# Identity Operator for Multiple Particles

• I
• acegikmoqsuwy
In summary, the two-particle state that respects Fermi antisymmetry is the Kronecker-product state that is orthonormalized to the space of single-particle states.

#### acegikmoqsuwy

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!

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##.

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.,
$$\langle p_{1j},p_{2k}|p_{1j'},p_{2k'} \rangle=\delta_{jj'} \delta_{kk'}.$$
$$\sum_{j,k} |p_{1j},p_{2k} \rangle \langle p_{1j},p_{2k}|=\mathbb{1}.$$

## What is the Identity Operator for Multiple Particles?

The Identity Operator for Multiple Particles is a mathematical operator used in quantum mechanics to describe the behavior of a system of multiple particles. It is represented by the symbol 'I' and acts as a unit operator, leaving the state of the system unchanged.

## How is the Identity Operator used in Quantum Mechanics?

In quantum mechanics, the Identity Operator is used to represent the concept of 'no change' or 'no effect' on a system. This is particularly useful when describing the behavior of a system of multiple particles, where the individual particles may interact with each other but the overall state of the system remains the same.

## What are the properties of the Identity Operator?

The Identity Operator has two main properties: it is a unit operator, meaning that when it acts on a state, it leaves the state unchanged, and it is also commutative, meaning that it can be applied in any order without changing the result.

## How is the Identity Operator related to the Pauli Exclusion Principle?

The Pauli Exclusion Principle states that no two identical fermions (particles with half-integer spin) can occupy the same quantum state simultaneously. The Identity Operator plays a role in this principle by ensuring that the state of a system remains unchanged when two fermions interact, preventing them from occupying the same state.

## Can the Identity Operator be used for particles with integer spin?

No, the Identity Operator is only applicable to particles with half-integer spin, such as electrons, protons, and neutrons. Particles with integer spin, such as photons, follow different rules in quantum mechanics and do not require the use of the Identity Operator.