Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Generalizing Born's Rule for 4-dimensions $x_4$, write

$$\langle a\rangle = \int\Psi A\Psi^* \mathrm{d}x_4$$

  1. Is this consistent with quantum mechanics?
  2. Is this a generalized form of the Born's Rule for the phase space of the wave function?

@DavidZas I was thinking of a four dimensional space where Time is not a parameter. There been recent work on Time Reversed light pulses. and the PHYSORG description

And this leads me to wonder about abstract phase space, and I mean phase both in the epoch angle of a wave function and the phase space of a Hamiltonian (which yes, is very confusing). The way I see it, the wave function implictly has the Action of phase space built into it.

$\psi = e^{i(\omega t - k x)} = e^{\frac{i}{\hbar }(\mathbb{E} t - \mathbf{p} x)}$

Where the phase and the action are connected by

$\frac{\mathbb{E} t - \mathbf{p} x}{\hbar }$

Imagine dividing the phases by the action of the particle It's a bit like there's a blob (note the very precise language) of phase space wobbling about like the liquid drop model of the nuclear physics. Only this is a more abstract phase space that's oscillating.

So, again, would Born's Rule apply to such a system?

Actually after rereading what I just written, I'll be surprised if anyone replies this time! :)

share|cite|improve this question
The PRL article is behind a paywall. Here is the free arXiv version – Qmechanic Jun 16 '11 at 12:12
Dear @metzgeer: What the PRL article seems to mean by "time reversal" is that the authors are able to take an arbitrary wave package smaller than approximately 30 $\mu m$ in size, and make it travel with its rear end first. As far as I can tell everything appears to be within the realm of standard physics where time is a parameter. – Qmechanic Jun 18 '11 at 13:43
phase and action are one and the same thing (modulus a $\hbar$ constant to make the action adimensional); the phase of a classical path is the integral of the lagrangian of the path, which is also the classical action. What do you want to achieve? – lurscher Jun 18 '11 at 17:31
Time and space are not on equal footing in non-relativistic quantum mechanics. If we for simplicity consider a particle with position $\vec{x}\in\mathbb{R}^3$, it is well-known that the Hilbert space of wave functions is isomorphic to $L^2(\mathbb{R}^3)$. The Hilbert space is not isomorphic to $L^2(\mathbb{R}^4)$, basically because temporal evolution of the wave function is not arbitrary but completely determined by the time-dependent Schroedinger equation. – Qmechanic Jul 6 '11 at 13:34

2 Answers 2

Of course this is consistent -- the calculation of an expectation value is defined in fact for operators in a totally generic Hilbert space. If you want that Hilbert space to describe 4 spatial dimensions, more power to you.

On the other hand, if you are imagining $x_4$ as a time dimension, it's important to remember that Q.M. treats space and time asymmetrically. (Or more concretely, time is a parameter and not an operator, which is really just to say that there's nothing special at all about time averaging in Q.M.!)

share|cite|improve this answer
+1 for the insight between time being a parameter in QM and being equivalent to the fact that time averages fail to be physically meaningul! – lurscher Jun 16 '11 at 4:27
@lurscher For free quantum fields, space and time can all be considered to be parameters, insofar as we introduce the operator-valued distribution $\hat\phi(x)$, parameterized by the position $x$ in space-time, to generate the algebra of observables. To say that time averages are not physically meaningful is not generally true of Hilbert spaces as a way to generate probabilities (which is what the Born rule does for us). The Hamiltonian+phase-space approach to dynamics may/can be replaced by an algebraic block-world approach (which is about as much as I can fit in a Comment). – Peter Morgan Jun 16 '11 at 12:28
Well, they can be meaningful -- we just have a tendency to focus our attention on stationary states of the hamiltonian, so $\psi(t) = \psi e^{iEt}$. – wsc Jun 16 '11 at 12:31
Or, @Peter beat me to it with a more correct and complete answer... ;) – wsc Jun 16 '11 at 12:32
-1: Only the standard nonrelativistic particle formalism treats time asymmetrically, and this answer has too many votes. This is a complete lie in 4d relativistic formalisms, and these are known since the 1940s. The proper time formulation of the Feynman propagator in the Schwinger representation gives a 4-d amplitude for travelling from point x to point y in proper time $\tau$ which adds up over all proper times to the normal propagator. – Ron Maimon Jan 3 '12 at 12:44

Your starting point is not a "generalization of the Born rule". Assuming that you properly defined a Hilbert space that contains your $\Psi$ as a vector, your starting point is an instance of the Born rule.

The Born rule takes a vector $V$, in a Hilbert space, $H$, on which there is an inner product $(\cdot,\cdot)$, by definition, and tells us that the expected value of an observable represented by the operator $\hat A$, which acts on the vector $V$ in that Hilbert space, is $\bigl<\hat A\bigr>=(V,\hat A V)$. There's no mention, as you see, of 3-dimensional space. I suspect you're thinking in terms of the Schrodinger equation, where the Hilbert space is the infinite-dimensional Hilbert space of complex-valued functions on $\mathbb{R}^3$, $L^2(\mathbb{R}^3)$, and seeking to generalize to space-time by introducing the Hilbert space $L^2(\mathbb{R}^4)$. Since this is just another Hilbert space, the Born rule applies to it.

One can do this as Mathematics, but to my knowledge this construction is not used in Physics. Instead, the usual route is to introduce operator-valued distributions as part of quantum field theory. The approach that I have suggested you are proposing could be seen as the one-particle subspace of the many-particle Hilbert space of a quantum field theory (I usually try to avoid all talk of particles, but I think putting this statement in field-theoretic terms wouldn't be helpful here). If you want to move to Minkowski space, I believe you will not be able to avoid moving to quantum field theory. Or, at least, that you will have to be able to present very explicitly what the relationship is between quantum field theory and whatever you move to.

share|cite|improve this answer
This is correct (no downvote), but irrelevant. The question is simply "can you find a wavefunction for 4-dimensional particle propagation where the amplitude to go from x to y obeys the 4 dimensional amplitude superposition rule". The answer is yes, and this is the Feynman Schwinger formalism. – Ron Maimon Jan 3 '12 at 12:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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