# Is the wave function the Radon-Nikodym derivative of a complex measure?

I read somewhere (latest version of a webcomic to be honest) that "superposition" means:

a complex linear combination of a $0$ state and a $1$ state... Quantum mechanics is just a certain generalization of probability... Quantum mechanics has probability amplitudes, and they can be positive or negative or even complex.

Question: Does this mean that quantum wave functions are the Radon-Nikodym derivatives of (normalized) complex measures? In particular, when the complex measure has no atoms (in the sense of measure theory), then is the wave function the density of a complex-valued measure?

In particular, that would seem to imply a certain analogy between the (time-dependent) Schrodinger equation, which would then involve the time derivative of the density of a complex-valued measure, and the Fokker-Planck equation, which involves the time derivative of the density of a (real-valued) probability measure.

(Just checking Wikipedia's article about the Fokker-Planck equation now, and apparently there is a formal analogy between the two equations -- is the analogy what I've described above?)

Discussion: Wikipedia's article on probability amplitude mentions that the complex modulus of the probability amplitude is the Radon-Nikodym derivative of a probability measure, i.e. $$\int_X | \psi |^2 d\mu = 1\,,$$ but that is not my question -- what I am asking about is $\psi$ itself (i.e. not $|\psi|^2$), and whether $\psi$ is the "Radon-Nikodym derivative" of a complex measure.

The Wikipedia page for the Radon-Nikodym theorem says that the theorem generalizes even to complex-valued measures, and apprently all Hilbert spaces have the Radon-Nikodym property.

Also please note that when I say a "normalized" complex measure, I just mean a complex measure whose variation is a probability measure. So perhaps another way to state my question is:

Does the fact that $|\psi|^2$ is a Radon-Nikodym derivative of a probability measure imply that $\psi$ is the Radon-Nikodym derivative of a complex measure whose variation is the probability measure defined by $|\psi|^2$? Is it at least the Radon-Nikodym derivative of some complex measure?

Note: I decided to major in math, not physics, so although I know some basic Lagrangian and Hamiltonian mechanics, I am very ignorant of quantum mechanics, but do know a lot of relevant math for the subject (e.g. functional analysis), so I would prefer answers which more heavily emphasize mathematical formulation of the situation and not physical intuition.

Let me know if this should be migrated to Math.SE -- since what motivated the question was my reaction to an attempt to describe physical intuition, I imagine it might not be well-received there.

Related but different question: Probability amplitude in Layman's Terms

• Quantum states are noncommutative probabilities, and not classical probabilities; so I am not sure that your question makes complete sense. – yuggib Dec 14 '16 at 17:46
• @yuggib Are you saying that $|\psi|^2$ is not a probability density? I don't understand what you mean (non-commutative probabilities, quantum states, classical probabilities), so I have no idea if it's relevant here or not. In any case, the condition $\int |\psi|^2 d \mu=1$ looks like a probability density to me (since $|\psi|^2 \ge 0$). – Chill2Macht Dec 14 '16 at 17:59
• If $\text d\nu = \psi \text d ^3{\mathbf r}$, then $\text d \vert \nu \vert = \vert \psi \vert \text d ^3 \mathbf r$. This is not a probability measure ($\vert \psi \vert ^2 \text d^ 3 \mathbf r$ is). – pppqqq Dec 14 '16 at 18:03
• @pppqqq This is a good point (I was hoping the square wouldn't come up as an issue, but oh well). Then at least is $\nu$ still a complex measure, even if $|\nu|$ isn't a probability measure (as I had hoped)? That would still be interesting to me, and make it easier for me to understand $\psi$ (mathematically). – Chill2Macht Dec 14 '16 at 18:06
• Well, $\nu$ is a well defined complex measure if (and only if) $\psi \in L^1$. However, in quantum mechanics, we do not require $\psi$ to be $L^1$ (and I honestly don't know if there's any use of doing this, nor to look at $\psi$ as a measure). The point of your first quote isn't an interpretation of $\psi$ as a measure, it is the fact that if the state $\psi=\psi _1 + \psi _2$, to compute the density at $x$ you can't simply add the separate densities $\vert \psi _1 \vert ^2$ and $\vert \psi _2 \vert ^2$, because the two amplitudes interfere. – pppqqq Dec 14 '16 at 18:08

The wavefunction $\psi$ of a nonrelativistic spinless particle is not to be interpreted as the Radon-Nikodym derivative of an absolutely continuous complex measure (i.e. a measure of the form $\text d \nu = \psi \text d^3 \mathbf x$).

First of all, in general $\psi \notin L^1(\mathbb R ^3)$, so that mathematically $\text d \nu = \psi \text d^3 \mathbf x$ doesn't make any sense.

On the other hand, we require $\psi \in L^2(\mathbb R ^3)$, so that $\vert \psi \vert ^2$ could be interpreted as a probability density in position space. Indeed, if we know that the particle is in the state $\psi$, and we want to compute, say, the expectation value of its position $\mathbf x$, we can compute it as an ordinary mathematical expectation (here and henceforth I assume that every integral is well defined): $$\langle \mathbf x \rangle =\intop \mathbf x \vert \psi (\mathbf x)\vert ^2\text d^3\mathbf x.$$

However, there are observables that are not of the form $F(\mathbf x)$. One of these is the momentum $\mathbf p = -i\hbar \frac{\partial}{\partial \mathbf x}$; to compute, it's expectation value, we need the whole $\psi$ (not just its modulus):$$\langle \mathbf p \rangle = \intop \psi^*(\mathbf x )( -i\hbar \frac{\partial}{\partial \mathbf x} )\psi(\mathbf x)\,\text d^3 \mathbf x.$$ We can perform a quantum Fourier transform $$\phi(\mathbf p )= \intop\frac{\text d ^3 \mathbf x}{(2\pi \hbar)^{\frac{3}{2}}}e^{-i\frac{\mathbf p \cdot \mathbf x}{\hbar}}\psi (\mathbf x)$$ so that the above expectation value becomes $$\langle \mathbf p \rangle = \intop \mathbf p \vert \phi(\mathbf p) \vert^2\text d^3 \mathbf p.$$ Note that, in this representation: $$\langle \mathbf x \rangle = \intop \phi ^* (\mathbf p ) (i\hbar \frac{\partial}{\partial \mathbf p})\phi (\mathbf p ) \,\text d ^3 \mathbf p$$

Moral. The state $\psi$ is not a complex measure and the square $\vert \psi \vert ^2$ is not a probability measure. This is ultimately rooted in the fact that there are propositions, like "the particle has position $\mathbf x$ and momentum $\mathbf p$" which simply don't make sense in quantum mechanics. This is also related to quantum interference, but I think I'm going a bit off-topic. I suggest you to read Feynman, if you want to get a better idea of why a quantum state is described by a probability amplitude and not a probability distribution.

• Dumb follow-up: so are we usually dealing with infinite measure spaces in regards to wave functions/quantum mechanics? Since square integrable implies integrable (by Cauchy-Schwarz) for finite measure spaces? (I think) – Chill2Macht Dec 15 '16 at 12:53
• Measure spaces are not really central in quantum mechanics. The general setting is an abstract Hilbert space. For example, the position space wave function $\psi\in L^2(\text d^3 \mathbf x)$ can be seen as a particular representation of a vector $\vert s \rangle$ in an abstract Hilbert space $\mathcal H$ (the momentum space wave function $\phi\in L^2 (\text d ^3 \mathbf p)$ is another representation; all these representation are unitarily equivalent). On the other hand, probability measures definitely play a role in quantum theory [continued] – pppqqq Dec 15 '16 at 13:03
• The link is provided by the spectral theorem, which allows us to assign to each self-adjoint operator (an observable) $A$ on $\mathcal H$ and each state $\psi\in \mathcal H$ a Borel probability measure $\text d P_{A,\psi}$ on $\mathbb R$ which, (very) loosely speaking, gives the probability distribution of the observable $A$ in the state $\psi$. Since you are interested in the mathematical setting, you might want to give a look to the (very lucid, IMO) presentation of QM given in Folland G.B., "Quantum Field Theory". – pppqqq Dec 15 '16 at 13:09