1. Dec 11, 2008

### KFC

For an opeation as follow

$$\langle \varphi | \hat{A} |\psi\rangle$$

where $$\hat{A}$$ is an operator.

It is no problem to have $$\hat{A}$$ directly operate on the ket state $$|\psi\rangle$$, but if I want $$\hat{A}$$ operates on the bra state $$\langle\varphi |$$, do I have to take the adjoint of A first? That is

$$\left(\langle \varphi | \hat{A}^\dagger\right) |\psi\rangle = \langle \varphi | \left(\hat{A} |\psi\rangle\right)$$

2. Dec 11, 2008

### Fredrik

Staff Emeritus
No. Lose the that dagger on the left-hand side. Se #7 in this thread. (In particular the stuff I quoted from another thread).

Last edited: Dec 11, 2008
3. Dec 12, 2008

### KFC

So, as you told in that thread. For

$${\Large\langle\varphi | A | \psi \rangle}$$

the opreator A can either operate on $$\psi$$ or $$\varphi$$ ?

4. Dec 12, 2008

### fractal_uk

Think of it this way, A acts on the thing to the right while $$A^{\dagger}$$ acts on the thing to the left. If you want to operate on the bra-state $$\langle \phi |$$ then you do need $$A^{\dagger}$$ unless A is self adjoint, in which case $$A = A^{\dagger}$$ and the operator can act either to the left or the right.

If A is some observable then it must be self-adjoint.

5. Dec 12, 2008

### KFC

Got it. In this sense, the following conclusion only true when $$A^\dagger = A$$, right? That is, if operator A is self-adjoint, it can either operate on the left or the right?

$$\left(\langle \varphi | \hat{A}^\dagger\right) |\psi\rangle = \langle \varphi | \left(\hat{A} |\psi\rangle\right)$$

6. Dec 12, 2008

### Hurkyl

Staff Emeritus
This is nothing new, nor special. Think about any functions at all:
$$f(g(x)) = (f \circ g)(x)$$
i.e. both of these give the same result:
1. Evaluating g at x, then evaluating f at the result
2. Composing f and g, and evaluating the result at x

And similarly, think about matrix arithmetic. I can evaluate the product wAv in any order I choose.

7. Dec 12, 2008

### Fredrik

Staff Emeritus
I suppose you can say that, if you mean $|\psi\rangle$ and $\langle\varphi|$.

The expression $\langle\varphi|A|\psi\rangle$ means $\langle\varphi|(A|\psi\rangle)$, and by definition of the bra, that's the scalar product of $|\varphi\rangle$ and $A|\psi\rangle$. By definition of the adjoint operator, that scalar product is equal to the scalar product of $A^\dagger|\varphi\rangle$ and $|\psi\rangle$, which in bra-ket notation takes the form $(\langle\varphi|A)|\psi\rangle)$. That's why you can drop the parentheses.

I don't see a way to interpret this as a correct statement.

Any operator X satisfies $\langle\varphi|(X|\psi\rangle)$=$(\langle\varphi|X)|\psi\rangle$, so yes, your equation is correct for all $\langle\varphi|$ and $|\psi\rangle$ if and only if A is self-adjoint, but I don't see why you would want to express that as "if operator A is self-adjoint, it can either operate on the left or the right".

Last edited: Dec 12, 2008
8. Dec 13, 2008

### KFC

Oh, there are too many comments here. It becomes more confusing. Fredrik, could you please ask me the following questions again. I read the thread you post and now let's start from

$$\langle f | A | g \rangle = (|f\rangle, A|g\rangle) = (A|g\rangle, |f\rangle)^* = \langle g | A^\dagger | f \rangle ^*$$

Is this right? In this case (A might not btself-adjoint), $$A^\dagger$$ should operate to the left or right?

My second question is : assuming $$A^\dagger$$ still operates to the right, considering the last term in above equation

$$\langle g | A^\dagger | f \rangle ^* = (|g\rangle, A^\dagger | f \rangle)^* = \left((A^\dagger | f \rangle, |g\rangle)^*\right)^* = (A^\dagger | f \rangle, |g\rangle)$$

Can we interpret $$\langle f | A | g \rangle$$ in this way? Firstly, let $$A^\dagger$$ operates to the ket $$|f\rangle$$ and then take the inner product with the ket $$|g\rangle$$, right? In a word, if I want an operator operates to a bra $$\langle f|$$, I can have its corresponding adjoint operator acting on the corresponding ket $$|f\rangle$$ to get another ket, after I get the result, I convert that ket back to a corresponding bra, is that right?

I am asking this question because sometimes I got an expression like $$\langle f | A | g \rangle$$, I need to keep $$|g\rangle$$ unchanged while I've already know the eigenvalue problem $$A|f\rangle$$, that's why I am asking how to make A operate to the left.

9. Dec 13, 2008

### Fredrik

Staff Emeritus
Suppose that $A|a\rangle=a|a\rangle$, then what is $A^\dagger|a\rangle$? Let's find out:

$$A^\dagger|a\rangle=\sum_{a'}|a'\rangle\langle a'|A^\dagger|a\rangle$$

$$\langle a'|A^\dagger|a\rangle=(|a'\rangle,A^\dagger|a\rangle)=(A|a'\rangle,|a\rangle)=(a'|a'\rangle,|a\rangle)=a'^*(|a'\rangle,|a\rangle)=a'^*\delta_{a'a}$$

$$A^\dagger|a\rangle=\sum_{a'}|a'\rangle a'^*\delta_{a'a}=a^*|a\rangle$$

You can use this to find out what you should do when the eigenstate appears in the form of a bra on the left.

$$\langle a|A|b\rangle=(|a\rangle,A|b\rangle)=(A^\dagger|a\rangle,|b\rangle)=(a^*|a\rangle,|b\rangle)=a(|a\rangle,|b\rangle)=a\langle a|b\rangle$$

Note that there's no need to ever talk about an operator acting to the left. We don't have to define the "product" of a bra and an operator, but we do it anyway because it's convenient. We can define $\langle f|A$ either by

$$(\langle f|A)|g\rangle=\langle f|(A|g\rangle)$$

or (equivalently) by

$$(\langle f|A)|g\rangle=(A^\dagger |f\rangle, |g\rangle)$$

This definition allows us to interpret the previous result as

$$\langle a|A=\langle a|a$$

Last edited: Dec 13, 2008
10. Dec 13, 2008

### KFC

Got you. Thank you so much!