Why psi^2?

1. Mar 1, 2006

pivoxa15

In 1926, Born suggested that the square of the wave at any point is proportional to the probability of finding the particle at that point. Born was led to this interpretation by relationship between the intensity of light and the square of its amplitude.

Since we can renormalise the function, why does it have to be psi^2 to get the probability? Why not have psi^4, 6, 100, 10000 ... ?

2. Mar 1, 2006

dicerandom

It's not really $\psi^2$, it's $\psi^*\psi$, or the complex conjugate of $\psi$ multiplied with $\psi$ itself. This is because we can view the wave functions as being vectors within a complex vector space and in order to take a dot product we need the complex conjugate. You can convince yourself of the latter fact by considering the dot product between the vectors representing the complex numbers $1+i$ and $1-i$.

3. Mar 1, 2006

ecolitan

Also it doesn't make sense for it to ever be a negative value.

4. Mar 1, 2006

Gokul43201

Staff Emeritus
What are you saying here ? Could you explain your problem more clearly ?

5. Mar 2, 2006

Perturbation

$$\psi\psi^*$$ is the probabilty density. The formulation of QM says that the probability density for the ensemble characterised by the state $$\rho$$ to take on some observable q is $$\langle q|\rho|q\rangle$$. For a pure state $$\rho=|\psi\rangle\langle\psi|$$ giving the probability density as $$|\langle q|\psi\rangle |^2$$.

6. Mar 2, 2006

pivoxa15

I understand that it is $\psi^*\psi$

$\psi$ = Ae^i(kx-wt)
conjugate psi= Ae^-i(kx-wt)

combine the above two psi and you get A^2 which after A is renormalised gives you the probability density.

if we had (psi)(psi)*(psi)*(psi) than we would have A^4 which we can then renormalise and get a different A than before but that is okay because all the probabilities will end up the same as before. Therefore it seems arbitary how many psi we multiply together as long as the number of conjugate psi matches the number of non conjugate psi.

Last edited: Mar 2, 2006
7. Mar 2, 2006

HallsofIvy

Then what you are saying is "we could use a much more complicated formula to get the same answer". Yes, we could. In fact, we could use $A= \sqrt{\pi^2r^4}$ to find the area of a circle.

Why would we want to?

8. Mar 2, 2006

vanesch

Staff Emeritus
Ok, but the normalisation of psi*^2 psi^2 will not be conserved under unitary evolution, will it ?

What I mean is: psi* psi = N(psi) has the special property that, for orthogonal vectors psi1 and psi2, we have N(psi1 + psi2) = N(psi1) + N(psi2), which is necessary for the probability interpretation.
So the interpretation of a state being the sum of orthogonal states in a certain basis, and being a superposition of those states corresponding to a probability measure associated with each of the terms individually, only holds under the quadratic measure, if we are going to allow for unitary evolution (= change of orthogonal basis).

Indeed, we want that N(psi1+psi2) = N(psi1) + N(psi2) and after unitary evolution: U(psi1) = psi3 and U(psi2) = psi4, so that N(psi1+psi2) = N(psi3 + psi4) = N(psi3) + N(psi4).

The first equality will hold for other powers (unitarity conserves hilbert norm, and hence all powers thereof). But the expansion in a sum of measures of each of the orthogonal components only holds for the quadratic norm.

Now, this is essential, because N(psi3) = probability to find psi3, N(psi4) = probability to find psi4.

Note:
You might now object that we could re-normalize after evolution (or change of basis). I perfectly agree with you ! That's actually what I'm claiming, that one cannot derive the Born rule from the postulates of unitary QM, if one does not require non-contextuality.
But if we require that the probability assignment to orthogonal components is directly derivable from the direction of the component and the quantum state vector and from nothing else (= requirement of non-contextuality) then the only possible probability measure is given by the quadratic rule (= the Born rule).

9. Mar 2, 2006

CarlB

In my opinion, the probability postulate is flawed, but not for the reason that it is special, as compared with, for example, |\psi|^4. The problem is in the assumption of complex numbers.

Probabilities appear naturally in a Banach space, which, to any vector, gives a positive number, called the norm, ||v||. If your space can be written as a linear vector space over the reals, then a natural norm is, for example:

||aA + bB + cC || = a^2 + b^2 + c^2,

where a,b,c are real (or complex) numbers and A, B, C is a basis for the vector space. Note that if the basis is defined, the natural Banach space defined by the above norm is unique.

This is a very natural place to do quantum mechanics, and in fact it is possible to do density matrix calculations this way. However, if one does this, one cannot have the useful ability of taking linear combinations of states. To take linear combinations of states, one can try to "factor" the Banach space norm into a Hilbert space "inner product". For the norm chosen above, it is always possible to factor the Banach space, unfortunately, there is no single unique Hilbert space that results from the factoring.

In a Hilbert space, the natural probability function takes two vectors as input instead of one. For example, let the above vector aA + bB + cC + ... be called "V", then the Banach norm was ||V|| = a^2+b^2+c^2+..., and thus one requires that the Hilbert space inner product satisfy:

<(a,b,c) , (a,b,c)> = a^2 + b^2 + c^2.

The natural solution for the inner product is therefore:

<(a,b,c) , (d,e,f)> = a*d + b*d + c*e,

where "a*" means the complex conjugate of a. It appears that the Hilbert space inner product so defined is unique, but this is not the case.

The problem is that one can multiply the basis of the Banach space by a set of arbitrary complex numbers and thus obtain a new basis that is equivalent in every way to the old one. The complex conjugation disappears in the Banach space, but pops out in the Hilbert space because the Hilbert space can compare distinct vectors. Consequently, there are a bunch of Hilbert spaces that are all equivalent factorizations of the same Banach space.

The complex phases in the choice of basis set for the Banach space amount to a "gauge" of the states.

Carl

10. Mar 2, 2006

Hurkyl

Staff Emeritus
The norm of that Banach space arises from an inner product, which turns it into a real Hilbert space. Are you basically saying we should use real Hilbert spaces instead of complex ones? Or did you mean that we should be using an arbitrary Banach space?

So what does the algebra of operators look like? Is it nothing more than a real Banach algebra? Or do we have additional constraints like $||A||^2 = ||A^2||$ by analogy with C*-algebras?

In one presentation I've seen, you start with a C*-algebra of "operators", and then form the state space as the set of all positive linear functionals with unit norm on that algebra. (And you can find a representation in the usual Hilbert space stuff)

Is there a corresponding construction here? What is the right sort of algebra to use?

11. Mar 3, 2006

Sherlock

This seems like a good question to me.

So, for my pedestrian understanding (that is, giving a physical interpretation to the OP's original question), is it ok to think in terms of the probability of detection at location x being proportional to the intensity of the incident wave at x --- and this intensity is simply the square of the amplitude of the incident wave at x?

That is, no matter how one might complicate one's calculation, eventually it should reduce to a statement of the intensity of the system being measured (the quantum system impinging on the measuring apparatus) at the point(s) in space-time where it contacts the measuring apparatus?

12. Mar 3, 2006

akhmeteli

One would think that a definition of probability density should pass the following test: is the integral of the probability density (over the entire 3d space) constant in time? Otherwise the (average) number of particles would vary with time. Therefore, one may suspect that there is no "one size fits all" definition of probability density, as the definition may depend on the equations of motion. If we use the Schroedinger equation, we may define probability density as \psi^*\psi (sorry, I have difficulties inserting TeX), and this definition will pass the test, and other polynomials will not. However, this definition does not pass the test for the Klein-Gordon equation.

13. Mar 5, 2006

CarlB

It's cool that someone else is interested in this esoteric subject.

The problem with using a real Hilbert space is that QM is very naturally complex. As far as I know, the only formulation of QM that has absolutely no arbitrary gauge freedom is the one that Schwinger dreamed up in the 1950s, now known as the Schwinger Measurement Algebra, and described in his book inexpensive "Quantum Kinematics and Dynamics".

Schwinger's formulation is based on the algebra of Stern-Gerlach filters. A Stern-Gerlach filter is a Stern-Gerlach apparatus that only allows passage of some type of particle. For example, it might only allow passage of spin-1/2 particles with +1/2 in the z-direction. Or it might only allow passage of right handed neutrinos. Or whatever. A filter of "0" passes no particles, while a filter of "1" passes all.

The SMA seems simple, but it is complicated enough to contain the Pauli algebra. That is, one can envisage putting together sequences of spin-1/2 filters. Addition means combining outputs from two filters, while multiplication means putting one filter after another so the output of one filter is the input of the other. This all works beautifully and simply as is explained in the opening chapter of Schwinger's book.

The Pauli algebra representation of a filter that allows only particles with spin-1/2 measurements of +1/2 in the u direction is given by the projection operator $$(1+\sigma_u)/2$$, where u is a unit 3-vector giving the direction, and $$\sigma_u = u_x\sigma_x+u_y\sigma_y+u_z\sigma_z$$.

In this algebra, one can consider the filter that corresponds to successive filters for particles in the +z, +x +y and finally +z direction. The Pauli algebra representation for this is:

$$\frac{1+\sigma_z}{2}\;\;\frac{1+\sigma_y}{2}\;\;\frac{1+\sigma_x}{2}\;\;\frac{1+\sigma_z}{2},$$

$$=\sqrt{\frac{1}{8}}\;\;e^{i\pi/4}\;\;\frac{1+\sigma_z}{2},$$

where $$i = \sigma_x\sigma_y\sigma_z$$, and I've written the result in a way that shows it as a complex multiple of a projection operator. The factor 1/8 corresponds to the reduction in amplitude from going through three consecutive 90 degree (i.e. independent) spin-1/2 measurements. And the factor of pi/4 is the phase induced from doing this.

Now the SMA is a very basic theory of QM. It doesn't even use spinors, for example, and because it is so simple and basic, and because it matches the results of experiment perfectly, it is a fact that any quantum mechanical theory must contain the SMA. So things that are true about the SMA are true about any quantum theory. (Note that the reverse is not true. For example, the spinor theory of electrons holds that electrons get multiplied by -1 when they are rotated by 360 degrees but this does not hold true in the SMA. The SMA corresponds to a density matrix formulation of QM. Of course it's well known that the arbitrary complex phases that appear in spinors, and cause that multiplication by -1, do not appear in the density matrix formalism.)

In the context of the SMA, one must associate a particular particle, or combination of particles, with any given set of filters. The measurement described above of consecutive filters in the +z, +x, +y, and +z directions shows that one must allow complex multiples of particles, and this makes a real Hilbert space impossible to use.

Yes, that's what I'm getting at, but I wouldn't use "arbitrary".

Carl

Last edited: Mar 5, 2006
14. Mar 5, 2006

CarlB

In the SMA, which as I mentioned above must exist at the heart of every quantum theory, the fundamental operators are projection operators and the fundamental equation that they satisfy is therefore:

$$M M = M$$

Addition is associative and commutative because it does not matter what order you combine the outputs of two or three Stern-Gerlach filters. Multiplication is not commutative because the order in which you measure a given particle is important, but because the order in which you hook up a sequence of three filters is not important, multiplication is associative.

These rules match the rules for multiplying and adding matrices, and also match the rules for a Clifford algebra. Consequently, one can model the opreator algebra by matrices or with a Clifford algebra. The advantage of using a Clifford algebra is that you can get geometric content into the theory that way.

So, is that the definition of a "real Banach algebra"?

I don't think that the constraint you mention here is compatible with the algebra of projection operators. Since any projection operator satisfies AA=A, the above would imply that ||A||=1 for any projection operator. Since all of 1 and $$(1\pm\sigma_z)/2$$ are projection operators, this would imply that the probability of a random (i.e. unoriented) particle surviving all these very different filters must be equal. In other words, what I'm saying is that this would an impractical restriction for a norm intended to be used as a probability measure.

Yes, there is a construction. Instead of starting with spinors and defining the density matrices as a sort of side show, you begin with the SMA and define the spinors from them. I'm busily writing it up and should be done in a matter of days.

The basic idea for how one obtains spinors from the SMA is (following Schwinger, more or less) to choose one element as the "vacuum", which we can call Z. Given projection operators R, G, and B, the associated bras and kets are defined by:

$$|R> = RZ/T_{RZ}$$

$$<G| = ZG/T_{GZ}$$

where $$T_{GZ}$$ is a real number that corresponds to the loss in amplitude associated with putting the two filters in order. For example, if there is an angle of theta between R and Z, and we are using the usual Pauli algebra, then $$T_{GZ} = \sqrt{(1+\cos(\theta))/2}$$.

The technical reason for having to come up with a Z state is that this is what is required to allow a Clifford algebraic analog of complex numbers to be used. In short, products of projection operators that begin and end with "Z", naturally commute with an exponential that acts just like a complex phase. Thus the complex numbers show up when you make spinors out of density matrices.

The above definition of spinors is subject to a disadvantge in that it blows up if R or G is antiparallel (i.e. annihilates) to Z. Now if R annihilates Z, then R does not annihilate -Z, so one can usually change the definition of the vacuum to avoid the issue. The usual spinor formalism seen in use in QM avoids this issue by keeping track of the above for both Z and -Z.

It should be noted that these difficulties that show up in the definition of spinors are purely mathematical. If you simply leave the theory in density matrix form none of this shows up. On the other hand, if you insist on the spinors as the basis for QM, you cannot avoid unphysical gauge freedoms.

I should be done with my paper in a few days. It is about 40 pages long and goes through numerous examples and has lots of exercises.

Carl

Last edited: Mar 5, 2006
15. Mar 5, 2006

Hurkyl

Staff Emeritus
Hah! I did a google search, and your post https://www.physicsforums.com/showthread.php?t=108622 came up as one of the references! I had even written a tentative reply, but I was unhappy with it and deleted it!

I'm going to go reply over there, so as not to hijack this thread any further.

Oh, I almost missed your next post (#14)!

Last edited: Mar 5, 2006
16. Mar 6, 2006

akhmeteli

17. Mar 6, 2006

CarlB

Thanks for the references. I took a look at them. I'm approaching the problem from the point of view of Schwinger's measurement algebra, which can be used to consider the effect on a beam of particles of successive Stern-Gerlach filters. In that situation, which is sufficiently basic that it has to appear in any quantum theory, the operators are naturally complex in that one can obtain operators that satisfy:

$$Q Q = -1$$

and Q commutes with the rest of the algebra.

I seem to recall that Landau and Lipsh itz, in their classic introduction to relativitistic quantum mechanics, show that the Dirac equation can be rewritten in real form. That is, the "i" can be removed from the equation. However, in so doing, they extend special treatment to one of the directions.

Hestenes factored the Dirac equation and got rid of the complex numbers, but he did it also by assuming a special direction and rewriting the equation with Clifford algebra. In terms of the Clifford algebra, he chose a somewhat arbitrary element, $$\gamma_1\gamma_2$$ to be the replacement for i. Note that this squares to -1 as it should. I go into this fact at great length in my just issued incomplete paper:
http://brannenworks.com/GEOPROB.pdf

But all these things amount to unphysical gauge freedoms, as the above paper shows at great length. I think that unphysical gauge freedoms are a bad idea. I think we should begin with a theory that is free of them, and when we introduce them for mathematical convenience, we should keep close track of them. This is opposite to the way that QM is contrived.

Carl

Oh, also note recent work by Brown & Hiley on density matrices as the fundamental of Bohmian mechanics instead of wave states. The references are in the above link, and I think you will find them interesting.

18. Mar 7, 2006

akhmeteli

Thank you for looking at the references. As for "has to appear in any quantum theory", I believe this statement should be qualified somewhat. I guess the Klein-Gordon-Maxwell electrodynamics is also a quantum theory (although not a final one), and there is no spin there, so I guess Stern-Gerlach filters are irrelevant. Furthermore (or, maybe, therefore:-) ), one can definitely do without complex numbers there, just with real ones, as Schroedinger demonstrated.

Actually, they just write the Dirac equation using the Majorana representation of the gamma-matrices. As a result, the equation obtains a real form, but there are at least two caveats. First, this can only be done for the free Dirac equation, without electromagnetic field, second, \psi-function in a given point is regarded as a set of four complex numbers. What I am trying to do, is to use a set of four real numbers instead. The two nontrivial theories that I consider are not equivalent to the Dirac theory. Certainly, this does not look inspiring, although it is not evident that they cannot explain existing experiments. There are however some advantages. First, one does not need quantum potentials in the Bohm interpretation of these theories, as their role is played by the ordinary electromagnetic potentials, and the electromagnetic field evolves independently (it is noteworthy that these two properties plus the reality of the wave function hold for the Klein-Gordon-Maxwell electrodynamics). Therefore, these theories may be regarded at least as attractive toy models of quantum mechanics with a natural deterministic interpretation. By the way, the requirement of reality naturally fixes the gauge.

Again, as far as I understand, Hestenes' formulation is equivalent to the Dirac equation with complex bispinors.

I started to read your draft article. Thank you for the reference to Brown-Hiley article.