)^2 \left[cos^2\left(\frac{\lambda_+}{\hbar}t\right)(a+c) + sin^2 \left(\frac{\lambda_+}{\hbar}t\right) (a-c)\right]$$

$$= \left(\langle-,z|+,B'\rangle\right)^2 \left[a + c*cos\left(\frac{2\lambda_+}{\hbar}t\right)\right]$$

$$= \left[\frac{sin^2\theta}{2(cos\theta+1)}\right]\left[\sqrt{\frac{1+cos\theta}{2}} - cos\left(\frac{2\lambda_+}{\hbar}t\right)\sqrt{\frac{1-cos\theta}{2}}\right]$$

This looks crazy, compared to the answer in part (b)i using perturbation theory.

 

[TSny]

Yeah I got these two eigenstates:

If you forget normalization for a moment, notice that $|+,B'\rangle \sim \left( \begin{array}{c} 1+\sec\theta\\ \tan\theta\\ \end{array} \right) \sim \left( \begin{array}{c} \cos\theta +1\\ \sin\theta\\ \end{array} \right)$

Use half-angle identities to express $1+\cos\theta$ and $\sin\theta$ in terms of $\theta/2$ and simplify.

At t = 0, the state of the particle is along z-axis, but we wish to express it in basis along net B field, B'.

$$|\psi_0\rangle = |+,z\rangle = a|+,B'\rangle + c|-,B'\rangle$$

This implies that $a = \langle+,B'|+,z\rangle = \frac{1}{\sqrt{2sec\theta(sec \theta + 1)}}(1+sec\theta)$ and $c = \langle-,B'|+,z\rangle = \frac{1}{\sqrt{2sec\theta(sec\theta-1)}}(1-sec\theta)$



That's true, because these are basis kets in the z-direction, and not in direction of tan θ. I suppose the above method can only be used.

Following from above, the ket at time t is:
$$|\psi_t\rangle = a*exp(\frac{-i\lambda_{+}}{\hbar}t)|+,B'\rangle + c*exp(\frac{-i\lambda_{-}}{\hbar}t)|-,B'\rangle$$

Note that the eigenvalue $\lambda_+$ corresponds to the higher energy and therefore belongs to the spin state that is anti-parallel to B'. This is due to the overall negative sign in H.

 

[unscientific]

If you forget normalization for a moment, notice that $|+,B'\rangle \sim \left( \begin{array}{c} 1+\sec\theta\\ \tan\theta\\ \end{array} \right) \sim \left( \begin{array}{c} \cos\theta +1\\ \sin\theta\\ \end{array} \right)$

Use half-angle identities to express $1+\cos\theta$ and $\sin\theta$ in terms of $\theta/2$ and simplify.



Note that the eigenvalue $\lambda_+$ corresponds to the higher energy and therefore belongs to the spin state that is anti-parallel to B'. This is due to the overall negative sign in H.

Yeah it gets pretty messy, at this stage its hard to believe that it will all turn out to be a simple sin²θ function. I will definitely work it out later today or tomorrow. But for now, is this concept/method right?

1. Express initial state |+,z> in terms of basis along B'.
2. Each ket evolves with time given by exp(-iEt/h)
3. Overlap with <-,z| to find probability amplitude

 

[TSny]

But for now, is this concept/method right?

1. Express initial state |+,z> in terms of basis along B'.
2. Each ket evolves with time given by exp(-iEt/h)
3. Overlap with <-,z| to find probability amplitude

Yes. That will solve it.