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Suppose we have a system of bosons represented by their occupation numbers $$\tag{1} | n_1, n_2, ..., n_\alpha, ... \rangle$$ Then we can define creation and annihilation operators $$\tag{2} a_\alpha^\dagger| n_1, n_2, ..., n_\alpha, ... \rangle = \sqrt{n_\alpha+1} | n_1, n_2, ..., n_\alpha+1, ... \rangle$$ $$\tag{3} a_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle = \sqrt{n_\alpha} | n_1, n_2, ..., n_\alpha-1, ... \rangle$$ This is nice because the number operator is just $a_\alpha^\dagger a_\alpha$. However, would it be sensible to define an alternate set of operators to work with? $$\tag{4} b_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle = | n_1, n_2, ..., n_\alpha+1, ... \rangle$$ $$\tag{5} c_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle = \begin{cases} | n_1, n_2, ..., n_\alpha-1, ... \rangle & n_\alpha>0 \\ 0 & n_\alpha=0 \end{cases}$$ $$\tag{6} N_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle = n_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle $$ Why don't we work with these operators? The bosonic creation and annihilation operators $a_\alpha^\dagger$ and $a_\alpha$ were defined to mimic the harmonic oscillator's raising and lowering operators ($x \pm i p$), but is there any compelling reason to keep the $\sqrt{n_\alpha+1}$ and $\sqrt{n_\alpha}$ factors?

I suppose $a_\alpha^\dagger$ and $a_\alpha$ obey nice properties such as $[a_\alpha,a_\alpha^\dagger]=1$ and the fact that they are Hermitian adjoints of each other. What are the analogous relationships that $b_\alpha$ and $c_\alpha$ would obey?

ChickenGod
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2 Answers2

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Here are three properties that would make your definitions awkward.

You can think of $a^\dagger\,a$ as the LU (lower triangular, upper triangular or Cholesky) decomposition of the number observable. Actually, it's not the unique Cholesky factorisation but it is the one found by the outer product version of the algorithm. Your definition would not have this property.

As a result, the Lie bracket between your two $b,\,c$ operators would be a bit nasty. In energy eigen co-ordinates (i.e. so that a quantum harmonic oscillators state is given by an infinite column vector of probability amplitudes to be in each of the number states) it would be:

$$[b,\,c]={\rm diag}[1,\,0,\,0,\,\cdots]$$

and the Hamiltonian would be complicateder:

$$\hat{H} = \hbar\,\omega\left(c\,b\,N + \frac{1}{2}\right) = \hbar\,\omega\left(N\,c\,b + \frac{1}{2}\right)$$

and there's no simple way to write the Hamiltonian in terms of $b\,c ={\rm diag}[0,\,1,\,1,\,1,\,\cdots]$.

All this would make rewriting operator and observable expressions in normal ordering very awkward indeed.

Lastly, there is a rather elegant way of writing down a general coherent state of the quantum harmonic oscillator, through the so-called displacement operator:

$$\left|\left.\alpha\right>\right. = \exp\left(\alpha\,a^\dagger - \alpha^* \, a\right)\left|\left.0\right>\right.$$

which shifts the quantum ground state to a coherent one with amplitude (i.e. displacement from the origin in $x-p$ phase space) $\alpha$. This highly useful formula would be much awkwarder in your notation.

Selene Routley
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The problem is that the physical states have positive occupation numbers $n_1, n_2,...$.

With your operators, you have, for instance :

$ c_\alpha| n_1, n_2, ..., 0, ... \rangle = | n_1, n_2, ..., -1, ... \rangle$.

This gives you a totally unphysical state, so you would have to add by-hand constraints like $n_1 \geq 0, n_2 \geq 0,...$, .

With the operators $a_{\alpha}$, you do not have this problem, the no-existence of states with negative occupation numbers is completely natural, because of the term $\sqrt{n_\alpha}$

Trimok
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