# Hermitization and a problem from Bohm's quantum Theory

1. Aug 28, 2012

### A_B

Hi,

Bohm discusses several ambiguities is the hermitization of operators that consist of products op p's and x's (page 184 and further)

He reaches the conclusion that an operator $p^m x^n$ should be replaced by
$$\frac{p^m x^n + x^n p^m}{2}$$
on the grounds that this way the expectation for the operator is always real. But this method then regard the DIFFERENT operators like p^2x, pxp and xp^2 as being the same.

This Hermitization seems to me a very shakey procedure with little ground in actual physics, or is there more to it?

Bohm goes on the describe another ambiguity. For the operator (px)^2, should we use
$$\frac{p^2 x^2 + x^2 p^2}{2}$$
or
$$\left( \frac{xp+px}{2} \right)^2$$

He than poses the problem to show that the above are not the same but differ by quantities of order $\hbar^2$.

I do obtain a difference of $3\hbar^2/4$ But then Bohm claims that this difference can be neglected and the two alternatives for expressing the operator seen als FAP equivalent because the difference is so small.

But The ENTIRE operators are of the order $\hbar^2$ (as can easily be seen since each term contains p twice) so I don't see how his argument holds.

Lastly, Bohm suggests that in the future experimental results might differentate between the two ways of expressing the operator. Well, this is the future, so is there any news on that?

Thanks,
A_B

2. Aug 28, 2012

### DrDu

That the expression for an operator is proportional to hbar does not mean that it is of order hbar. In fact, p is an unbounded operator and e.g. it's expectation value can be arbitrary large. On contrast, the commutator [x,p] is a bounded operator so that it makes sense to say that it is of order hbar.

3. Aug 28, 2012

### A_B

Ok so, is this correct?

The terms that occured in both operators contained x, as wel as derivatives of psi w. respect to x. Therefore, by choosing an appropriate wavefunctions, their expectations can be made arbitrarily large.

The term where the two operators differ is a constant, the expectation of wich is simple the constant itself and is therefore bounded.

In the classical limit the wavefunction will always be such that the agreeing terms have a fairly large expectation value in comparison with the constant term, so that this constant term may be neglected.

A_B

4. Aug 28, 2012

### DrDu

Sounds ok.
If you want more people to comment on your questions maybe it would be a good idea to give a full citation of the book ("Bohm") you are referring to.

5. Aug 28, 2012

### A_B

6. Aug 28, 2012

### Jazzdude

The simple answer is that the product order and hermitization procedure is not defined in the standard classical to quantum transition with means of canonical quantization. So it's really to be determined by experiment, which method is the correct one. In quantum field theory you deal with annihilation and creation operators, and there the Wick- or normal-ordering is what you take in order to avoid arbitrary constant terms in the vacuum energy.

However, there's a longer and better answer, but it requires some deeper look into the relationship between classical systems and quantum systems and goes beyond canonical quantization, to geometric quantization.

Eugene Wigner tried to understand the notion of phase space and its associated geometry in the context of quantum theory and invented a map that that takes a Hilbert space operator to its Wigner distribution, which lives in a classical phase space. Quantum states can be mapped as projection operators and therefore have a corresponding Wigner distribution too. Classical states (both single and ensemble) are non-negative valued distributions on the position-momentum plane, whereas quantum states (also both pure and ensemble) are real valued distributions on that same plane. That allows for negative values and forbids a direct correspondence between classical and quantum distributions. However, if you have a quantum ensemble with certain properties then the phase space distribution becomes non-negative and you can treat it like a classical ensemble in many aspects.

Now the states are only half of the story. As I said before you can also map quantum operators to functions on phase space, and the story is roughly the same as for the states. What is still missing are the operations that allow an operator to act on a quantum state (or another operator) or a phase space functional to act on a classical state (or another functional). Both quantum and classical domains have their standard operation: The operator product in the quantum domain and the pointwise multiplication of the distributions in the classical domain. And this is where you might get disappointed: The Wigner map does not take operator products to pointwise products or vice versa. They define two different algebras, each defined naturally in its own domain but with no obvious connection. One is even non-commutative while the other is trivially abelian.

And this is where we can finally start to understand your question in the context of geometric quantization. If there is a commutative algebra on classical phase space, why is there not one for the Hilbert space? The obvious answer is that there must be one, induced by the Wigner map. We simply lift the classical algebra to the quantum space using the Wigner map and have a new algebra on the Hilbert space that has all the classical properties we desire. This is the heart of the geometric quantization procedure. The resulting operator product in the Hilbert space is called the fully symmetric operator product and is defined roughly like follows. For an operator in the universal enveloping algebra of {p,x,1} we can expand this operator in a power series of the {x,p,1} operators. We can now define a projection onto the fully symmetric algebra by replacing each term in that power series with the sum over all of its factor permutations divided by the number of permutations. The fully symmetric product is then just the result of an ordinary operator product projected onto the fully symmetric sub algebra. It's straight forward to see that this product is entirely agnostic of the order of its factors.

The other direction of taking the ordinary operator product to classical phase space results in the so called Moyal product, which is a locally deformed pointwise product. That means if the two distributions you want to multiply are sufficiently smooth and blurry (on a scale where the phase space area of interest is comparable to hbar) then the result will look mostly like a pointwise product. This allows to understand the conditions for when quantum systems act mostly classically quite well, and it helps to understand the result of operator products and their associated non-commutative geometry.

So, I hope this little detour helped you to understand how geometric quantization delivers a commutative operator product as the best approximation to classical systems. If that really is the what nature has realized is a totally different question though!

Cheers,

Jazz

7. Aug 28, 2012

### A_B

Unfortunately, since I'm only starting to learn QM and have yet to study statistical mechanics in any depth, and I feel also because of some big holes in my knowledge of the relevant mathematics, I could not follow your argument. But I'll try to remeber your post as a starting point for further study, when the time is right.

Thanks again
A_B

8. Aug 28, 2012

### Jazzdude

That's ok, it will all come to you one day. Maybe I should simply give you an example. So let's denote the symmetric operator product by (.) and evaluate p^2(.)x. First, list all the (distinguishable) permutations of the involved factors of p and x:

ppx, pxp, xpp

As expected we get (2+1)! /(2! 1!) = 6/2 = 3 terms

and the resulting operator product is then p^2(.)x = (p^2 x + x p^2 + p x p)/3

This is the fully symmetric product as induced by geometric quantization. It's well defined and the only order definition with some serious theoretic background (as far as I'm aware of).

Don't worry if you didn't get the details. It took me many years to see the light ;-)

Cheers,

Jazz