1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Quantum states are noncommutative probabilities, and not classical probabilities; so I am not sure that your question makes complete sense. $\endgroup$ – yuggib Dec 14 '16 at 17:46
  • $\begingroup$ @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 $|\psi|^2 \ge 0$). $\endgroup$ – Chill2Macht Dec 14 '16 at 17:59
  • 1
    $\begingroup$ 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). $\endgroup$ – pppqqq Dec 14 '16 at 18:03
  • $\begingroup$ @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). $\endgroup$ – Chill2Macht Dec 14 '16 at 18:06
  • 1
    $\begingroup$ 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. $\endgroup$ – pppqqq Dec 14 '16 at 18:08
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – Chill2Macht Dec 15 '16 at 12:53
  • 1
    $\begingroup$ 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] $\endgroup$ – pppqqq Dec 15 '16 at 13:03
  • 1
    $\begingroup$ 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". $\endgroup$ – pppqqq Dec 15 '16 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.