Identity Operator for Multiple Particles

  • #1
Hi,

For a particle in a box (so that the momentum spectrum is discrete), we can write the identity operator as a sum over all momentum eigenstates of a projection to that eigenstate: $$I=\displaystyle\sum\limits_{p} |p\rangle\langle p|.$$

I was wondering what the corresponding form of the identity operator would be if we had multiple particles? In particular, a sum over both momenta appears to not lead to normalized states (when the particles are identical): $$A=\displaystyle\sum\limits_p\sum\limits_q |p,q\rangle\langle p,q|.$$ For example, given two possible momentum states for each particle 0 and 1, $$A=\sum\limits_{p=0}^1\sum\limits_{q=0}^1 |p,q\rangle\langle p,q|,$$ we find that $$A|0,1\rangle = |0,1\rangle +|1,0\rangle\langle 1,0|0,1\rangle=|0,1\rangle\pm|1,0\rangle$$ where the + sign is taken if the particles are both bosons and the minus sign if they are both fermions.

This looks like what we're supposed to get up to a normalization constant. I'm trying to figure out exactly how to modify the operator A so that the normalization constant works out. In particular, just dividing A by root 2 to fix this problem, $$A = \dfrac 1{\sqrt 2} \sum\limits_{p=0}^1\sum\limits_{q=0}^1 |p,q\rangle\langle p,q|,$$ we find that $$A|0,0\rangle = \dfrac 1{\sqrt 2}|0,0\rangle$$ which is not normalized.

Any help would be appreciated. Thanks!
 

Answers and Replies

  • #2
stevendaryl
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I don't know what the most elegant way is to deal with Fermions and Bosons, but one approach is to say that the complete basis does not respect the statistics, and require an additional constraint on physical states that they must be symmetric or antisymmetric.

To make it simple, let's only consider spin degrees of freedom. A complete basis for a single particle is ##|U\rangle## and ##|D\rangle## (spin-up and spin-down along the z-axis, say). A complete basis for a two-particle state is the four states:
  1. ##|U,U\rangle##
  2. ##|U,D\rangle##
  3. ##|D,U\rangle##
  4. ##|D,D\rangle##
The inner product of two states is component-wise:

##\langle X, Y|X, Y\rangle = 1##
##\langle X, Y|X', Y'\rangle = 0## if ##X \neq X'## or ##Y \neq Y'##

So obviously, the identity for two-particle states is:

##I = |U,U\rangle \langle U,U| + |U,D\rangle \langle U,D| + |D,U\rangle \langle D,U| + |D,D\rangle \langle D,D| ##

With this basis, the only two-particle state that respects Fermi antisymmetry is:

##|0\rangle \equiv \frac{1}{\sqrt{2}} (|U,D\rangle - |D,U\rangle)##

And it's clear that ##I |0\rangle = |0\rangle##

In terms of the unsymmetrized basis, ##\langle U, D|D, U\rangle = 0##, not ##-1##.
 
  • #3
vanhees71
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If you have a complete orthonormalized set of vectors ##|u_n \rangle##, then
$$\sum_{n=1}^{\infty} |u_n \rangle \langle u_n|=\mathbb{1}.$$
For your case you have single-particle states ##|p_n \rangle## with ##p_n=2 \pi n \hbar/L##, ##n \in \mathbb{Z}## (for the case of particles in a box with periodic boundary conditions; for rigid boundary conditions this makes no sense at all, because you cannot even define a momentum observable to begin with), forming a complete set of one-particle states.

If you consider two identical particles the Hilbert space is spanned by the totally symmetrized (antisymmetrized) Kronecker-product states,
$$|p_{1j},p_{2k} \rangle=\frac{1}{\sqrt{2}} (|p_{1i} \rangle \otimes |p_{2j} \rangle \pm |p_{2j} \rangle \otimes |p_{1i}).$$
These states are orthonormal, i.e.,
$$\langle p_{1j},p_{2k}|p_{1j'},p_{2k'} \rangle=\delta_{jj'} \delta_{kk'}.$$
The completeness relation then reads
$$\sum_{j,k} |p_{1j},p_{2k} \rangle \langle p_{1j},p_{2k}|=\mathbb{1}.$$
 

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