# What to understand by $\langle \phi | \hat{A}|\psi \rangle$?

What to understand by $$\langle \phi | \hat{A}|\psi \rangle$$? Should it be understood as $$(\langle \phi | \hat{A})|\psi \rangle$$ or as $$\langle \phi | (\hat{A}|\psi \rangle)$$?

EDIT:

I could not understand answer by @mike stone below perhaps because I have not studied functional analysis and know only little bit of linear algebra. So I am rephrasing my question.

What I get from Mike's answer is that $$(\langle \phi | \hat{A})|\psi \rangle$$ and $$\langle \phi | (\hat{A}|\psi \rangle)$$ are the same thing. So my question now is, how can we prove using methods of linear algebra that $$(\langle \phi | \hat{A})|\psi \rangle = \langle \phi | (\hat{A}|\psi \rangle)$$ Perhaps it would get easier answers for me. :)

• "How can we prove using methods of linear algebra..." - Mike Stone did use methods of linear algebra. Dual spaces are very useful and I recommend that you embrace them. Sep 28, 2022 at 9:16
• Yes, Mike did use linear algebea; I don't deny. But his use of both functional analysis and linear algebra, at least mixing of symbols, is something I didn't get as I have pointed in my comments below his answer. Perhaps someone can answer my rephrased question in a different way I may be able to understand.
– t2m
Sep 28, 2022 at 10:33

Both.

Recall that a linear map $$A:V\to W$$ automatically induces a map $$A^*:W^*\to V^*$$ where $$V^*$$ is the dual space of $$V$$.

Given $$f\in W^*$$ we can evaluate $$f(A(x))$$ for any $$x$$ in $$V$$, and so $$f(A(\phantom x))$$ is an element of $$V^*$$ that we may denote by $$A^*(f)$$. Thus, $$A^*(f)(x) = f(A(x)). \qquad (\star)$$

In function analysis the operator $$A^*$$ is usually called the conjugate of $$A$$. (Warning: $$A^*$$ is not the "Hermitian conjugate" of $$A$$. The Hermitian conjugate is a map $$V\to V$$ and needs an inner product for its definition).

The word "conjugate" and the symbol $$A^*$$ is rather unfortunate as it has the potential for generating confusion --- not least because the $$(\ldots)^*$$ map is linear. No complex conjugation is involved. Thus $$(\lambda A+\mu B)^*= \lambda A^*+\mu B^*.$$

Dirac notation deftly sidesteps this notational problem by writing $$\langle {\psi}|A$$ for the action of the conjugate of the operator $$A:V\to V$$ on the bra vector $$\langle {\psi}|\in V^*$$. After setting $$f\to \langle\psi|$$ and $$x \to| \chi\rangle$$, thje equation $$(\star)$$ therefore reads $$\left(\langle{\psi}|A\right)|{\chi}\rangle = \langle{\psi}|\left(A|{\chi}\rangle \right).$$

This shows that it does not matter where we place the parentheses, so Dirac simply drops them and uses one symbol $$\langle{\psi}|{A}|{\chi}\rangle$$ to represent both sides.

Dirac notation thus avoids the non-complex-conjugating "$$*$$" by suppressing the distinction between an operator and its conjugate. If, therefore, for some reason we need to make the distinction, we cannnot use Dirac notation.

• I am unable to follow your argument owing to which $\star$ implies $\left(\langle{\psi}|A\right)|{\chi}\rangle = \langle{\psi}|\left(A|{\chi}\rangle \right)$. Can you please help by elaborating little more? I'm a beginner in QM with elementary understanding of linear algebra.
– t2m
Sep 26, 2022 at 15:05
• @Aarone. It's just a rewriting of the same thing in a different notation. All you need is to make sure that you understand what the definition of $A^*$ is saying. Sep 26, 2022 at 17:59
• You are right, I did not understand the meaning of conjugate $A^*$. By saying $A^*$ as 'conjugate' of $A$, do you mean that they are like dual vectors?
– t2m
Sep 27, 2022 at 15:58
• Unfortunately I couldn't fully appreciate your answer. But I picked up few things from yours and could rephrase my question, which brought answer I am presently able to understand, because of you. Thank you very much! May I'll be able to understand and appreciate your answer some day. :)
– t2m
Sep 28, 2022 at 14:26

The equality $$$$\tag{1} (\langle \phi | A)|\psi \rangle = \langle \phi | (A|\psi \rangle)$$$$ is trivial if we recall how the LHS and the RHS are defined:

• $$(\langle \phi | A)|\psi \rangle$$ is the image of $$|\psi\rangle$$ under the function $$\langle\phi|\circ A$$ and
• $$\langle \phi | (A|\psi \rangle)$$ is the image of $$[$$the image of $$|\psi\rangle$$ under the function $$A]$$ under the function $$\langle\phi|$$.

In other words, $$(1)$$ is simply the definition of the composition of two functions. Mike Stone simply pointed out that $$\langle\psi|\circ A$$ equals the image of $$\langle\psi|$$ under the dual of $$A$$...

• May be Mike's answer is more rigours, your heuristic approach is what I could understand. I'm, therefore, accepting your answer. You should in turn 'accept' Mike's. ;)
– t2m
Sep 28, 2022 at 14:20
• @Aarone Thank you :) I want to emphasize that my approach is not heuristic at all and that it is equivalent to Mike's answers (see the last sentence in my answer). Btw, to be precise we have to distinguish between the algebraic dual space and the continuous dual space: Given an Hilbert space $H$, the function$$H\ni |\psi\rangle\mapsto\langle\psi|$$is a bijection onto its continuous dual space. Sep 28, 2022 at 14:40
• '... my approach is not heuristic at all and that it is equivalent to Mike's answers'. In that case, it's even better! :)
– t2m
Sep 28, 2022 at 14:48
• @Aarone I am glad to hear that! Here is a comment to see why the distinction is relevant: Since $A$ is linear and the composition of linear functions is linear, the function $\langle\phi|\circ A$ is an element of the algebraic dual (i.e. linear), but not necessarely in the continuous dual. If you require that $A$ is additionally continuous, then $\langle\phi|\circ A$ is also continuous (i.e. an element of the continuous dual), because the composition of continuous functions is continuous. Sep 28, 2022 at 14:53