Adjoint Operator: Use & Benefits

[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)

 

[Fredrik]

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

 

[KFC]

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

So, as you told in that thread. For

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

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

 

[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.

 

[KFC]

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.

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)

 

[Hurkyl]

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.

 

[Fredrik]

So, as you told in that thread. For

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

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

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.

Think of it this way, A acts on the thing to the right while A^{\dagger} acts on the thing to the left.

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

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)

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".

 

[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.

 

[Fredrik]

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

 

[KFC]

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, |g\rangle)

This definition allows us to interpret the previous result as

\langle a|A=\langle a|a

Got you. Thank you so much!