# Homework Help: Two Ag atoms in magnetic field

1. Jan 25, 2015

### skrat

1. The problem statement, all variables and given/known data
Two Ag atoms wits spin $s=1/2$ are at $t=0$ in state $|\uparrow >|\uparrow >$. The first atom is in homogeneous magnetic field in $y$ direction while the second atom is in homogeneous magnetic field in $z$ direction. The strength of the two magnetic fields are the same. Magnetic moment of the silver atom is equal to Bohr's magneton $\mu _B$. What are the wave functions of each atom at time $t$?
Write combined wavefunction of both atoms at time $t$ in basis with squared value of aggregate spin and its projection on $z$ axis?

2. Relevant equations

3. The attempt at a solution
I haven't got a single clue here. :(

2. Jan 25, 2015

### Staff: Mentor

You need to make some attempt at a solution before we can help you. For a start, you might think about what equations are relevant; for example, what equation governs the time evolution of wave functions?

3. Jan 26, 2015

### skrat

Ok, we know that $$\psi (t=0)=|\uparrow >|\uparrow >$$ and we also know that $$\psi (t)=\psi (0)e^{-i\frac H \hbar t}.$$ So what I need is to calculate $H=2\frac {\mu _B}{\hbar}\vec S \vec B =\mu _B \vec \sigma \vec B$ where $\vec \sigma$ is a vector of Pauli matrices.

So for the first particle in $\vec B=(0,B,0)$ I have to than calculate $H=\mu _BB_y\sigma _y |\uparrow >|\uparrow >$. Now how do I do that? I don't know how to write $|\uparrow >|\uparrow >$ in matrix form, so I can multiply it with Pauli matrices.

4. Jan 26, 2015

### Staff: Mentor

States aren't matrices, they're vectors. Vectors multiplied by matrices give other vectors, so multiplying a state vector by a matrix representing the Hamiltonian operator (or any other operator) gives another state vector.

5. Jan 26, 2015

### skrat

Yesyes, I got confused because if you multiply this $$\begin{bmatrix} 0 &-i \\ i&0 \end{bmatrix}\begin{bmatrix} 1\\ 0 \end{bmatrix}\begin{bmatrix} 1\\ 0 \end{bmatrix}$$ from right to left, you don' get much. Of course if one multiplies this from left to right, gets the right result.

So $|\psi _1, t>=|\uparrow >|\uparrow >e^{\frac{\mu _B B_y}{\hbar}t}$ and $|\psi _2, t>=|\uparrow >|\uparrow >e^{-i\frac{\mu _B Bz}{\hbar}t}$.

Hmm, what is a combined wavefunction of both atoms? :/ This part, I really don't know.

EDIT: I am supposed to make a linear combination of both wavefunctions and than transform them to another basis using Clebsch Gordan coefficients?

6. Jan 26, 2015

### Staff: Mentor

That matrix should only multiply one of the state vectors, not both, shouldn't it? The matrix represents the B field in the y direction, but that field is only acting on the first particle. The second particle is acted on by the B field in the z direction, so to multiply its state vector you need a matrix that represents the B field in the z direction.

This doesn't look right. The wave function $\psi_1$ only describes the first particle, and the wave function $\psi_2$ only describes the second particle. So each only should only include one $| \uparrow >$, correct?

You know what it is at $t = 0$, correct? (It's what's given in the problem statement.) That $t = 0$ wave function has a particular simple form; so the question is whether that simple form is preserved by time evolution (since, from the above, you know how each atom's individual wave function evolves in time). It should be if the atoms are not interacting with each other; is that the case?

7. Jan 26, 2015

### skrat

I mean, the problem states that both of the particles at $t=0$ are in state $|\uparrow>|\uparrow >$. I understood this as if $|\psi _1,0>=|\psi _2,0>=|\uparrow>|\uparrow >$. I guess this is wrong?

8. Jan 26, 2015

### Staff: Mentor

Yes. $| \uparrow >$ describes a single particle in a particular state (question: which state?). $| \uparrow > | \uparrow >$ describes two particles, both in the same state.

9. Jan 26, 2015

### skrat

This should be the case yes. My Hamiltonian doesn't describe any interaction between the two particles $H=\mu _B \vec \sigma \vec B=\mu _B (\sigma _y B_y+\sigma _z B_z)$.
Right?

10. Jan 26, 2015

### Staff: Mentor

Yes. (A note on notation: what you are writing as $\vec{\sigma} \vec{B}$ should properly be written as $\vec{\sigma} \cdot \vec{B}$.)

11. Jan 26, 2015

### skrat

So... What I should do is simply multiply the wave functions together, like $$|\psi ,t>=|\psi _1,t>|\psi_2,t>=|\uparrow >|\uparrow >e^{\frac{\mu _B}{\hbar}(B_y-iB_z)t}$$ which using Clebsch Gordan coefficients means $$|\psi ,t>=|1,1>e^{\frac{\mu _B}{\hbar}(B_y-iB_z)t}$$ Or is that a complete nonsense?

12. Jan 26, 2015

### Goddar

This last expression makes sense given what you get for the individual wave functions. Unfortunately these are not correct, you need to go back to your time-evolution:
For particle 1, the Hamiltonian is: H1 = –μBσy,
For particle 2, the Hamiltonian is: H2 = –μBσz
Now you need to solve the Schrodinger equation for each of these particles, given the initial conditions:
ψ1(0) = ψ2(0) = (1; 0) (in vector notation)

13. Jan 26, 2015

### skrat

That is exactly what I did. $$H_1|\psi _1,0>=\mu _B \vec \sigma \cdot \vec B |\uparrow >=\mu _BB_y \sigma _y|\uparrow>$$ where $$\sigma _y|\uparrow>= \begin{pmatrix} 0 &-i \\ i & 0 \end{pmatrix}\begin{pmatrix} 1\\ 0 \end{pmatrix}=i \begin{pmatrix} 0\\ 1 \end{pmatrix}.$$ Aha, I can see what might be wrong... $\sigma _y$ also changes my wave function to $|\downarrow >$. Hmm... ???

14. Jan 26, 2015

### Staff: Mentor

Yes.

Yes; but will that happen immediately? Or will it take some time for the wavefunction to go from $| \uparrow >$ to $| \downarrow >$? If it will, how would you capture that time dependence in the wave function?

15. Jan 26, 2015

### Goddar

Initial conditions come into play only after you solved the Schrodinger equation. To solve it, use arbitrary coefficients (a ; b)

16. Jan 26, 2015

### skrat

All my answers to these questions would be simply guessing and not based on knowledge. Sorry. :/ I have no idea. I assume it takes some time but I have no idea how would one capture that in the wave function. Maybe adding a phase in the exponent?

17. Jan 26, 2015

### Goddar

i hate to insist but this is precisely what you do when you solve the Schrodinger equation...

18. Jan 26, 2015

### skrat

Please do insist if necessary! :D

I ignored you the first time you suggested this because I had no Idea what you are trying to say - or how this changes anything. But now I tried it and I hope I didn't make this problem a lot more complicated than necessary : $$H_1(a|\uparrow >+b|\downarrow>)=E(a|\uparrow >+b|\downarrow>)$$ $$\mu _B B_y \begin{pmatrix} 0 &-i \\ i & 0 \end{pmatrix}(a \begin{pmatrix} 1\\ 0 \end{pmatrix}+b \begin{pmatrix} 0\\ 1 \end{pmatrix})=E(a|\uparrow >+b|\downarrow>)$$ $$i\mu _BB_y(a|\downarrow >-b|\uparrow >)=E(a|\uparrow >+b|\downarrow>)$$
This gives me a set of equations: $$i\mu _BB_y\begin{pmatrix} 1 & -E\\ E & 1 \end{pmatrix}\begin{pmatrix} a\\ b \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}$$ Solving this (looking for eigen values and eigen vectors) gives me for
$\lambda _1=i\mu _BB_y(1-iE) \Rightarrow |\psi _1>=\frac{1}{\sqrt 2}(|\uparrow >+i|\downarrow >)$ and
$\lambda _2=i\mu _BB_y(1+iE) \Rightarrow |\psi _1>=\frac{1}{\sqrt 2}(|\uparrow >-i|\downarrow >)$.

Now I am confused what my $H_1$ in $\psi (t)=\psi (0) e^{-i\frac{H_1}{\hbar} t}$ is?

Or maybe this isn't what you meant? :/

19. Jan 26, 2015

### Staff: Mentor

This looks like an area that you need to review, then. It's a very basic part of QM.

The basic idea is this: the states $| \uparrow >$ and $| \downarrow >$ form a basis of the Hilbert space for the spin of a single Ag atom. So the most general spin state for a single Ag atom will be $| \psi > = a | \uparrow >| + b \downarrow >$, where $a$ and $b$ are complex numbers such that $|a|^2 + |b|^2 = 1$. As Goddar said, when you are solving the Schrodinger equation, you should use the general form of $| \psi >$ that I just wrote down, and then use the Schrodinger equation to tell you what $a$ and $b$ are as functions of time. Try doing that and see what you come up with.

20. Jan 26, 2015

### Staff: Mentor

That tells you the stationary states of the Hamiltonian you are working with; but this problem isn't asking for stationary states. The state of at least one of the Ag atoms in this problem is changing with time, so it isn't stationary. So you can't just look for eigenvalues and eigenvectors of the Hamiltonian; you have to use the more general procedure I described in my last post.

21. Jan 26, 2015

### Goddar

You're trying to solve the eigenvalue problem, or in other terms the time-independent Schrodinger equation, which is already solved because you presumably already know the eigenvalues and eigenfunctions of the Pauli matrices... What i'm talking about is the time-dependent equation, which gives for H1 for example:
ih da/dt = μB b
ih db/dt = – μB a
You can solve these differential equations simultaneously to get a(t) and b(t), then match them to your initial conditions.
The same goes for H2, with a different result...
I would express everything in terms of matrices and vectors if i were you and avoid the Dirac notation for the moment, by the way.

22. Jan 26, 2015

### Goddar

Sorry i didn't see the last posts from PeterDonis, i'll leave it to him from now on

23. Jan 26, 2015

### skrat

Uh, ok, I hope I got it right this time.
$$H\psi=i\hbar \frac{\partial }{\partial t} \psi$$
$$i\mu _BB_y(a|\downarrow >-b|\uparrow >)=i\hbar (\dot a|\uparrow >+\dot b |\downarrow >)$$ This gives me a system of two equations $$\mu _BB_ya-\hbar \dot b=0$$ and $$\hbar \dot a+\mu _B B_yb=0$$ Solving this hopefully gives me $$a(t)=Acos(\omega t)-Bsin(\omega t)$$ and $$b(t)=Asin(\omega t)+Bcos(\omega t)$$ for $\omega =\frac{\mu _B B_y}{\hbar}$. Using inital condition $\psi (t=0)=|\uparrow >$ should set my $B$ to zero if I am not mistaken. Meaning $$a(t)=Acos(\omega t)$$ and $$b(t)=Asin(\omega t).$$ I can determine the value of $A$ with normalization of my wave function

So $\psi (t)=A(cos(\omega t)|\uparrow >+sin(\omega t)|\downarrow >)$. Where $A=1$. But.. What is my $H_1$ now? $i\frac \omega \hbar$ ?

24. Jan 26, 2015

### Staff: Mentor

With your given value of $\omega$, yes.

$H_1$ doesn't change; it's what it was before. The only thing to check is that the form of $H_1$ you wrote down before is consistent with the value of $\omega$ that you obtained.

25. Jan 26, 2015