Schwinger's model of angular momentum

[BobaJ]

Homework Statement

Consider two pairs of operators X_α, P_α, with α=1,2, that satisfy the commutation relationships [X_α,P_β]=ihδ_αβ,[X_α,X_β]=0,[P_α,P_β]=0. These are two copies of the canonical algebra of the phase space.
a) Define the operators $$a_\alpha = \frac{1}{\sqrt{2\hbar}}(X_\alpha+ip_\alpha)$$ and $$a_\alpha^\dagger=\frac{1}{sqrt{2\hbar}}(X_\alpha-iP_\alpha)$$.
Show that those satisfy the commutation relationships [a_α,a_β^†]=δ_αβ. These are the creation and annihilation operators for two independent harmonic oscillators.

b) The number operators are N_α=a_α^†a_α, with a=1,2 (there is no implicit sum). Considere the states $$|n_1,n_2\rangle_H=\frac{(a_1^\dagger)^{n_1}}{\sqrt{n_1!}}\frac{(a_2^\dagger)^{n_2}}{\sqrt{n_2!}}|0,0\rangle_H$$, where |n₁,n₂) is the state of excitement n₁ in the first oscillator and n₂ in the second oscillator, while |0,0) is the vacuum of both oscillators. Show that n₁ and n₂ are the eigenstates of N₁, N₂. What are their eigenvalues?

c) Define the operators $$J^a=\frac{\hbar}{2}\sum_{\alpha,\beta}a_\alpha^\dagger\sigma_{\alpha\beta}^aa_\beta$$ and $$J^0=\frac{\hbar}{2}\sum_\alpha a_\alpha^\dagger a_\alpha$$, with a=1,2,3 and where σ_α,β^a is the (α,β) entry of the Pauli matrix σ^a. Show that the three matrixes J^a satisfy de commutation relationships [J^a,J^b]=ih∑_cε^abcJ^c

d) Show that $$J^2=\sum_a (J^a)^2=J^0(J^0+1)$$

e) If the operators J^a satisfy the algebra of angular momentum, a base of the space of states has to consist of states of the form |J,M)_S, simultaneously eigenstates of J²,J₃, with M=-J,-J+1,...,J-1,J and possibly various values of J. Consider the states $$|J,M\rangle_S=|J+M,J-M\rangle_H=\frac{(a_1^\dagger)^{J+M}}{\sqrt{n_1!}}\frac{(a_2^\dagger)^{J-M}}{\sqrt{n_2!}}|0,0\rangle_H$$. These are the Schwinger states of angular momentum. That's why we use the sub indexes H and S to distinguish between the states of the harmonic oscillators and those of Schwinger. Show that these are eigenstates of J² and J₃. What are the eigenvalues?

Homework Equations

$$a_\alpha = \frac{1}{\sqrt{2\hbar}}(X_\alpha+ip_\alpha)$$
$$a_\alpha^\dagger=\frac{1}{sqrt{2\hbar}}(X_\alpha-iP_\alpha)$$
$$N_\alpha=a_\alpha^\dagger a_\alpha$$
$$J^a=\frac{\hbar}{2}\sum_{\alpha,\beta}a_\alpha^\dagger\sigma_{\alpha\beta}^aa_\beta$$
$$J^0=\frac{\hbar}{2}\sum_\alpha a_\alpha^\dagger a_\alpha$$

The Attempt at a Solution

Ok, so I think that I already managed to get a) and c). I just put them her for the sake of completeness.

For points b) and e) I honestly have no idea where to get started.

And for point d) I started trying to substitute the definition of J^a into the middle part of the equation. So I get $$\sum_a (J^a)^2=(J^1)^2+(J^2)^2+(J^3)^2 \\ = (\frac{\hbar}{2}\sum_{\alpha\beta}a_\alpha^\dagger \sigma^1 a_\beta)^2+(\frac{\hbar}{2}\sum_{\alpha\beta}a_\alpha^\dagger \sigma^2 a_\beta)^2+(\frac{\hbar}{2}\sum_{\alpha\beta}a_\alpha^\dagger \sigma^3 a_\beta)^2$$. But I don't know how to go on from here.

Any help would be appreciated.

 

[DrDu]

In b) use the commutation relations to bring all a+ to the left of the a operators when operating with n on a state.

 

[BobaJ]

So, in b) …. I assume that I hace to use the eigenvalue equations $$N_1|n1,n2\rangle=n_1|n1,n2\rangle$$ and $$N_2|n1,n2\rangle=n_2|n1,n2\rangle$$, is that correct? So I could substitute the definition of |n₁,n₂) in there …. but I'm still not sure how to do it. If got a big blackout

By the way, I finally managed to do d). So there is just b) and e) left.

 

[DrDu]

start from $| 0,0\rangle$. Show that $N_1| 0,0\rangle=0$. Next consider $N_1 a^\dagger_1 | 0,0\rangle$. Use the commutation relationship to bring all $a^\dagger$ to the left of the a. Next try to work out the commutator $[ a,(a^\dagger)^j]$.

 

[BobaJ]

Ok, So I've managed to show that $N_1|0,0\rangle=0$ using the definition of the creation and annihilation operators.

I'm a little stuck on the $N_1a_1^\dagger |0,0\rangle$ part. It would be $$a_1^\dagger a_1 a_1^\dagger |0,0\rangle = a_1^\dagger (1+a_1^\dagger a_1)|0,0\rangle \\ = a_1^\dagger |0,0\rangle +(a_1^\dagger)^2 a_1 |0,0\rangle \\ = a_1^\dagger |0,0\rangle + a_1^\dagger |0,0\rangle \\ = 2a_1^\dagger |0,0\rangle$$ Is that right?

I've managed to show that $[a,(a^\dagger)^j]=j(a^\dagger)^{j-1}$.

 

[George Jones]

$$a_1^\dagger |0,0\rangle +(a_1^\dagger)^2 a_1 |0,0\rangle $$

$a_1 \left| 0,0 \right> = ?$

 

[BobaJ]

I think it's $a_1|0,0\rangle = \sqrt{0+1}|0+1,0\rangle \\ = |1,0\rangle$

 

[George Jones]

I think it's $a_1|0,0\rangle = \sqrt{0+1}|0+1,0\rangle \\ = |1,0\rangle$

I had assumed, possibly incorrectly, that $a_1$ is a lowering/annihilation operator, not a raising/creation operator.

 

[BobaJ]

I had assumed, possibly incorrectly, that $a_1$ is a lowering/annihilation operator, not a raising/creation operator.

Yes, as I understand the exercise $a_1$ is the creation operator.

But I still don't know how to go on after to solve the problem.

 

[George Jones]

Yes, as I understand the exercise $a_1$ is the creation operator.

Are you sure? This would be very non-standard notation. Also, if $a_1$ is taken to be an annihilation operator, then the problem works out.

 

[BobaJ]

Well, I stated the problem the exact way they gave it to me. I thought that because of the order they wrote it $a_1$ would be the creation operator. But now that you point it out, revising the books $a_1$ always is the annihilation operator.

then what would $a_1|0,0\rangle$ be? Can I lower $|0,0\rangle$ even more? Or would it just be 0?

 

[George Jones]

Or would it just be 0?

It is zero, which is different than $\left|0,0\right>$, i.e., $\left|0,0\right> \ne 0$.

 

[BobaJ]

Ok, so then I would just be left with $N_1a_1^\dagger|0,0\rangle = a_1^\dagger |0,0\rangle$

 

[George Jones]

Ok, so then I would just be left with $N_1a_1^\dagger|0,0\rangle = a_1^\dagger |0,0\rangle$

Yes, and $a_1^\dagger|0,0\rangle = ?$

 

[BobaJ]

Yes, and $a_1^\dagger|0,0\rangle = ?$

Right, now as $a_1^\dagger$ is the creation operator $a_1^\dagger |0,0\rangle = |1,0\rangle$. And with this $N_1a_1^\dagger=|1,0\rangle$.

 

[George Jones]

Right, now as $a_1^\dagger$ is the creation operator $a_1^\dagger |0,0\rangle = |1,0\rangle$.

Yes.

And with this $N_1a_1^\dagger=|1,0\rangle$.

Not quite. On the left, you have the product of two operators, which is an operator, while on the right, you have a ket. An operator is not equal to a ket.

 

[BobaJ]

Yes.
Not quite. On the left, you have the product of two operators, which is an operator, while on the right, you have a ket. An operator is not equal to a ket.

You are right, I think it should read $N_1 a_1^\dagger |0,0\rangle = |1,0\rangle$.

 

[George Jones]

Use

$a_1^\dagger |0,0\rangle = |1,0\rangle$.

on both sides.

You are right, I think it should read $N_1 a_1^\dagger |0,0\rangle = |0,1\rangle$.

 

[BobaJ]

So, I could apply $a_1^\dagger$ multiple times and using the commutation relationship $[a_1,(a_1^\dagger)^n]=n(a_1^\dagger)^{n-1}$ I would get $N_1(a_1^\dagger)^{n_1}|0,0\rangle=n_1\sqrt{n_1!}|n_1,0\rangle$
would that be correct?

 

[George Jones]

I would get $N_1(a_1^\dagger)^{n_1}|0,0\rangle=n_1\sqrt{n_1!}|n_1,0\rangle$
would that be correct?

Replacing the $(a_1^\dagger)^{n_1}|0,0\rangle$ in $N_1(a_1^\dagger)^{n_1}|0,0\rangle$ by ... gives?

 

[BobaJ]

I suppose I could replace it with $\sqrt{n_1!}|n_1,0\rangle$ to get $N_1\sqrt{n_1!}|n_1,0\rangle=n_1\sqrt{n_1!}|n_1,0\rangle$.
So that would get me $N_1|n_1,0\rangle=n_1|n_1,0\rangle$

 

[George Jones]

So that would get me $N_1|n_1,0\rangle=n_1|n_1,0\rangle$

Yes. What does this mean?

 

[BobaJ]

Yes. What does this mean?

Does it mean that $n_1$ is an eigenstate of $N_1$?

 

[nrqed]

Does it mean that $n_1$ is an eigenstate of $N_1$?

$n_1$ is not a state, it is a number! So it cannot be an eigenstate of $N_1$. You mean that $n_1$ is an eigenvalue of $N_1$.

 

[BobaJ]

$n_1$ is not a state, it is a number! So it cannot be an eigenstate of $N_1$. You mean that $n_1$ is an eigenvalue of $N_1$.

I thought that, but as the problem states that I have to show that n1 and n2 are the eigenstates of N1 and N2 and after that to determine their eigenvalues, I got really confused.

 

[nrqed]

I thought that, but as the problem states that I have to show that n1 and n2 are the eigenstates of N1 and N2 and after that to determine their eigenvalues, I got really confused.

Oh, I see. You are right to be confused! is that from a textbook or is it a problem presented by your instructor?

 

[BobaJ]

Oh, I see. You are right to be confused! is that from a textbook or is it a problem presented by your instructor?

It is a problem presented by my instructor. I don't know from where he got it