# Hermitian conjugate

1. Nov 16, 2011

### v_pino

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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?

2. Nov 16, 2011

### 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).

3. Nov 16, 2011

### v_pino

(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.

4. Nov 17, 2011

### 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.

5. Nov 17, 2011

### v_pino

Is this correct?

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

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.

6. Nov 17, 2011

### 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 ?

7. Nov 17, 2011

### dextercioby

Now it finally looks OK.