Solving Hermitian Conjugate Homework

[v_pino]

Homework Statement

a.) Show $\hat {(Q^\dagger)}^\dagger=\hat Q$, where $\hat {Q^\dagger}$ is defined by $<\alpha| \hat Q \beta>= <\hat Q^ \dagger \alpha|\beta>$.

b.) For $\hat Q =c_1 \hat A + c_2 \hat B$, show its Hermitian conjugate is $\hat Q^\dagger =c_1^* \hat A^\dagger + c_2^* \hat B^\dagger$.

Homework Equations

a.) I found an example that might be related to this problem. It says that $|T^\dagger \alpha> = T^\dagger |\alpha>$ and $<T|=(|T>)^\dagger$ .

The Attempt at a Solution

For part (a), I'm thinking that I might be rewrite the right hand side of the second equation. From the relevant equations I gave, do you think $<\hat Q^\dagger \alpha| \beta> = \hat Q^\dagger <\alpha| \beta>$ is permitted? And if so, how do I proceed from here?

 

[dextercioby]

For both a) and b) assume the operators are bounded, hence domains issues do not appear. So for point a) both the adjoint and the adjoint of the adjoint exist and share the same domain,

$$\langle \psi, Q\phi\rangle = \langle Q^{\dagger}\psi,\phi\rangle = ...$$

For point b), use the definition of adjoint used at a).

 

[v_pino]

Here is my attempt after your advice:

(a) $<\alpha|\hat Q \beta>=<\alpha| \hat Q^{\dagger \dagger} \beta>$

(b) $<\hat Q^* \alpha| \beta>=<\hat Q^{\dagger} \alpha|\beta>$

I'm not exactly sure about the intermediate steps i.e. from 'first principle' to derive these equations.

 

[dextercioby]

Ok, in my writing above instead of ... there's what you've written (with other vectors, and with LaTex write \langle and \rangle to get nicely looking eqns)

So

$$\langle \psi,Q\phi\rangle = \langle \psi,Q^{\dagger\dagger}\phi\rangle$$ from where you have that both the range and the domain of Q and Q double dagger are equal, hence the 2 operators are equal. q.e.d.

For point b) pay more attention with your writings and redo your calculations.

 

[v_pino]

Is this correct?

$\langle\alpha|\hat Q \beta \rangle=\hat Q \langle \alpha| \beta \rangle <br /> = \langle\hat Q^* \alpha| \beta \rangle$

Therefore,

$\langle\hat Q^* \alpha| \beta \rangle = \langle \hat Q^\dagger \alpha |\beta \rangle$

If the above is correct, I get:

$\hat Q^\dagger = \hat Q^* = c_1^* \hat A + c_2^* \hat B$

But I don't get the $\dagger$ above the A and the B.

 

[dextercioby]

Not really.

$$\langle \psi,(c_1 A + c_2 B)\phi\rangle = \langle (c_1 A + c_2 B)^{\dagger}\psi, \phi\rangle = \langle \psi,c_1 A \phi\rangle + \langle \psi,c_2 B \phi\rangle$$.

Can you go further with the sequence of equalities ?

 

[dextercioby]

Now it finally looks OK.