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In QM, the state vector $|\psi\rangle$ seem to have various dimensions under different representations: (only in space of continuous dimension)

$$\langle x|\psi\rangle = [\frac{1}{\sqrt{Length}}]$$ $$\langle p|\psi\rangle = [\sqrt\frac{time}{{Mass Length}}]$$ etc, and $|\psi\rangle$ can equal to $|x'\rangle$ or $|p'\rangle$ in special cases (System in an eigenstate of a certain observable), therefore

$$\langle x|\psi\rangle = \langle x|x'\rangle = [\frac{1}{L}]$$

So what is the physical dimension of $|\psi\rangle$? Is this a valid question?

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up vote 8 down vote accepted

The dimensionality of $| \psi \rangle$ doesn't have any physical significance (though you can formally set it to be whatever you want). The space of states is not the Hilbert space of the theory itself, but the projective Hilbert space. That is to say that two nonzero vectors in the Hilbert space $| \psi \rangle, | \psi' \rangle$ define the same physical state if $| \psi \rangle = c | \psi' \rangle$ for some nonzero (complex) scalar $c$.

When we talk about dimensionality in physics, it's talking about how quantities scale when we change the units that we use to describe the problem. That is to say, for example, if under a change from kilometers to meters (keeping all other units the same), some quantity $q$ changes numerically as $q \rightarrow 10^9 q$, then $q$ has dimensions of volume since it scales as length cubed.

Under the same change, how $| \psi \rangle$ changes is not a measurable quantity. It could change to $| \psi \rangle$ or $10^9 | \psi \rangle$ or even $ i| \psi \rangle$, but all of these are exactly the same physical state. So in quantum mechanics, the dimensionality of $| \psi \rangle$ is a formally meaningless quantity. It has no physical significance. There is no experiment that can measure it even in principle. You could choose a particular convention, but whatever choice you make has absolutely no impact on physics. If you are going to choose a dimension, the most natural one is dimensionless.

With that said, it is often conventional to adopt some normalization condition for $| \psi \rangle$. This is done to facilitate explicit calculation, though it's not exactly physical. Such conditions can lead to some constraints on dimensionalities. One example is $\langle \psi | \psi \rangle = 1$. This requires $[\langle \psi |] = - [| \psi \rangle ] = -[D]$ where $[ \cdot ]$ denotes the total dimensionality (some might argue that this is true by definition of the dual space, but I won't go that far) and $[D]$ is just some dimensionality. If we require the 1-dimensional wavefunctions $\psi^*(x) = \langle \psi | x \rangle$ and $\psi(x) = \langle x | \psi \rangle$ to satisfy $\int \psi^*(x) \psi(x) dx =1$ then this gives us conditions for the position basis and dual basis, namely that $[\psi(x)] = [\psi^*(x)] = [L^{-1/2}]$, where $[L]$ denotes length. This gives $[|x \rangle] =-\frac12[L]+[D]$ and $[\langle x |] = -\frac12[L]-[D]$. We still have this $[D]$ floating around though, which is totally arbitrary, and the only way to get rid of it is to just pick some dimensionality for one of these quantities (or impose some equivalent criterion, such as requiring that $[\langle \psi |] = [| \psi \rangle ]$, giving $[D]=0$). No physical argument can give us a dimensionality $[D]$ for the state vector itself, as it has no physical meaning.

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Thank you for the crystal clear answer! Just one minor typo to point out $\psi^*(x) = \langle x | \psi \rangle$ should be $\psi(x) = \langle x | \psi \rangle$, and I tried to correct it but it's less than 6 characters... – REX Sep 8 '13 at 12:07
@REX Thanks for catching that. I've now corrected it. I make silly typos like that all the time. – Logan M Sep 8 '13 at 23:53

In quantum mechanics any physical state $ \psi $ satisfies a normalisation condition (in momentum space, or coordinate space or some other), from this normalisation condition it is easy to see what the physical dimensions are in each context. The coordinate wave function has always dimensions $1/\sqrt{V}$ for a volume $V$.

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Thanks for answering, but I wonder if there is a dimension for $|\psi>$ that is INDEPENDENT of contexts? – REX Sep 8 '13 at 0:34

First of all, the state |ψ> does not have any dimension. You are correct to point out that these constants only appears when one ore more of the observables of the complete commuting set take on a continuous range of possible values. Normalization constants and weight functions are used to properly construct the identity operator for a given orthogonal eigenbasis in terms of an indiscrete sum.

Let me take 2 cases to illustrate. Let p be an observable and form a complete commuting set all on its own.

Case 1:

p takes on only integral values like 1,-5 etc. So Σ|p>< p| = I, where I is the identity operator sum is taken over all integers. In this case you can see that there will obviously have no dimension. So we don't need constants.

Case 2:

p takes on all real numbers between a and b. The sum of all values of a continuous function in [a,b] has infinitely dense number of terms and is thus proportional to ∫ab fdx, regardless of whether the summation value diverges or not, it is proportional to the integral.

Therefore ∫ab |p> dp*W < p| = I. In some cases with more than one observable, the terms of the the indiscrete sum need to be weighted to get consistency across all co-ordinate systems chosen which is why we have the weight function W(p).

As <ψ|ψ> = 1 due to condition of normalization, putting I in between should give the same result, we can determine the constant to be present in W from this condition. dp*W(p) is thus effectively the 'volume' element dV divided by V. For convenience we multiply the square root of 1/V into , in which case we don't need to explicitly write it out in the integral.

You might also ask what if the limits of p tend to infinity and the integral diverges? In that case V diverges, so we cannot evaluate it.

In a nutshell, there is no physical dimension for |ψ> at all.

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