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  Why particle number operator ˆN is ˆaˆa rather than ˆaˆa?

+ 3 like - 0 dislike
4141 views

Both ˆaˆa and ˆaˆa 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
asked Oct 10, 2014 in Mathematics by LePtC (15 points) [ no revision ]

4 Answers

+ 5 like - 0 dislike

Since you define e.g. in the bosonic case

cj:HSNHSN+1,cj|nj:=nj+1|nj+1 cj:HSNHSN1,cj|nj:=nj|nj1

it makes more sence to use aa which will give you nj (instead of nj+1) as prefactor when acting on a state |nj.

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user pawel_winzig
answered Oct 10, 2014 by pawel_winzig (70 points) [ no revision ]
But you are using the eigen state of ˆaˆa, maybe we can redefine the eigen state |ni=|ni+1 to avoid that problem?

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user LePtC
@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
+ 3 like - 0 dislike

The vacuum should have particle number 0. In some detail: we would like ˆN |0=0 and ˆN=aa is the only ordering that does that. It follows from the usual commutation relations that ˆN|n=n |n which is in sync with interpretation of  |n 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

answered Oct 10, 2014 by suresh (1,545 points) [ revision history ]
edited Oct 12, 2014 by suresh
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
I think he means that aa kills the vacuum while aa doesn't.

This post imported from StackExchange Physics at 2014-10-11 09:52 (UTC), posted by SE-user elfmotat
+ 2 like - 0 dislike

It just depend on the definition of the number operator. In a Hilbert space H, one starts from the algebra [a,a]=1

for a pair of operators a,a defined in some invariant domain DH also assuming  that there is a unique vector,  in D, denoted by |0 such that a|0=0.

Only exploitong (i) the commutation rules above, (ii) the definition of |0 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:=(a)nn!|0n=0,1,2,

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

From the given definition, it turns out that

a|n=n|n1anda|n=n+1|n+1.

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 En=ω(n+12). In QFT there is a more sophisticated construction and, in fact np denotes the number of particles with a certain value of the four momentum p and there are operatores ap,ap 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=n|n

whatever is the meaning of n. As  the vectors |n form a Hilbert basis in 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=aa|n.

In fact aa|n=an|n1=na|n1=n(n1)+1|(n1)+1=n|n

=N|n.

Since the vectors |n form a basis, essentially by linearity, we can equivalently write N=aa.

answered Oct 12, 2014 by Valter Moretti (2,155 points) [ revision history ]
edited Oct 12, 2014 by Arnold Neumaier
+ 1 like - 0 dislike

We require that the number operator have the following property:

ˆn|0=0

We know that

ˆa|0=0

and we know that

ˆa|1=|0

and we know that

ˆa|0=|1

thus, it follows that

ˆaˆaˆn

since

ˆaˆa|0=ˆa|10

Now, it remains to be shown that

ˆaˆa=ˆn

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
answered Oct 10, 2014 by Alfred Centauri (110 points) [ no revision ]
Thank you! You clearly solved my problem, but now I have another question: what's the difference between 0 and |0? (should I open a new question?)

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

@LePtC Uh, |0 is the ground state while "0" is just the number zero?

|0 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 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/)

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

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