Schwinger's model of angular momentum

In summary: But I still don't know how to go on after to solve the ##N_1a_1^\dagger |0,0\rangle## part. I can't get rid of the ##a_1^\dagger |0,0\rangle## part, so I can't get the equation to hold. I've tried using the commutation relationship to move the ##a_1^\dagger## to the right, but that just gives me an extra term which I don't know how to get rid of.Any further hints?Try using the commutation relationship again to move the ##a_1^\dagger## to the left, and see if you can use the eigenvalue equations for ##N_
  • #1
BobaJ
37
0

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.
 
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  • #2
In b) use the commutation relations to bring all a+ to the left of the a operators when operating with n on a state.
 
  • #3
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 :headbang:

By the way, I finally managed to do d). So there is just b) and e) left.
 
  • #4
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]##.
 
  • #5
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}##.
 
  • #6
BobaJ said:
$$a_1^\dagger |0,0\rangle +(a_1^\dagger)^2 a_1 |0,0\rangle $$

##a_1 \left| 0,0 \right> = ?##
 
  • #7
I think it's ##a_1|0,0\rangle = \sqrt{0+1}|0+1,0\rangle \\ = |1,0\rangle##
 
  • #8
BobaJ said:
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.
 
  • #9
George Jones said:
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.
 
  • #10
BobaJ said:
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.
 
  • #11
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?
 
  • #12
BobaJ said:
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##.
 
  • #13
Ok, so then I would just be left with ##N_1a_1^\dagger|0,0\rangle = a_1^\dagger |0,0\rangle##
 
  • #14
BobaJ said:
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 = ?##
 
  • #15
George Jones said:
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##.
 
  • #16
BobaJ said:
Right, now as ##a_1^\dagger## is the creation operator ##a_1^\dagger |0,0\rangle = |1,0\rangle##.

Yes.

BobaJ said:
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.
 
  • #17
George Jones said:
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##.
 
  • #18
Use

BobaJ said:
##a_1^\dagger |0,0\rangle = |1,0\rangle##.

on both sides.

BobaJ said:
You are right, I think it should read ##N_1 a_1^\dagger |0,0\rangle = |0,1\rangle##.
 
  • #19
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?
 
  • #20
BobaJ said:
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?
 
  • #21
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##
 
  • #22
BobaJ said:
So that would get me ##N_1|n_1,0\rangle=n_1|n_1,0\rangle##

Yes. What does this mean?
 
  • #23
George Jones said:
Yes. What does this mean?
Does it mean that ##n_1## is an eigenstate of ##N_1##?
 
  • #24
BobaJ said:
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##.
 
  • #25
nrqed said:
##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.
 
  • #26
BobaJ said:
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?
 
  • #27
nrqed said:
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
 

1. What is Schwinger's model of angular momentum?

Schwinger's model of angular momentum is a mathematical framework developed by physicist Julian Schwinger in the 1950s to explain the intrinsic angular momentum, or spin, of elementary particles.

2. How does Schwinger's model describe angular momentum?

Schwinger's model describes angular momentum as a vector quantity that can have discrete values in quantum systems. It also takes into account the spin-statistics theorem, which states that particles with half-integer spin must follow Fermi-Dirac statistics, while particles with integer spin must follow Bose-Einstein statistics.

3. What are the key features of Schwinger's model?

The key features of Schwinger's model include the use of unitary operators to describe spin transformations, the concept of spinors to represent spin states, and the inclusion of spin in the Hamiltonian operator to account for its effects on energy levels.

4. What are the applications of Schwinger's model?

Schwinger's model has been applied in various areas of physics, including quantum mechanics, quantum field theory, and particle physics. It has been used to explain the properties of fundamental particles and to develop theories such as quantum electrodynamics and quantum chromodynamics.

5. Are there any limitations to Schwinger's model?

While Schwinger's model has been successful in explaining many phenomena, it has some limitations. For example, it does not account for the spin-orbit interaction, which is the interaction between a particle's spin and its motion in a magnetic field. Also, it is limited to describing spin in non-relativistic systems and cannot be applied to particles moving at high speeds.

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