2
$\begingroup$

Okay so this has been bugging me for a while now because everyone I talked to so far didn't know the answer. I'm an A level student but I got quite interested in quantum physics and so I decided to buy Griffiths introductory QM book and just try to understand the maths and how things fit together (I'm only on the first few pages, maybe 24 I think). So far the only thing I didn't understand was that why the mod squared of the wave function is the probability density. I mean I understand that it has to be real but then why don't we square root it afterwards? Is it something that just works or there is an actual reason behind the whole thing?

$\endgroup$
2
$\begingroup$

The main reason is that the probability density, constructed as you indicated, leads to a CONSERVED overall probability over all space.

In most QM books it is shown that the modulus square of the wave function, integrated over all space, does not depend on time. It can then be normalized to 1. The derivation of the conservation of the probability over all space being conserved is from the Schrodinger equation. I've not used Griffith but if it's a standard book it should have that derivation. The 'trick' is to find a current j in terms of the wave function and its complex conjugate such that the divergence of j plus the partial of the probability density with time is zero. By Gauss's theorem then the derivative wrt time of the integral of the density over space is zero.

You can then use the prob. density to get the average (I.e., expectation) values of dynamical valuable, or observables, like position and momentum, and get the Newton equations for those expectation values, and appropriate conserved quantities.

You can in more advanced books get the current density j and probability density from the Lagrangian for the quantum theory (Schrodinger or later relativistic quantum field theories) In a continuity of probability equation that conserves the probability integrated over the appropriate space.

There are also hysical reasons such as the similarity with the energy density in electromagnetism. It is also the the wave variables, E and B, squared, and then the energy is conserved when integrated over space.

The square root of the modulus square has been shown to not obey a conservation equation, so that was ruled out.

You can see a little of the math and essentially the same answer, stated with some differences, at Born Interpretation of Wave Function

$\endgroup$
  • $\begingroup$ Ah yes, his book has that derivation also the expectation values of momentum and everything else and it makes sense but so is it the main reason that it is the mod squared because this way we can show that the probability doesn't depend on time hence the wave function stays normalised for any future time once you normalised it lets say at t=0 because otherwise the whole statistical interpretation would fall apart? Griffiths book talks about this but it didn't say anything about why it is the modulus squared. And so they discarded the square rooting idea because it didn't work? $\endgroup$ – Peter Fazekas Dec 22 '16 at 23:58
  • $\begingroup$ The first question: the total probability of finding the particle at some x, or some p or some other variable, has to be 1. If it's so it means that some prob. Current flows in or out of a certain box then the prob. Increases or decreases the same way. It means you have a consistent interpretation. This is used also for the density of a fluid for instance, to mathematically state matter conservation. All physics conservation laws are this way. The second question, right, it is not conserved. The square is, the modulus is not $\endgroup$ – Bob Bee Dec 23 '16 at 5:22
0
$\begingroup$

In my opinion, the more natural function is that a state is a positive linear functional of norm 1 on the observable algebra. The value of this functional at an observable gives the expected value of the observable.

If we follow the intuition that observables can be represented as something like a matrix algebra, then all linear functionals should all expressible as linear combinations of basic functions of the form

$$ \psi_{v,w}(A) = v^* A w $$

where $v$ and $w$ are vectors. Think making $v$ and $w$ standard basis vectors, in which case $\psi_{v,w}$ is simply picking out a single matrix component of $A$.

If we restrict ourselves to Hermitian matrices real-valued functionals, then these can be expressed in terms of the quadratic functionals

$$ (v^* A w) + (v^* A w)^* = (v+w)^* A (v+w) - v^* A v - w^* A w $$

That is, we expect all real-valued linear functionals on Hermitian matrices to be linear combinations of functionals of the form $\phi_v(A) = v^* A v$.

If $A$ is a "proposition" — i.e. a Hermitian matrix with eigenvalues $1$ (corresponding to when an event happens) and $0$ (corresponding to when it doesn't happen), then after making a change of basis to diagonalize $A$, $v^* A v$ is simply the sum of $v_i^* v_i = |v_i|^2$ over all of the components of $v$ corresponding to the event happening.

So that's where the modulus squared comes from: it's drawn from the abstract case of converting a matrix into a scalar by multiplying it on both sides by a vector.

With the idea in hand, there are general constructions and theorems: e.g. the GNS construction.


The bit I find somewhat surprising is that you can get away with doing so much with a single Hilbert space and not having to appeal to the greater theory that describes what's actually going on with everything.

$\endgroup$

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.