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From what I currently understand given a general state vector $|\psi\rangle$ the wave function: $$\psi(x) = \langle x|\psi\rangle$$ represent the vector $|\psi\rangle$ in the base of the eigenvalues of the position operator. Similarly the wave function $$\psi(p)=\langle p|\psi\rangle$$ represent the same vector but in the base of momentum. In practice we can think of wave functions as column vectors with an infinite number of entries, one for every real number.

So when we write $|\psi\rangle$ do we mean to represent the abstract vector $|\psi\rangle$ without referring to a specific base? Why do we do this? In friendly 3D linear algebra we almost always think of vectors in the context of a specific representation of them in some base. Wouldn't be easier to always represent states vectors in some specific base, so as wave functions? I am saying this because using this double way of representing vectors sometimes tends to make things confusing; for example: in QM lectures happens often that a certain operator is described as acting on ket vectors: $$A|\psi\rangle$$ and then after a bit the same operator, without any further explanation, is shown as acting on functions: $$A\psi(x)$$ But there are some operations that make sense only if applied on functions and not on ket vectors. Why do we represent things in such a way? Why don't we only use wave function representation of vectors in some specific base?

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  • $\begingroup$ Wavefunctions $\psi(x)$ represent the components of the vectors $|\psi \rangle$. When we are using on operator acting on $|\psi \rangle$, one can use the same operator but in a different representation on $\psi(x)$. For example the momentum operator $-i \hbar \frac{d}{dx}$ is acting on $\psi(x)$, but it is a special representation of the general operator $\hat{p}_x$ acting on $|\psi \rangle$. $\endgroup$ Sep 4, 2020 at 16:14
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    $\begingroup$ I disagree with your assertion that choosing a representation of a vector is easier. In fact, I try to avoid it as much as I can. The specific representation is superfluous information, and superfluous things are superfluous at best, confusing at worst. $\endgroup$ Sep 4, 2020 at 16:15
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    $\begingroup$ I would like to answer this, but I am reluctant to spend time on it because I wrote a pretty extensive answer on your other question that has not yet been resolved. I offered to follow up on your confusions in chat, and I would be happy to improve the answer to the point that it can be accepted. Please let me know if you would be interested in that. $\endgroup$
    – DanielSank
    Sep 4, 2020 at 16:19
  • $\begingroup$ I will say one thing though: this whole issue has absolutely nothing to do with quantum mechanics. This whole issue is just a question of whether we're working in a specific basis or with a basis-independent representation of the vectors. Notations such as $A \psi(x)$ are not consistent and shouldn't be used. I'll write a more comprehensive answer once we sort out your previous question. $\endgroup$
    – DanielSank
    Sep 4, 2020 at 16:20
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    $\begingroup$ In friendly 3D linear algebra we almost always think of vectors in the context of a specific representation of them in some base. Didn’t you ever abstractly write a matrix as $\mathbf{M}$ and a vector as $\mathbf{v}$? $\endgroup$
    – G. Smith
    Sep 4, 2020 at 16:43

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In QM lectures happens often that a certain operator is described as acting on ket vectors $A|\psi\rangle$ and then after a bit the same operator, without any further explanation, is shown as acting on functions $A\psi(x)$.

This is not correct. You may have seen it somewhere, but the author was being sloppy or abusing notation.


Let $|\psi\rangle$ be an abstract ket vector. If we wish to represent it in the continuous position basis, we can insert the identity operator $\mathbb 1=\int|x\rangle\langle x| dx$ and obtain

$$|\psi\rangle = \int|x\rangle\langle x|\psi\rangle dx = \int |x\rangle \psi(x) dx$$ Loosely speaking, $\psi(x)$ is the component of $|\psi\rangle$ along the basis vector $|x\rangle$. If you want to think of something as an infinitely long column vector, then it should be $|\psi\rangle$, not $\psi(x)$ (which is just a complex number).

Similarly, if $A$ is an abstract operator, then we can let it act on abstract kets $|\psi\rangle$ as $A|\psi\rangle$. Expanding $|\psi\rangle$ out in the position basis, we find

$$A|\psi \rangle = \int A|x\rangle \psi(x) dx$$

$A$ is still an abstract operator which acts on a ket (in this case, $|x\rangle$), not a function. If we insert another identity operator $\int |y\rangle\langle y| dy$, we find

$$A|\psi\rangle = \iint |y\rangle\langle y|A|x\rangle \psi(x) dy \ dx$$

The object $\langle y|A|x\rangle \equiv A_{yx}$ is the $yx$ component of the abstract operator $A$. This object is what acts on functions. The result is that

$$A|\psi\rangle = \iint |y\rangle A_{yx} \psi(x) dy\ dx$$

For example, the position operator $Q$ has components $Q_{yx} \equiv \langle y|Q|x\rangle = \delta(y-x) \cdot x$ while the momentum operator has components $P_{yx} \equiv \langle y |P|x\rangle = -i\hbar \delta(y-x) \frac{d}{dx}$. We would therefore have

$$Q|\psi\rangle = \int |x\rangle x\cdot \psi(x) dx$$ $$P|\psi\rangle = \int |x\rangle (-i\hbar)\psi'(x) dx$$


If we are being very strict, we would say that the position operator $Q$ eats a ket with position-space wavefunction $\psi(x)$ and spits out a ket with position-space wavefunction $x\psi(x)$. However, we often relax a bit and say that $Q$ eats a wavefunction $\psi(x)$ and spits out $x\psi(x)$.

The reason we use kets in the first place is that it can be quite convenient to not restrict yourself to a particular basis. I find it very difficult to believe that you've never used the vector notation $\vec r$ as opposed to the index notation $r_i$, and this is precisely the same thing. The only difference is that the index $i$ in $r_i$ runs over $\{1,2,3\}$, while the index $x$ in $\psi(x)$ runs over $\mathbb R$.

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  • $\begingroup$ I think that your answer may create a lot of confusion. Don't misunderstand me, I think your answer is completely correct but I have a problem with it. $\phi(x)$ is a number if $x$ is fixed, however we often interprete $\phi(x)$ not as a number but as a function, functions indeed form a complex vector space and can be thought as a vector with infinite entries. All you answer is based on the notion that $\phi(x)$ is a number in my question, but I intended to say $\phi(x)$ as a function. I think that this would be better to use instead of $|\phi\rangle$. Do you get my point? If not let me know. $\endgroup$
    – Noumeno
    Sep 6, 2020 at 17:16
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    $\begingroup$ @Noumeno I think you are misunderstanding my answer. You are free to work exclusively in the position representation if you wish, but if you want to use bra-ket notation, then you need to understand that there's a difference between an abstract operator (which acts on abstract kets) and the position-space representation of the operator (which acts on wavefunctions). $\endgroup$
    – J. Murray
    Sep 6, 2020 at 17:30
  • $\begingroup$ Yes, I understood that, and I think that you are completely right. But I also think that we can think of function as complex vectors in infinite dimension, in your answer you seem to imply that we can only think this about ket vectors, also I find that $f(x)$ is not a number as you state, because it can be a number if $x$ is fixed, but it can also represent a function if $x$ is not fixed. So your answer has some problematic parts, do you agree or am I missing something regarding this? But the main message of the answer is correct in my opinion of course. $\endgroup$
    – Noumeno
    Sep 6, 2020 at 17:40
  • $\begingroup$ @Noumeno Either you're working in bra-ket notation or you aren't. If you are, then $\phi(x)$ should be considered to be the component of your vector along the basis vector $|x\rangle$. If you are not working in bra-ket notation, then the vector is the function $\phi$. It seems to me that you're trying to mix both conventions, which is not a good idea. $\endgroup$
    – J. Murray
    Sep 6, 2020 at 17:48
  • $\begingroup$ I am trying to argue that the second convention is the better one, anyway I get your point. $\endgroup$
    – Noumeno
    Sep 6, 2020 at 19:51

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