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Let $H$ be a hilbert space and let $\hat{A}$ be a linear operator on $H$.

My textbook states that $|\hat{A} \psi\rangle = \hat{A} |\psi\rangle$. My understanding of bra-kets is that $|\psi\rangle$ is a member of $H$ and that $\psi$ alone isn't defined to be anything, so $|\hat{A}\psi\rangle$ isn't defined.

Is $|\hat{A} \psi\rangle = \hat{A} |\psi\rangle$ just a notation or is there something deeper that I am missing?

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A substantive answer is impossible without a broader context. – Incnis Mrsi Oct 22 '14 at 6:19
up vote 6 down vote accepted

This should be understood as a mere definition, i.e. a new label for the state you get when you apply the operator A to the ket psi.

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Yes, it is a notation...and not a particularly good notation if you ask me. (This question is proof!) – Steve B Feb 2 '12 at 18:44

I'm not 100% sure about this, but I believe this notation stems from the practice of treating $\psi$ as a wavefunction. In a typical introductory quantum mechanics class, the theory is expressed in terms of continuous functions $\psi(x)$ in the space $\mathbb{R}\to\mathbb{C}$, and operators that act on these functions to produce other functions. So intro-level QM students get used to thinking of $\psi(x)$, not $|\psi\rangle$, as the fundamental object in the theory.

Given this viewpoint, when Dirac notation is introduced, $|\psi\rangle$ looks like nothing more than a convenient notation for $\psi(x)$. In other words, a beginner to quantum mechanics interprets the thing inside the ket as a function, not a label. At this point in the class, the action of an operator on a function is well defined, whereas the action of an operator on the ket (an abstract object) is not. I would guess that your textbook is trying to define the action of an operator on a ket by relating it to the action of the operator on the function, something which can easily be understood by a student who is used to thinking in terms of wavefunctions.

To frame it more precisely: let $F$ be the (Hilbert) space of all possible wavefunctions $\psi:\mathbb{R}\to\mathbb{C}$, and $K$ be the (Hilbert) space of all possible kets $|\psi\rangle$. Then let $\mathcal{O}_F$ be the space of all operators acting on $F$, and let $\mathcal{O}_K$ be the space of all operators acting on $K$. At this point in the book, I'm guessing $F$ and $\mathcal{O}_F$ have already been discussed, $K$ is relatively new, and $\mathcal{O}_K$ has literally just been introduced. It's known from the given definition of the ket that every wavefunction corresponds to a ket, i.e. that there is a mapping $F\to K$ where the image of $\psi$ is written $|\psi\rangle$. But in order to work with kets the same way you work with wavefunctions, you also need a way to obtain the ket-operator corresponding to any given function-operator, i.e. a mapping $\mathcal{O}_F\to\mathcal{O}_K$. The book defines that mapping as follows: any given operator $\hat{A}_F\in\mathcal{O}_F$ maps to the operator $\hat{A}_K\in\mathcal{O}_K$ which satisfies $\hat{A}_K|\psi\rangle \equiv |\phi\rangle$, where $\phi = \hat{A}_F\psi$.

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Continuous functions do not form a Hilbert space but on finite discrete spaces. Although this case is the most intuitively appealing one for understanding bra and ket, ironically, it is not the case of ℝ. Could you fix the ambiguity? – Incnis Mrsi Oct 21 '14 at 19:44
@IncnisMrsi I don't see the ambiguity you're referring to. Could you clarify your objection? – David Z Oct 21 '14 at 21:33
Rubbish is “To frame it more precisely: let $F$ be the (Hilbert) space of all possible wavefunctions $\psi:\mathbb{R}\to\mathbb{C}$…”. What mean parentheses? Is David Z not sure that all continuous fucntions (see the first paragraph) form a Hilbert space? Is he not sure that “introductory quantum mechanics” textbooks, operating with continuous functions on ℝ, are on good terms with mathematical rigour of “Hilbert space”? Both statements are false. – Incnis Mrsi Oct 22 '14 at 5:54
@IncnisMrsi I'm having a very difficult time understanding your writing. There are many things that are not clear about your comments. (Who are you addressing?) If you're asking about the parentheses, that simply means that $F$ is a Hilbert space, but the fact that it is a Hilbert space (as opposed to just a vector space) is not relevant to the rest of the argument. I still don't see any evidence of any ambiguity that needs to be fixed. – David Z Oct 22 '14 at 17:16
@IncnisMrsi If you are referring to the equivalence class thing: Please stop being pedantic. In fact, Reed & Simon, which is somewhat of a standard work in this topic, defines $L^p$ as the space of equivalence classes, so we should identify elements of $L^p$ as functions, whereas the `primitive' space is denoted by $\mathscr L^p$. Furthermore, your condescending tone is starting to get to me, so it may be that I stop responding soon, so we may both happily continue our lives. – Danu Oct 22 '14 at 19:19

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