Symmetry of the particle creation and annihilation operators in quantum physics a+ and a-

Fundamental “operators” in quantum physics

a⁺ and a^-

They started out as the particle number operator a⁺a^-.

It is a hermitian operator:

a⁺=((a^-)’)^T,

For a state of n particles, <n|a⁺a^-|n>=n.

More fundamental is the operator commutation relationship a^-a⁺-a⁺a^-=1, for bosons.

It means that if <n|a⁺a^-|n>=n, then <n+1|a⁺a^-|n+1>=n+1.

The proof is:

<n|a⁺a^-|n>=n and a^-a⁺-a⁺a^-=1 means <n|a^-a⁺|n>=n+1,

meaning a⁺|n>=(n+1)^0.5e^iθn|n+1>,

thus |n+1>=(n+1)^-0.5e^-iθna⁺|n>,

thus <n+1|=<n|a^-(n+1)^-0.5e^iθn,

thus <n+1|a⁺a^-|n+1>=<n|a^-(n+1)^-0.5e^iθna⁺a^-(n+1)^-0.5e^-iθna⁺|n>

=1/(n+1)<n|a^-(a⁺a^-)a⁺|n>.

We specify that these operators obey (AB)C=A(BC), in other words the sequence of grouping does not matter. Then

<n+1|a⁺a^-|n+1>=<n|a^-(a⁺a^-)a⁺|n>/(n+1)

=<n|(a^-a⁺)(a^-a⁺)|n>/(n+1)

=<n|(a⁺a^-+1)(a⁺a^-+1)|n>/(n+1)

=[<n|(a⁺a^-)(a⁺a^-)|n>+2n+1]/(n+1)

Since |n> is the eigen vector of a⁺a^-, <n|(a⁺a^-)(a⁺a^-)|n> = <n|(a⁺a^-)n|n>=n².

Therefore, <n+1|a⁺a^-|n+1> = n+1.

Fermi statistics comes from the alternative operator commutation relationship

a^-a⁺+a⁺a^-=1.

The vacuum state |0> is defined as <0|a⁺a^-|0>=0,

then <0|a^-a⁺|0>=1,

meaning a⁺|0>=eⁱ^θ|1>,

meaning <1|a⁺a^-|1>=<0|a⁺a^-a⁺a^-|0>=1.

Thus <1|a^-a⁺|1>=0, meaning

a⁺|1>=0.

Physically, it means there cannot be two identical particles. In order for this to be true for any state in the hilbert space, one needs to have

a⁺(x)a⁺(y)+a⁺(y)a⁺(x)=0 for any x and y states, and equivalently

a^-(x)a^-(y)+a^-(y)a^-(x)=0.

With a^-a⁺+a⁺a^-=1 there is a full symmetry between the |0> vs |1> states and a⁺ vs a^- operators

The equations

<0|a⁺a^-|0>=0, <1|a^-a⁺|1>=0, <0|a^-a⁺|0>=1, <1|a⁺a^-|1>=1

are symmetrical with respect to switching |0> with |1> and a⁺ with a^-. This is the basis of the Fermi sea idea.

There is no such symmetry with the bosons.

Thoughts