B Understanding the Dagger Notation in Quantum Mechanics

[Mayan Fung]

let say I have a vector |a> and |b> and a transformation matrix A
What is the difference between <a|A|b> and <a|Ab>?
And also, I don't quite understand why <a|Ab> = <A+a|b>. Where does this identity come from?

Thanks!

 

[andrewkirk]

None. They mean the same thing.

 

[Mayan Fung]

None. They mean the same thing.

Oh thanks! Do you have any idea about the identity <a|Ab> = <A+a|b>?

 

[Mayan Fung]

None. They mean the same thing.

Does that also mean that <Aa| = A<a|?

 

[andrewkirk]

Let me backtrack a bit. They denote the same value, which is an element of the field F over which the vector space is set.

However, they can have different interpretations.

$\langle a|A|b\rangle$ can be interpreted to mean either $\langle a|Ab\rangle$ or $\langle A^\dagger a|b\rangle$. Those two items are different inner products, but they have the same value.

Also, $\langle a|A|b\rangle$ can be interpreted as the result of applying the element $\langle a|A$ of the dual space to the element $|b\rangle$ of the vector space. There is an associative law in operation whereby:

$$\langle A^\dagger a|b\rangle=(\langle a|A)|b\rangle=\langle a|(A|b\rangle)=\langle a|Ab\rangle$$
so we can drop the parentheses and just write $\langle a|A|b\rangle$.

Does that also mean that <Aa| = A<a|?

No. In fact $A\langle a|$ cannot not mean anything, as $\langle a|$ is a function from $V$ to $F$ and $A$ is a function from $V$ to $V$, so we cannot apply $A$ to the result of $\langle a|$.

What you can write is $\langle a|A=\langle A^\dagger a|$ where $A^\dagger$ denotes the adjoint of $A$.

Oh thanks! Do you have any idea about the identity <a|Ab> = <A+a|b>?

The right side of that identity has no meaning in general, because $A$ is a linear operator and $a$ is just a label for a vector, and the two cannot be added. In a particular context that expression might be able to have a meaning, but you'd need to say what that context is, and what the label $a$ means.

 

[Mayan Fung]

Let me backtrack a bit. They denote the same value, which is an element of the field F over which the vector space is set.

However, they can have different interpretations.

$\langle a|A|b\rangle$ can be interpreted to mean either $\langle a|Ab\rangle$ or $\langle A^\dagger a|b\rangle$. Those two items are different inner products, but they have the same value.

Also, $\langle a|A|b\rangle$ can be interpreted as the result of applying the element $\langle a|A$ of the dual space to the element $|b\rangle$ of the vector space. There is an associative law in operation whereby:

$$\langle A^\dagger a|b\rangle=(\langle a|A)|b\rangle=\langle a|(A|b\rangle)=\langle a|Ab\rangle$$
so we can drop the parentheses and just write $\langle a|A|b\rangle$.

Thanks, but I still don't quite understand why⟨A^†a|=⟨a|A. I learned that (AB)^† = B^†A^†
So isn't ⟨A^†a|= (⟨a|A)^† ?
um... I interpret ⟨A^†a| as A^†<a|

 

[Mayan Fung]

Oh! I think I got that.
⟨A^†a| = (A^†a)^† = a^†A = <a|A
Am I correct?

 

[andrewkirk]

So isn't ⟨A†a|= (⟨a|A)† ?

No, at least not in my experience. The 'dagger' $\dagger$ denotes taking the adjoint of an operator, so it can only be applied to operators, which are functions from V to V. But $\langle a|A$ is a function (map) from V to F. That is not an operator. It is an element of the dual space V*. So we cannot use the dagger symbol on it.

By the way, the notation $\langle A^\dagger a|$ denotes the dual vector of $|A^\dagger a\rangle$.

I find it really helps when working with these things to keep a clear view of what sort of a map each element is. We have (1) vectors in V, (2) dual vectors, which are maps from V to F, and (3) linear operators from V to V. We also often have (4) scalars in F, which can be applied multiplicatively to any of the other objects, on the left or the right.

If in doubt about how to interpret an expression, work through and identify what type of object each symbol represents.

 

[Igael]

@andrewkirk : nice explanation. But, some authors use the dagger much like a generalization of complex conjugate. For example in $\Big(A^\dagger\big\vert f\big\rangle \Big)^\dagger \big\vert g \big\rangle = \big\langle f\big\vert A \big\vert g\big\rangle$