Why particle number operator $\hat{N}$ is $\hat{a}^\dagger\hat{a}$ rather than $\hat{a}\hat{a}^\dagger$?

Question

Both $\hat{a}^\dagger\hat{a}$ and $\hat{a}\hat{a}^\dagger$ are Hermitian, how do we know which one represents the particle number?

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user LePtC

Answer 1

Since you define e.g. in the bosonic case

$c_j^\dagger: H_N^S \rightarrow H_{N+1}^S,\quad c_j^\dagger | \ldots n_j \ldots \rangle := \sqrt{n_j+1} |\ldots n_j+1 \ldots \rangle$ $c_j: H_N^S \rightarrow H_{N-1}^S,\quad c_j | \ldots n_j \ldots \rangle := \sqrt{n_j} |\ldots n_j-1 \ldots \rangle$

it makes more sence to use $a^\dagger a$ which will give you $n_j$ (instead of $n_j+1$) as prefactor when acting on a state $| \ldots n_j \ldots \rangle $.

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user pawel_winzig

Comments

But you are using the eigen state of $\hat{a}^\dagger\hat{a}$, maybe we can redefine the eigen state $|n'_i\rangle=|n_i+1\rangle$ to avoid that problem?

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user LePtC

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@LePtC: This would only "shift" your problem...

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user pawel_winzig

Answer 2

The vacuum should have particle number $0$. In some detail: we would like $\hat{N}\ |0\rangle=0$ and $\hat{N}=a^\dagger a$ is the only ordering that does that. It follows from the usual commutation relations that $\hat{N} |n\rangle =n\ |n\rangle$ which is in sync with interpretation of  $|n\rangle $ as an $n$-particle state in second-quantisation.

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user suresh

Comments

You want to elaborate on this a bit?

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user Pranav Hosangadi

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I think he means that $a^\dagger a$ kills the vacuum while $a \, a^\dagger$ doesn't.

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user elfmotat

Answer 3

It just depend on the definition of the number operator. In a Hilbert space ${\cal H}$, one starts from the algebra $$[a,a^*]=1$$ for a pair of operators $a,a^*$ defined in some invariant domain $D\subset {\cal H}$ also assuming  that there is a unique vector,  in $D$, denoted by $|0\rangle$ such that $a|0\rangle = 0$.

Only exploitong (i) the commutation rules above, (ii) the definition of $|0\rangle$ and (iii) the fact that $a^*$ is the adjoint of $a$, at least when working in $D$, one easily sees that $D$ must contains an infinite orthonormal set of vectors of the form

$$|n\rangle := \frac{(a^*)^n}{\sqrt{n!}}|0\rangle \quad n=0,1,2,\ldots$$

$n$ here just denotes how many times $a^*$ acts on $|0\rangle$ in the formula above to produce $|n\rangle$ up to normalization coefficients.

From the given definition, it turns out that

$$a|n\rangle = \sqrt{n}|n-1\rangle \quad \mbox{and}\quad a^*|n\rangle = \sqrt{n+1}|n+1\rangle\:.$$

The meaning of that $n$ depends on physical context. In elementary QM, this machinery is used to compute the spectrum of the Hamiltonian operator of the harmonic oscillator. In that case $n$ denotes an eigenvalue and $E_n = \hbar\omega (n+ \frac{1}{2})$. In QFT there is a more sophisticated construction and, in fact $n_{p}$ denotes the number of particles with a certain value of the four momentum $p$ and there are operatores $a_p, a^*_p$ for each $p$. (This construction, in QFT, can be readapted to a generic state not necessarily with defined momentum and relies upon the notion of Fock-Hilbert space space.)

The number operator $N$ is just defined as the operator such that $$N|n\rangle = n|n\rangle$$ whatever is the meaning of $n$. As  the vectors $|n\rangle$ form a Hilbert basis in ${\cal H}$ (or in a closed subspace which can be considered the true physical Hilbert space of the system), $N$ turns out to be self-adjoint with pure point spectrum and thus is a proper quantum observable whose (eigen)values are the numbers $n$.

Using the commutation rules of $a$ and $a^*$ as well as the definition of $|0>$ one easily sees that $$N|n\rangle = a^*a|n\rangle\:.$$

In fact $$a^*a|n\rangle = a^* \sqrt{n}|n-1\rangle = \sqrt{n} a^* |n-1\rangle = \sqrt{n}\sqrt{(n-1)+1}|(n-1)+1\rangle = n|n\rangle$$ $$= N|n\rangle\:.$$

Since the vectors $|n\rangle$ form a basis, essentially by linearity, we can equivalently write $$N = a^*a\:.$$

Answer 4

We require that the number operator have the following property:

$$\hat n |0\rangle = 0$$

We know that

$$\hat a |0\rangle = 0$$

and we know that

$$\hat a |1\rangle = |0\rangle$$

and we know that

$$\hat a^{\dagger} |0\rangle = |1\rangle $$

thus, it follows that

$$\hat a \hat a^{\dagger} \ne \hat n$$

since

$$\hat a \hat a^{\dagger} |0\rangle = \hat a|1\rangle \ne 0$$

Now, it remains to be shown that

$$\hat a^{\dagger}\hat a = \hat n $$

Can you take it from here?

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user Alfred Centauri

Comments

Thank you! You clearly solved my problem, but now I have another question: what's the difference between 0 and $|0\rangle$? (should I open a new question?)

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user LePtC

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@LePtC Uh, \(\vert 0\rangle\) is the ground state while "0" is just the number zero?

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\(|0\rangle \) represents (in the corresponding Hilbert space) the state which has 0 one-particles while 0 is a number and in the case you apply the number operator N to  \(|0\rangle \) it gives you the operator's eigenvalue, which for that case it is zero. Check out Sakurai's book  (http://www.amazon.com/gp/product/0805382917/)

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0 is the zero vector (not a number!!) in the Fock space (with norm 0, the number zero), while $|0\rangle$ is the vacuum state (containing zero particles but having norm 1)