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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?

1)

[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?

1)

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