Title: Is $∣1 \rangle$ an abuse of notation?
Tags: quantum-mechanics;soft-question;hilbert-space;notation
Question: In introductory quantum mechanics it is always said that $∣ \rangle$ is nothing but a notation. For example, we can denote the state $\vec \psi$ as $∣\psi \rangle$. In other words, the little arrow has transformed into a ket.
But when you look up material online, it seems that the usage of the bra-ket is much more free. Example of this usage: http://physics.gu.se/~klavs/FYP310/braket.pdf pg 17
A harmonic oscillator with precisely three quanta of vibrations is
described as $|3\rangle$., where it is understood that in this case we
are looking at a harmonic oscillator with some given frequency ω, say.
Because the state is specified with respect to the energy we can
easily find the energy by application of the Hamiltonian operator on
this state, H$|3\rangle$. = (3 + 1/2)$\omega h/2\pi |3 \rangle$.
What is the meaning of 3 in this case? Is 3 a vector? A scalar? If we treat the ket symbol as a vector, then $\vec 3$ is something that does not make sense.
Can someone clarify what it means for a scalar to be in a ket?
Comment: @IllegalImmigrant You're imposing unnecessarily harsh restrictions on the use of symbols as labels.  If I write $|\mathrm{stuff}\rangle$, where $\mathrm{stuff}$ can be any symbols whatsoever you can write, then $\mathrm{stuff}$ is simply being used as a label.  It could be a number, a sequence of numbers, a happy face, a portrait of your grandmother...

Comment: @JamalS I know :(.  Teaching has been taking up all of my time.  Hopefully I can add more soon.

Comment: @joshphysics Hey, how's phermi coming along? I've been checking daily and there are no new problems.

Comment: $|\rangle$ = arrow then $\vec {scalar}$ doesn't make sense

Comment: Okay, I thought the ket is identically equivalent to the vector symbol.

Comment: It can be a vector OR a function. Or honestly, anything that makes sense. The Ket notation seems to be more mathematical than anything else. As long as the inner product is defined for what you're representing with the Ket, it should be fine. i.e. As long as there is a bra, for your ket and you can do something like  to get a magnitude there can be anything in there. (First year tho so pinch of salt)

Comment: https://www.youtube.com/watch?v=fIYIFCVICcA&list=PLdgVBOaXkb9Bv466YnyxslT4gIlSZdtjw&index=4 cleared up a lot for me.

Comment: What about the definition given is unclear to you? It says specifically that $\lvert 3 \rangle$ shall denote the HO state with three excited quanta, i.e. $\lvert 3 \rangle = (a^\dagger)^3\lvert \text{vacuum} \rangle$.

Here is the accepted answer: What is the meaning of 3 in this case?
In this case, the character "3" is a convenient, descriptive label for the state with three quanta present.
It is often the case that an eigenstate is labelled with its associated eigenvalue.
In the harmonic oscillator case, the number operator commutes with the energy operator (Hamiltonian) so a number eigenstate is also an energy eigenstate.
Thus, the state with three quanta present satisfies
$$\hat N |3\rangle = 3\,|3\rangle$$
But, it also satisfies
$$\hat H |3\rangle = (3 + \frac{1}{2})\hbar \omega\, |3\rangle = \frac{7}{2} \hbar \omega\,|3\rangle$$
So we would be justified in labelling this state as
$$|\frac{7}{2} \hbar \omega\rangle $$
though that's not typical.

Here is another answer: In introductory quantum mechanics it is always said that $∣ \rangle$ is nothing but a notation. For example, we can denote the state $\vec \psi$ as $∣\psi \rangle$. In other words, the little arrow has transformed into a ket.
In Euclidean $n$-space, if I have a vector $\mathbf{v}$, I can decompose it with respect to some orthonormal basis $\{\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_n\}$:
$$\mathbf{v} = \sum_k \mathbf{e}_k(\mathbf{e}_k\cdot\mathbf{v})\text{,}$$
where the numbers $(\mathbf{e}_k\cdot\mathbf{v})$ are the components of $\mathbf{v}$ with respect to a given basis. If instead of $\mathbf{e}_k$, I write $|k\rangle$ and use bra-ket notation for the dot product, this turns into
$$|v\rangle = \sum_k |k\rangle\langle k|v\rangle\text{.}$$
So yes, this is just notation, but not in the sense of "turn little arrows into kets."
Can someone clarify what it means for a scalar to be in a ket?
Generally, it's simply a label, though commonly, the meaning is more specific: it means that we have a basis indexed by scalars and we're picking the one corresponding to the particular scalar.
In the typical case, we're talking about some particular observable and are labeling its eigenstates by the corresponding eigenvalues... which is exactly what's going on in your quote (except shifted over a bit): the energy eigenstates form an orthonormal basis, and we're labeling vectors in that basis.
If you think about what a wavefunction $\psi(x)$ means (assume one dimension), then you will realize it is actually a representation of a state in a particular basis, the position basis:
$$\psi(x) = \langle x|\psi\rangle\text{,}$$
where $|x\rangle$ means the state of definite position $x$. It is exactly the same thing, with a basis indexed by scalars corresponding to the position, and it is implicit in every use of a one-dimensional wavefunction. So we can also write
$$|\psi\rangle = \int |x\rangle\langle x|\psi\rangle\,\mathrm{d}x\text{,}$$
and similarly for other observables, though we have to watch out for degeneracy.

Here is another answer: The notation $\lvert \rangle$ is meant to imply that $\lvert \text{anything here you want to put here} \rangle$ is a vector in a Hilbert space.
If you have got some wavefunction $\psi(x)$, then you often denote the abstract vector (instead of the concrete realisation in a basis like $\psi(x)$) it represents by $\lvert \psi \rangle$.
If you have got only a 2D space on which spin operators live, then you denote the two eigenstates of one of them by $\lvert \uparrow \rangle$ and $\lvert \downarrow \rangle$.
Whatever you put between the $\lvert$ and the $\rangle$ is just a label that should uniquely identify the vector $\lvert \text{something} \rangle$ is supposed to be.
Comment for this answer: In case of  $\lvert \uparrow \rangle$  it is clear because $\uparrow$ is defined as a vector [1 0]. But in the case of  $\lvert 3 \rangle$ it is unclear what 3 is equivalent to.
Comment for this answer: There is nothing wrong with $|3\rangle$ for problems where some set of states can be labeled with integers and you have established which set you mean and how the labels are assigned. At that point it "uniquely identifies the vector". For instance the time-independant states of the harmonic oscillator.
Comment for this answer: @IllegalImmigrant switch basis, and $\left|\uparrow\right\rangle$ is no longer $[1\;0]$.

Here is another answer: It is just a label. More conventional notation uses indices for the same purpose, but the latter gets unwieldy if you need more elaborate qualifiers.
One particular application is labeling states by occupation number (cf second quantization).

Here is another answer: What they're saying is that $|3\rangle$ represents the third energy eigenstate of the oscillator.  So, it replaces something like $\psi_3$.
Writing $|3\rangle$ requires context - you would have to explain that you were going to number the nth energy eigenstate of the harmonic oscillator as $|n\rangle$ before using that notation.  It's not an abuse of notation, it's just not very self-descriptive.
You could use this notation too - the nth energy eigenstate of the harmonic oscillator is $|N_{energy}^{harm.~osc.} = 3\rangle$, but it would be pretty tedious to write.
Comment for this answer: @AlfredCentauri As someone who codes in C and works with relativity, I see nothing wrong with counting "zeroth," "first," "second," ... :)
Comment for this answer: @ChrisWhite From Conway and Guy's *Book of Numbers*: Wacław Sierpiński, the great Polish mathematician... was worried that he'd lost one piece of his luggage. "No, dear!" said his wife. "All six pieces are here." "That can't be true," said Sierpiński, "I've counted them several times: zero, one, two, three, four, five."
Comment for this answer: Isn't $|3\rangle$ the *fourth* energy eigenstate?  Remember, the ground state has zero quanta.

Here is another answer: Representing the third excited energy state by the symbol $\left\vert 3 \right\rangle$ is both (i) an established convention in quantum mechanics and (ii) explicitly defined in your book. Using numbers as labels is unambigous in the given context, so no, I wouldn't call it an abuse of notation.
As for your comment about the notation $\vec 3$ not making sense: although unconventional, there would be nothing wrong with defining such a notation if you felt like it. If a textbook on linear algebra decided to name the Cartesian unit vectors $\vec 1 \equiv (1,0,0)$, $\vec 2 \equiv (0,1,0)$, $\vec 3 \equiv (0,0,1)$ instead of $\hat{\mathbf x}$, $\hat{\mathbf y}$, $\hat{\mathbf z}$ or $\hat{\mathbf e}_x$, $\hat{\mathbf e}_y$, $\hat{\mathbf e}_z$, then nothing of importance would change. As long as you explicitly define your notation, you can express the mathematics itself in whatever notation you prefer.
With that said, I personally prefer the more verbose notation $\left\vert n=3 \right\rangle$ for ket vectors (at least in final results), since it specifically points out that it is the energy quantum number $n$ that is equal to three. This avoids confusion with the similar notations for momentum eigenstates $\left\vert p\right\rangle$, position eigenstates $\left\vert x \right\rangle$, and so on. It also generalizes nicely to systems with more quantum numbers, since it makes it easy to distinguish different representations of the same state space (like $\left\vert j_1, j_2, m_1, m_2 \right\rangle$ and $\left\vert j_1, j_2, j, m\right\rangle$ for the angular momentum of a composite system). There are also other conventions in the literature; for example, some authors use the notation $\left\vert \psi_3 \right\rangle$ or $\left\vert \phi_3 \right\rangle$ for the third excited energy state.

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