Linearity of momentum operator

1. Jun 26, 2008

nughret

Is there a theory or any physical reasoning for the momentum operator being a linear, rather than an antilinear, operator?

2. Jun 26, 2008

scorpion990

If I understand the basics correctly, all operators in quantum mechanics are linear. Not quite sure why, though.

I guess that raises more questions than answers, though.

3. Jun 26, 2008

Domnu

The momentum operator is linear because the derivative operator is linear. The derivative operator is linear because addition is linear .

4. Jun 27, 2008

nughret

An antilinear derivative operator will respect addition.
In other words lets say the differential operator, D, obeys:

D(f + g) = D(f) + D(g),
D(cf) = cD(f), where c is a constant

I define a differential operator B, and B obeys,
B(f + g) = B(f) + B(g),
B(cf) = c*B(f), where c* denotes complex conjugate of c

My question is if there is a mathematical or physical reason why the operator P cannot be described by -iB.

5. Jun 27, 2008

Fredrik

Staff Emeritus
The translation operator is $e^{-ipr}$ where p is the momentum operator and r the translation distance. The exponential is defined as a series, but the square of an anti-linear operator is linear, so the translation operator would be neither linear nor anti-linear. I think this might violate the theorem proved by Wigner that says that operators representing symmetry transformations must be either linear and unitary or anti-linear and anti-unitary, but I don't know if he proved that this must be the case, or just that it's possible to define the operators that way. (I think it's the latter actually).

Also note that if we require the function T defined by $T(r)=e^{-ipr}$ to be continuous, then T(r) can't be anti-unitary for any r since it's unitary for r=0.

If you want to know more, check out the appendix of chapter 2 of volume 1 of Weinberg's QFT book.

6. Jun 27, 2008

nughret

Thanks Fredrik I had a quick look at Weinberg's version of Wigner's proof but it was a bit a long! I like you argument on continuity but you have already exponentiated a linear operator and then setting it equal to an anti-linear operator will obviously lead to some difficulties. You must begin earlier, considering the infintesimal unitary operator, just for translations for simplicity, and then choosing the P's to be anti-linear rather than linear.
I have just began to look at this and so far it appears we get the constraint that the P's are antihermitian rather than hermitian but I will do some more work on this first.

7. Jun 27, 2008

strangerep

The crucial point is that anti-linear transformations are discrete, since
they involve a complex conjugation operation. Therefore, such
transformations cannot be of the identity-connected type (i.e., they
cannot be continuous with the identity transformation). This is
similar to how a parity inversion is discrete, and not identity-connected.

Examples of anti-linear transformations are charge conjugation
and time-reversal for a Dirac particle (most RQM textbooks
discuss this).

The waters are muddied further, however, if you consider
transformations that mix the annihilation and creation
operators in QFT (since these implicit mix a field with
its complex-conjugated counterpart). Look up
"Bogoliubov transformations" (used in superconductivity
and elsewhere) for examples of this. A subtle point here
is that these transformations generally take you between
inequivalent Hilbert spaces -- which is why the Wigner
theorem mentioned by Fredrick may (superficially) seem
to be in contradiction with such things.

8. Jun 28, 2008

per.sundqvist

Its because we WANT operators to be linear, and that its meanigful. From the beginning starting from de Broglie and Schrödinger you describe a particle as a plane wave with $$k=p/\hbar, \omega=p/\hbar$$, and then you could easilly identify an operator that gives you p. Also Dirac made this wish of lineaity when he formulated the reativistic quantum mechanics.