Two Ag atoms in magnetic field

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Homework Statement


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?

Homework Equations




The Attempt at a Solution


I haven't got a single clue here. :(
 

Answers and Replies

  • #2
PeterDonis
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I haven't got a single clue here.
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
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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
PeterDonis
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I don't know how to write ##|\uparrow >|\uparrow >## in matrix form, so I can multiply it with Pauli matrices.
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
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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
PeterDonis
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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.
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.

So ##|\psi_1,t>## = ##|\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}## .
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?

what is a combined wavefunction of both atoms?
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
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This doesn't look right. The wave function ψ1\psi_1 only describes the first particle, and the wave function ψ2\psi_2 only describes the second particle. So each only should only include one |↑>| \uparrow >, correct?
Not if you ask me. o_O
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
PeterDonis
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I understood this as if ##|\psi _1,0>=|\psi _2,0>=|\uparrow>|\uparrow >##. I guess this is wrong?
Yes. ##| \uparrow >## describes a single particle in a particular state (question: which state?). ##| \uparrow > | \uparrow >## describes two particles, both in the same state.
 
  • #9
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You know what it is at t=0t = 0, correct? (It's what's given in the problem statement.) That t=0t = 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?
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
PeterDonis
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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?
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
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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
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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?
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
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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
PeterDonis
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What I should do is simply multiply the wave functions together
Yes.

I can see what might be wrong... ##\sigma _y## also changes my wave function to ##|\downarrow >##.
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
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Initial conditions come into play only after you solved the Schrodinger equation. To solve it, use arbitrary coefficients (a ; b)
 
  • #16
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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?
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
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i hate to insist but this is precisely what you do when you solve the Schrodinger equation...
 
  • #18
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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
PeterDonis
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I assume it takes some time but I have no idea how would one capture that in the wave function.
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
PeterDonis
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Solving this (looking for eigen values and eigen vectors)
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
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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 : $$H_1(a|\uparrow >+b|\downarrow>)=E(a|\uparrow >+b|\downarrow>)$$ $$\mu _B B_y
\begin{pmatrix}
0 &-i \\
i & 0
\end{pmatrix}(a
\begin{pmatrix}
1\\
0
\end{pmatrix}+b
\begin{pmatrix}
0\\
1
\end{pmatrix})=E(a|\uparrow >+b|\downarrow>) $$ $$i\mu _BB_y(a|\downarrow >-b|\uparrow >)=E(a|\uparrow >+b|\downarrow>) $$
:/
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
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Sorry i didn't see the last posts from PeterDonis, i'll leave it to him from now on
 
  • #23
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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
PeterDonis
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So ##\psi (t)=A(cos(\omega t)|\uparrow >+sin(\omega t)|\downarrow >)##. Where ##A=1##.
With your given value of ##\omega##, yes.

What is my ##H_1## now? ##i\frac \omega \hbar## ?
##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
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Solving this hopefully gives me
$$a(t)=Acos(\omega t)-Bsin(\omega t)$$ and $$b(t)=Asin(ωt)+Bcos(ωt)$$
for ##\omega =\frac{\mu _B B_y}{\hbar}##.
 

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