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Creation/annihilation operators

by jdstokes
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jdstokes
#1
Apr11-08, 11:14 PM
P: 527
For a system of N non-interacting bosons we start with the tensor product of single particle states [itex]\otimes_{n=1}^N | \alpha_i \rangle[/itex] and then, due to the indistinguisability of the particles, symmetrize to obtain the occupation number state

[itex]| n_1,n_2,\ldots,n_k\rangle = \frac{1}{\sqrt{N! n_1!\cdots n_k!}} \sum_{\Pi\in S_N} \otimes_{i=1}^N | \alpha_{\Pi(i)} \rangle[/itex]

where [itex]n_1,\ldots,n_k[/itex] denote the multiplicities of each single-particle state in the system and [itex]S_N[/itex] denotes the N! permutations of {1,...,N}.

So far so good. In order to describe variable numbers of particles we introduct creation and annihilation operators [itex]\hat{a}_\lambda^\dag,\hat{a}_\lambda[/itex] such that

[itex]\hat{a}^\dag_\lambda | n_1,n_2,\ldots,n_k\rangle \propto| n_1,n_2,\ldots, n_\lambda+1 ,\ldots,n_k\rangle, \hat{a}_\lambda | n_1,n_2,\ldots,n_k\rangle \propto| n_1,n_2,\ldots, n_\lambda-1 ,\ldots,n_k\rangle[/itex]

with as yet to be determined proportionality constants.

I'm trying to figure out what is the minimum information needed to determine the proportionality constants ([itex]\sqrt{n_\lambda +1},\sqrt{n_\lambda}[/itex] respectively.)


Clearly [itex]| n_1,n_2,\ldots,n_k\rangle[/itex] is an eigenstate of [itex]a^\dag_\lambda a_\lambda [/itex]. Suppose now that the [itex]a_\lambda,a^\dag_\lambda[/itex] were constructed such that the corresponding eigenvalue is [itex]n_\lambda[/itex].

Then by further assuming that the operators obey some commutation relations we can determine the proportionality constants in the first two relations.

Can somebody correct if I am mistaken:

In order to determine the action of [itex]a^\dag_\lambda[/itex] and [itex]a_\lambda[/itex] on occupation number states we must assume the following defining relations:

[itex]\hat{a}^\dag_\lambda | n_1,n_2,\ldots,n_k\rangle \propto| n_1,n_2,\ldots, n_\lambda+1 ,\ldots,n_k\rangle[/itex]
[itex] \hat{a}_\lambda | n_1,n_2,\ldots,n_k\rangle \propto| n_1,n_2,\ldots, n_\lambda-1 ,\ldots,n_k\rangle[/itex]
[itex]a^\dag_\lambda a_\lambda | n_1,n_2,\ldots,n_k\rangle = n_\lambda | n_1,n_2,\ldots,n_k\rangle[/itex]
[itex][a_\lambda a_{\lambda'}^\dag ] = \delta_{\lambda\lambda'}[/itex]

Is there any simpler way of looking at this?



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jdstokes
#2
Apr12-08, 05:56 AM
P: 527
Anyone got any suggestions about this?
DavidWhitbeck
#3
Apr12-08, 10:43 AM
P: 352
You have what you need, you must prove that the states are orthogonal with unit norm, and then you can use that to show that that the coefficients are what they are.

The trick is based on saying that

[tex]\langle n + 1 | n + 1 \rangle = |C|^2(\langle n | a) (a^{\dagger} | n \rangle)[/tex]
[tex]\langle n + 1 | n + 1 \rangle = |C|^2\langle n | aa^{\dagger} | n \rangle[/tex]
[tex]\langle n + 1| n + 1\rangle = |C|^2\langle n | (1 + a^{\dagger}a) | n \rangle[/tex]
[tex]\langle n + 1| n + 1\rangle = |C|^2(n + 1)\langle n | n \rangle[/tex]
Now say that [tex]\langle k | k \rangle = 1[/tex] for all k to arrive at
[tex]1 = |C|^2(n + 1)[/tex]
[tex]C = 1/\sqrt{n + 1}[/tex] up to phase constant.

jdstokes
#4
Apr12-08, 06:07 PM
P: 527
Creation/annihilation operators

Hi David,

I know how to do the calculation, it's exactly the same as for the 1D harmonic oscillator.

But that's not what I'm asking. I want to know if any of the four relations I gave can be derived form the others?
lbrits
#5
Apr12-08, 07:21 PM
P: 410
Everything can be derived from [tex][a, a^\dagger] = 1[/tex] and knowledge of a single state (usually, the ground state)
DavidWhitbeck
#6
Apr12-08, 07:26 PM
P: 352
Well geez you should have just asked that in the first place!

Shouldn't #3 at least be partially dependent on #1, 2 and 4?

#3 is really two statements-- (1) those states are eigenvectors of [tex]a^{\dagger}a[/tex] and (2) the associated eigenvalues are [tex]n_\lambda[/tex]. #1 and 2 imply that the operator you see in #3 has those states as eigenvectors, but I don't think that you can use #4 to deduce what the eigenvalues must be.

It seems all backwards though-- defining the operators from the states. Shouldn't you actually be defining the states from the operators? After all the original starting place is the Hamiltonian, and the states are from the spectral decomposition theorem, that is to say they are the eigenstates of the Hamiltonian, and that is how they are defined. And that's why I really can't say that #3 depends on the others, because that was really the decomposition that defined the states in the first place.
jdstokes
#7
Apr12-08, 07:29 PM
P: 527
Quote Quote by lbrits View Post
Everything can be derived from [tex][a, a^\dagger] = 1[/tex] and knowledge of a single state (usually, the ground state)
For the 1D harmonic oscillator, certainly. I'm not sure whether this is true for general occupation number states, in fact I suspect it is false.
jdstokes
#8
Apr12-08, 07:35 PM
P: 527
Sorry David I thought that was implicit.

That's right, the operators are _defined_ in terms of their action on occupation number states, which are themselves the (anti)symmetrized tensor products of single particle states.

(1) and (2) imply that the occupation number states are eigenstates of a-dagger a, but in order to get the right eigenvalue, I think you also need to assume (3). The commutation relations [itex][a_\lambda^\dag,a_{\lambda'}]=\delta_{\lambda\lambda'},[a_\lambda,a_{\lambda'}]=[a_\lambda^\dag,a^\dag_{\lambda'}]=0[/itex] give the proportionality constants in (1) and (2).

I think we need to assume all of these.
jdstokes
#9
Apr12-08, 07:38 PM
P: 527
Quote Quote by DavidWhitbeck View Post
It seems all backwards though-- defining the operators from the states. Shouldn't you actually be defining the states from the operators?
I suppose you could, but that's not the way I've seen it done in quantum many-body theory. One first starts with the tensor product of single particle states, then applies symmetrization (or anti-symmetrization for fermions) to get the occupation number states. The raising and lowering operators are then defined by their action on the occupation number states.
jdstokes
#10
Apr12-08, 07:49 PM
P: 527
Quote Quote by lbrits View Post
Everything can be derived from [tex][a, a^\dagger] = 1[/tex] and knowledge of a single state (usually, the ground state)
For the 1D harmonic oscillator you start with the above commutation relation and then derive that eigenspectrum of [itex]a^\dag a[/itex] is the non-negative integers. This allows you to derive the action of a-dagger and a on the number states.

In our case the number states are not just abstrat state vectors: they have a concrete realization in terms of (anti)symmetrized tensor products of single particle states. So it is necessary to define a and a-dagger by their action on these states.
jdstokes
#11
Apr12-08, 08:03 PM
P: 527
I think that maybe you derive the coefficient in (3) using just (1), (2) and the commutation relations, but I don't have time to do it right now. So I'll come back to this thread on Monday.
nrqed
#12
Apr13-08, 09:27 PM
Sci Advisor
HW Helper
P: 3,025
Quote Quote by jdstokes View Post
I think that maybe you derive the coefficient in (3) using just (1), (2) and the commutation relations, but I don't have time to do it right now. So I'll come back to this thread on Monday.
It's easy to obtain the commutation relation from (1), (2) and (3).
lbrits
#13
Apr13-08, 10:04 PM
P: 410
The commutation relation is canonical. You don't really derive it.
kdv
#14
Apr13-08, 10:59 PM
P: 329
Quote Quote by lbrits View Post
The commutation relation is canonical. You don't really derive it.
Don't you agree that the cr follows from the first three relations?
lbrits
#15
Apr13-08, 11:17 PM
P: 410
Quote Quote by kdv View Post
Don't you agree that the cr follows from the first three relations?
Yes... but those relations have no physical meaning on their own. I mean, sure, given the matrix elements of those operators, you can derive their commutation relations. But usually one derives the commutation relations from [tex][x,p]= i\hbar[/tex] and then derives the matrix elements by diagonalizing the number operator. The order doesn't really matter though, it's all heuristic.
jdstokes
#16
Apr16-08, 06:44 AM
P: 527
After thinking about this a bit more I think I've resolved it.

I'm assuming here that we already know how to construct the number states, ie we use as our starting point the Fock space (note this is typically not what is done in quantum field theory!)

[itex]
\mathcal{H} = \bigoplus_{N=1}^{\infty} (H^{\otimes N})_{\mathrm{sym}}
[/itex]

where [itex](H^{\otimes N})_{\mathrm{sym}}[/itex] denotes the Hilbert space of symmetrized states for [itex]N[/itex] indistinguishable bosonic particles, where a typical such state is given by

[itex]|n_1,n_2,\ldots \rangle = \frac{1}{\sqrt{N! n_1!\ldots n_k!}} \sum_{\pi \in S_N}\bigotimes_{i=1}^{N}|\alpha_{\pi(i)}\rangle[/itex].

Now define a family of maps [itex]\hat{a}_\lambda : \mathcal{H} \longrightarrow \mathcal{H}[/itex]

such that [itex] \forall \lambda \geq 1[/itex]:

[itex]\hat{a}_\lambda | n_1,n_2,\ldots, n_\lambda,\ldots \rangle \propto |n_1,n_2,\ldots,n_\lambda -1 ,\ldots \rangle [/itex]

and

[itex][\hat{a}_\lambda,\hat{a}^\dag_{\lambda'} ] = \delta_{\lambda\lambda'},\quad [\hat{a}_\lambda,\hat{a}_{\lambda'}]=[\hat{a}^\dag_{\lambda},\hat{a}^\dag_{\lambda'}] = 0[/itex].

By the theory of the 1-dimensional harmonic oscillator, the eigenspectrum of [itex]\hat{a}^\dag_\lambda \hat{a}[/itex] is the non-negative integers.

Now consider an arbitrary element of the [itex]N[/itex]-eigenspace of [itex]\hat{a}^\dag_\lambda \hat{a}[/itex] (say [itex]|n_1,n_2,\ldots\rangle[/itex]). We want to prove that in fact [itex]N = n_\lambda[/itex].

Repeated application of [itex]\hat{a}_\lambda[/itex] lowers the eigenvalue in steps of 1, but by our first hypothesis it also lowers the occupation number of the [itex]\lambda[/itex]-th single-particle state in steps of 1.

Suppose [itex] N > n_\lambda[/itex]. Then this implies that the null ket is an eigenstate of [itex]\hat{a}^\dag_\lambda \hat{a}[/itex] with eigenvalue 0, which is false. Now suppose [itex] N < n_\lambda[/itex], then this implies that [itex]|n_1,n_2,\ldots, n_\lambda - n_\lambda = 0,\ldots\rangle[/itex] is the null ket. Also false.

Therefore [itex]N = n_\lambda[/itex].

Note that I have assumed only two conditions rather than my previous four. The commutation relations alone are not enough because although it guarantees that the eigenvalues of [itex]a^\dag_\lambda a_\lambda[/itex] are the non-negative integers, it does not guarantee they are the occurpation numbers nor does it guarantee that a and a-dag only affect the occupation number of one single particle state, we must postulate this.

From what I've read about quantum field theory, we start by introducing abstract operators which satisfy the commutation relations I listed above and then _interpret_ the eigenvalues as the occupation numbers. I personally find this unsatisfactory, however, for the reasons I gave above.
reilly
#17
Apr16-08, 02:39 PM
Sci Advisor
P: 1,082
Once again, a little history is in order. The origin of the creation and destruction operators (CDO) sometimes called step operators, is Heisenberg's original matrix-theory of the oscillator. Rather than use Q, and P, one uses the combinations, a =Q+iP and a' = Q-iP, apart from constants, and with a'= a adjoint. (Usually this is covered in a beginning QM course.) V. Fock took the CDO and built particle states, instead of oscillator states; basically a change in interpretation. But the key player was Dirac, who used the CDO for photons to develop QED in 1927, and thus set the stage and basic structure of modern QFT. It is absolute necessary to read Dirac's discussion of this in his QM book, if you want to understand the basics of QFT.

See also Chap 1 of Vol 1 of Weinberg's QFT books, and Schweber's QED and the Men Who Made It.

One important property of CDO is its ability to describe states with indefinite numbers of particles.

A very elegant and complete discussion of CDO and canonical variables is given in Optical Coherence and Quantum Optics, Mandel and Wolf when dealing with the quantization of the E&M field --Chap 10.

The usual approach in E&M comes from the form of the free Hamiltonian, E*E + B*B, which with suitable Q and P becomes a harmonic oscillator Hamiltonian. Then it's back to Heisenberg. The usual approach builds from 1. the CDO commutation rules, AND 2. the assumption of a ground state with zero particles -- this is evident in the oscillator solutions. In my opinion, starting with multiparticle states and working backwards is too algebraically complicated and formal, even though it can be made to work. Matters of taste and style.
Regards,
Reilly Atkinson


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