Wave mechanics : self-adjoint problem

[noblegas]

Homework Statement

The operators P and Q are self-adjoint and satisfy the commutation relation [P,Q]=ic where c is a real number. Show that the operator [P,Q] is anti-self-adjoint, that is , that the adjoint of the operator is the negative of the operator, consistent with the right-hand side of the equation.

Homework Equations

The Attempt at a Solution

I know a self-adjoint operator Q*=Q where * represents the dagger. [P,Q]=(PQ-QP)*phi=ic=0=PQphi-QPphi= PQ*phi-Q*Pphi=0?

 

[gabbagabbahey]

Why are you setting $[P,Q]$ equal to zero? and what is 'phi'...an example wavefunction?...There is no need to introduce any wavefunctions here.

By definition, $[P,Q]=PQ-QP$...you are told that this is equal to $ic$.

What is the definition of $[P,Q]^\dagger$?

 

[noblegas]

Why are you setting $[P,Q]$ equal to zero? and what is 'phi'...an example wavefunction?...There is no need to introduce any wavefunctions here.

By definition, $[P,Q]=PQ-QP$...you are told that this is equal to $ic$.

What is the definition of $[P,Q]^\dagger$?

Right... matrices don't commute. By definition , $[P,Q]^\dagger$=$[Q,P]=QP-PQ$

I think I got it . PQ=ic+QP => [Q,P]=QP-PQ=QP-(ic+QP)=-ic ?

 

[gabbagabbahey]

While its true that $[P,Q]^{\dagger}=[Q,P]$ (but only since $Q$ and $P$ are Hermitian), that's not really the definition of $[P,Q]^{\dagger}$. In fact, it's basically the very thing you are trying to prove, so I wouldn't use it as the basis of your proof!

By definition,

$$[P,Q]^{\dagger}=(PQ-QP)^{\dagger}=(PQ)^{\dagger}-(QP)^{\dagger}$$

Now, use the fact that $(AB)^{\dagger}=B^{\dagger}A^{\dagger}$ for any two operators $A$ and $B$, along with the fact that $Q$ and $P$ are Hermitian...

 

[noblegas]

While its true that $[P,Q]^{\dagger}=[Q,P]$ (but only since $Q$ and $P$ are Hermitian), that's not really the definition of $[P,Q]^{\dagger}$. In fact, it's basically the very thing you are trying to prove, so I wouldn't use it as the basis of your proof!

By definition,

$$[P,Q]^{\dagger}=(PQ-QP)^{\dagger}=(PQ)^{\dagger}-(QP)^{\dagger}$$

Now, use the fact that $(AB)^{\dagger}=B^{\dagger}A^{\dagger}$ for any two operators $A$ and $B$, along with the fact that $Q$ and $P$ are Hermitian...

$$[P,Q]^{\dagger}=(PQ-QP)^{\dagger}=(PQ)^{\dagger}-(QP)^{\dagger}$$=$$Q^{\dagger}P^{\dagger}-P^{\dagger}Q^{\dagger}$$; since $$Q^{\dagger}=Q$$ , then $$P^{\dagger}=P$$, therefore $$Q^{\dagger}P^{\dagger}-P^{\dagger}Q^{\dagger}=QP-PQ=[Q,P]$$

therefore $$[P,Q]^{\dagger}=[Q,P]$$

 

[gabbagabbahey]

Right, but your problem statement doesn't actually tell you the operators are Hermitian, so you won't want to use that.

Instead, just take the adjoint of both sides of the equation you are given...

$$[P,Q]=ic\implies [P,Q]^{\dagger}=(ic)^{\dagger}=-ic=-[P,Q]$$