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I'm trying to understand more intuitively what bras and kets are, but some aspects of them remain a mystery to me.

We usually think of $\psi (x)$ as having dimension of $[1/\sqrt{L}]$ so that squaring it and multiplying it with a distance differential would result in a dimensionless quantity. An example of this is:

$$\int_{-\infty}^{\infty} \mid\psi(x)\mid^2 dx= 1$$ for normalized wavefunctions. I also know it is possible to write a ket in position basis as:

$$\mid \psi \rangle = \int_{-\infty}^{\infty} \psi(x) \mid x \rangle dx$$

I would like to believe that $ \mid \psi \rangle$ has no units, it can be represented in the position or momentum base, so it having units doesn't make a whole lot of sense, but this leads me to the conclusion that $\mid x \rangle$ must have the same units as the wavefunction in order to cancel the lenght units of $dx$!

Is this correct? If so, what is its physical interpretation? do position and momentum kets have units after all?

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  • $\begingroup$ I'm open to correction on this, definitely, but to me it's just shorthand, and the units only come in when you do the integration, operation....whatever you use them as placeholders for. $\endgroup$ – user81619 May 25 '15 at 20:33
  • $\begingroup$ Just another point, the kets and bras, for the stuff I have used them for so far, are usually involved in calculating probabilies, so no units there either. Anyway, best of luck getting a definitive answer $\endgroup$ – user81619 May 25 '15 at 21:42
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    $\begingroup$ In short: $|\psi\rangle$ has no units. However, since you want to keep $$\int|x\rangle\langle x|\mathrm dx=1,$$ you end up having to swallow $|x\rangle$ having dimensions $L^{-1/2}$. This is mostly OK, since $|x\rangle$ is not really inside Hilbert space - it is the dual of $\langle x|$, which is a functional from $\mathcal H$ into what you might call $\mathbb C\times L^{-1/2}$. $\endgroup$ – Emilio Pisanty May 26 '15 at 2:09
  • $\begingroup$ @EmilioPisanty: You seem to be assuming that bras and kets have the same dimensions, but as far as I can see (and I said as much in my answer), there's no a priori reason to do that. Am I missing something? $\endgroup$ – Javier May 26 '15 at 3:19
  • $\begingroup$ Mostly an aesthetic sense of symmetry. However, if one (square integrable) ket has a nontrivial dimension, then all kets must have the same dimension, as the superposition principle demands that you be able to add them. This dimension then becomes irrelevant and can be factored away. $\endgroup$ – Emilio Pisanty May 26 '15 at 6:29
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This is a very interesting question. I don't know if there is a general and definitive answer, but I'll try to make some comments. I apologize if this ends up rambling; I'm finding this out as I write this answer.

Operators have dimensions, since their eigenvalues are physical quantities. For bras and kets it gets more complicated. First, you cannot in general say that they are dimensionless. To see why, consider a state with a certain position $|x\rangle$. Since $\langle x | x' \rangle = \delta(x-x')$ and the Dirac delta has the inverse dimension of its argument, it must be that $[ \langle x | ] \times [ | x \rangle ] = 1/L$. A similar relationship holds for momentum eigenstates. Of course, there are higher powers of $L$ in higher dimensions.

However, consider an operator with discrete spectrum, such as the energy in an atom or something like that. Then the appropriate equation is $\langle m | n \rangle = \delta_{mn}$, and since this delta is dimensionless, bras and kets must have inverse dimensions. This gets even weirder when you consider that the Hamiltonian for a hydrogen atom has both discrete and continuous eigenvalues, so the relationship between the bras and the kets' dimensions will be different depending on the energy (or whatever physical quantity is appropiate).

We have the equation $\langle x | p \rangle = \frac1{\sqrt{2\pi\hbar}} \exp(ipx/\hbar)$. I at first thought that this combined with $[\langle x |] \times [| p \rangle ] = [\langle p |] \times [| x \rangle ]$ would allow us to find the dimensions of $|x\rangle$ (and everything else), but it turns out that the normalization conditions of $|x\rangle$ and $|p\rangle$ force the dimensions of $\langle x | p \rangle$ to come out right. We can find that $[|p\rangle] = \sqrt{T/M} [|x \rangle]$, but we can't go any further. Similar relationships will apply for the eigenstates of your favourite operator.

Any given ket is a linear combination of eigenkets, but again there are subtleties depending on whether the spectrum is discrete or continuous. Suppose we have two observables $O_1$ and $O_2$ with discrete spectrum and eigenstates $|n\rangle_1$ and $|n\rangle_2$. Any state $|\psi\rangle$ can be expressed as a dimensionless linear combination of the eigenstates (dimensionless because since $\langle n | n \rangle = 1$, the squares of the coefficientes make up probabilities): $|\psi\rangle = \sum_n a_n |n\rangle_1 = \sum_n b_n |n\rangle_2$. This implies that the eigenkets of all observables with discrete spectrum have the same dimensions, and likewise for the eigenbras.

It gets trickier for observables with continuous spectrum such as $x$ and $p$, because of the integration measure. We have $|\psi\rangle = \int f(x) |x\rangle\ dx = \int g(p) |p\rangle\ dp$. $\langle \psi | \psi \rangle = 1$ implies $\int |f(x)|^2\ dx = 1$, so that $[f] = 1/\sqrt{L}$ and likewise $[g] = \sqrt{T/ML}$. This should be no surprise since $f$ and $g$ are Fourier transforms of each other, with an $1/\sqrt{\hbar}$ thrown in. From this we can deduce $[|p\rangle] = \sqrt{T/M} [|x \rangle]$, which we already knew, and $\sqrt{L} [|x \rangle] = [|n \rangle]$.

The conclusion seems to be the following. All eigenkets with discrete eigenvalues must have the same dimensions, but it looks like that dimension is arbitrary (so you could take them to be dimensionless). Furthermore, normalized states have that same dimension. Eigenstates with continuous spectrum are more complicated; if we have an observable $A$ (with continuous eigenvalues) with eigenvalues $a$, then we can use the fact that $|\psi\rangle$ can be written either as an integral over eigenstates of $A$ or as a sum over discrete eigenstates to find that $\sqrt{[a]} [|a\rangle] = [|n\rangle]$, where $|n\rangle$ is some discrete eigenket. So once you fix the dimensions of one ket, you fix the dimensions of every other ket.

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  • $\begingroup$ Great answer, I would love to see this written somewhere in the literature just to be extra sure, but it makes a lot of intuitive sense to me, thanks! I guess the idea of bras and kets having arbitrary dimensions for a discreet spectrum and some other specific dimensions for a continuous one is kind of strange, but probably it just needs some time to sink in. $\endgroup$ – Ignacio May 26 '15 at 3:24
  • $\begingroup$ @Ignacio: I haven't seen any author discuss this topic (then again, I haven't looked very hard), probably because after all it doesn't really matter very much. Usually in QM we use bras and kets to arrive at some formula which we can compare to experiment, and that formula can be analyzed with regular dimensional analysis. In fact, you could do everything with wavefunctions and row/column vectors and never worry about this. $\endgroup$ – Javier May 26 '15 at 3:27
  • $\begingroup$ Thank you for an amazing answer. Isn't it a bit weird on physical grounds that eigenstates of operators with continuous spectrum have to be dimensionful? In particular, a state is a state--it is not a quantity. What does it even mean to (having to) ascribe a dimension to it? $\endgroup$ – Dvij Mankad Apr 7 at 21:15
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    $\begingroup$ Ok, I think I was too quick to comment. It seems clear that for an eigenstate of a discrete spectrum to be written as linear combination of eigenstates of a continuous spectrum, the latter eigenstates would need to have the dimensions to balance the dimensions of the relevant measure of integration. In particular, such eigenstates can be seen as density of (usual) states as distributed among the continuous eigenvalues of the said continuous spectrum. $\endgroup$ – Dvij Mankad Apr 8 at 2:12
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    $\begingroup$ @Dvij that's a great interpretation, I hadn't thought about that. $\endgroup$ – Javier Apr 8 at 17:48
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Really good question.

Measurements have unit but in quantum mechanics, a measurement is the "evaluation" of an observable on a state (or a state on an observable) something like $$ \left\langle \psi | A | \psi \right\rangle ,\quad \psi\in\mathcal{H},\ A\in\mathcal{B}(\mathcal{H}) \ \text{self-adjoint}$$ A priori, there seem to be an arbitrariness in the choice of unit for $|\psi\rangle$ and $A$ but usually the measurement is a convex combination of the eigenvalues of $A$ (or an integral of a density of probability over the spectrum) so that the eigenvalues really have unit that of the observable (length it the observable is the position, Mass times speed if it is the momentum observable)

If $\left\lVert \psi\right\rVert^2 $ has interpretation of density of probability of presence it has unit $\frac{1}{Volume}$ (or length in 1 dimension) which cancels the $dx$ of integration (as you have written in your first equation). So if $|\psi\rangle$ has no dimension, then $|x\rangle$ indeed has dimension $\frac{1}{\sqrt{L}}$ (L length).

OOOhhhh, big possible confusion coming from quantum field theory (and also the case in second quantization of schrödinger eq. ) when one write the action as a functional of the fields, then these have dimensions, but they are operators and play a role that is more similar to the observable $A$ rather than a state.

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