# Schwinger's model of angular momentum

## Homework Statement

Consider two pairs of operators Xα, Pα, with α=1,2, that satisfy the commutation relationships [Xα,Pβ]=ihδαβ,[Xα,Xβ]=0,[Pα,Pβ]=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 [aα,aβ]=δαβ. These are the creation and annihilation operators for two independent harmonic oscillators.

b) The number operators are Nα=aαaα, 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 σα,βa is the (α,β) entry of the Pauli matrix σa. Show that the three matrixes Ja satisfy de commutation relationships [Ja,Jb]=ih∑cεabcJc

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

e) If the operators Ja 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 Ja 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.

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]##.

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
Staff Emeritus
Gold Member
$$a_1^\dagger |0,0\rangle +(a_1^\dagger)^2 a_1 |0,0\rangle$$
##a_1 \left| 0,0 \right> = ?##

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

George Jones
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
Gold Member
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##.

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

George Jones
Staff Emeritus
Gold Member
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 = ?##

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
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
Gold Member
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##.

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
Staff Emeritus
Gold Member
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?

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
Staff Emeritus
Gold Member
So that would get me ##N_1|n_1,0\rangle=n_1|n_1,0\rangle##
Yes. What does this mean?

Yes. What does this mean?
Does it mean that ##n_1## is an eigenstate of ##N_1##?

nrqed