What Happens to Two Spin-1/2 Particles in a Magnetic Field Over Time?

[gasar8]

Homework Statement

Two particles with spin $S_1={1 \over 2}$ and $S_2={1 \over 2}$ are at t=0 in a state with $S=0$.
a) Find wave function at t=0 in $S_{1z},$, $S_{2z}$ basis.
b) Second particle is in a magnetic field $B = (\sin\theta,0,\cos\theta)$, the Hamiltonian is $H=\lambda \vec{S_2} \cdot \vec{B}.$ Find the probability that we find particles at time t in $S=1$ state.

The Attempt at a Solution

a) $|10\rangle ={1 \over \sqrt{2}} \big(|{1 \over 2}-{1 \over 2}\rangle + |-{1 \over 2}{1 \over 2}\rangle\big)$

b) If I write Hamiltonian with Pauli matrices, I get:
$$\lambda \vec{S_2} \cdot \vec{B}=\frac{\lambda \hbar B}{2}<br /> \left(<br /> \begin{array}{cc}<br /> \cos\theta & \sin\theta\\<br /> \sin\theta & -\cos\theta<br /> \end{array}<br /> \right)$$

Then I wrote Schrödingers equation and got eigenvalues $E=\pm \frac{\lambda \hbar B}{2}$ and eigenvectors ($\cot\theta \pm {1 \over \sin\theta},1)$.
I should be right to this point, right? But now, I have some problems. Is the wave function really: $$|\psi,0\rangle= |\uparrow\rangle+ \bigg(\cot\theta \pm {1 \over \sin\theta} \bigg) |\downarrow\rangle.$$
How do I normalize it? And after that, how do I find its time evolution? I tried to put Hamiltonian on both states, so $H|\uparrow\rangle$ and $H|\downarrow\rangle,$ but is it right? Because H is a matrix and $|\uparrow\rangle=|{1 \over 2}{1 \over 2}\rangle$ is vector, so I get another vector?

 

[blue_leaf77]

a) |10⟩=1√2(|12−12⟩+|−1212⟩)|10⟩=12(|12−12⟩+|−1212⟩) |10\rangle ={1 \over \sqrt{2}} \big(|{1 \over 2}-{1 \over 2}\rangle + |-{1 \over 2}{1 \over 2}\rangle\big)

No, that state corresponds to one of $S=1$ states.

Second particle is in a magnetic field B=(sinθ,0,cosθ)B=(sin⁡θ,0,cos⁡θ)B = (\sin\theta,0,\cos\theta), the Hamiltonian is H=λ→S2⋅⃗B.H=λS2→⋅B→.H=\lambda \vec{S_2} \cdot \vec{B}. Find the probability that we find particles at time t in S=1S=1S=1 state.

The evolution of the state is calculated by applying the operator $e^{-iHt/\hbar}$ to the initial state $|00\rangle$, where that time evolution operator only acts on the second particle. So, first find the correct initial state.

 

[gasar8]

Ok, so the $S=0$ must be $|00\rangle = {1 \over \sqrt{2}} \big(|{1 \over 2}-{1 \over 2}\rangle - |-{1 \over 2}{1 \over 2}\rangle\big)$

Time evolution operator only acts on the second particle.

So I must apply Hamiltonian to $|S_2 S_{z2}\rangle$ like $H |{1 \over 2}-{ 1 \over 2}\rangle$ and $H |{1 \over 2} {1 \over 2}\rangle$?

 

[blue_leaf77]

So I must apply Hamiltonian to |S2Sz2⟩|S2Sz2⟩|S_2 S_{z2}\rangle like H|12−12⟩H|12−12⟩H |{1 \over 2}-{ 1 \over 2}\rangle and H|1212⟩H|1212⟩H |{1 \over 2} {1 \over 2}\rangle?

Not the Hamiltonian, instead the time evolution operator.

 

[gasar8]

So the time evolution operator is something like $U=e^{-i{H \over \hbar}t}=e^{\pm i {\lambda B \over 2}t}.$ But I am not sure how to apply it to an initial state, because there is also first particle in $|00\rangle.$ How do I know it applies only on the second? Is it just because there is a $S_2$ in Hamiltonian and I need to randomly choose one to be second?
So the time evolution must be:
$$|\psi(t)\rangle = c_{\uparrow} \ e^{- i {\lambda B \over 2}t} |\uparrow\rangle + c_{\downarrow} \ e^{ i {\lambda B \over 2}t}|\downarrow\rangle.$$
Do I have to pick these two coefficients from $|00\rangle= {1 \over \sqrt{2}} \big(|{1 \over 2}-{1 \over 2}\rangle - |-{1 \over 2}{1 \over 2}\rangle\big)$, so the $|S_2,S_{z2}\rangle = {1 \over \sqrt{2}} \big(|{1 \over 2}-{1 \over 2}\rangle - |{1 \over 2}{1 \over 2}\rangle \big).$ Is this correct? So the wave function would look like:
$$|\psi(t)\rangle = - {1 \over \sqrt{2}} e^{- i {\lambda B \over 2}t} |\uparrow\rangle + {1 \over \sqrt{2}} e^{ i {\lambda B \over 2}t}|\downarrow\rangle.$$

But where is the connection back to both particles and $S=1$ state? Btw, how do I even know which $S=1$ state do we need?

 

[blue_leaf77]

So the time evolution operator is something like U=e−iHℏt=e±iλB2t.U=e−iHℏt=e±iλB2t.

No, you cannot write the last equality. The most right expression is just the eigenvalue of the middle one while the middle one is an operator. There is no way a scalar can be equal to an operator.

How do I know it applies only on the second? Is it just because there is a S2S2S_2 in Hamiltonian and I need to randomly choose one to be second?

It's because we know that the Hamiltonian $H = \lambda \mathbf B \cdot \mathbf S_2$ applies to one particle only. Only particle two is placed in a magnetic field, the other particle is not associated with any Hamiltonian. It might be strange that a particle does not have Hamiltonian and hence its time evolution is static but that's the consequence of neglecting the spatial wavefunction and not having this particle interact with magnetic field.
For the purpose of this problem, it's more convenient to separate the kets of the two particle. Thus
$$
|\uparrow \uparrow\rangle \rightarrow |\uparrow\rangle |\uparrow\rangle
$$
If I were to write the complete time evolution operator, it will be $I\otimes e^{-iHt/\hbar}$ where $I$ only acts on the first particle (the first particle is static in time). For example if you apply to $|11\rangle$, it will be $(I\otimes e^{-iHt/\hbar})|11\rangle = (I\otimes e^{-iHt/\hbar})(|\uparrow\rangle |\uparrow\rangle) = (I|\uparrow\rangle )(e^{-i\lambda \mathbf B \cdot \mathbf S_2t/\hbar} |\uparrow\rangle)$.
Now, in order to calculate $e^{-i\lambda \mathbf B \cdot \mathbf S_2t/\hbar} |\uparrow\rangle$, I suggest that you expand the time evolution operator into Taylor series and simplify it using the properties of Pauli matrices. For this matter, you might be interested in the question part of this http://physics.stackexchange.com/questions/41697/what-is-the-spin-rotation-operator-for-spin-1-2.