Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 5 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.

share|improve this answer
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 $\phi = \hat{A}_F\psi$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.