Quantum mechanics, ha ha

L|ama

This should actually be pretty easy for a QM question.

I've got a wavefunction with a bunch of stuff but the important bit is that sin(x) is the only bit with x in it (ground state of infinite square well if you really want to know), and the question is about the parity operator R, defined as Rf(x)=f(-x)
a) Show that the parity operator is linear
-My guess for this: chuck in R[f(x)+f(a)] and Rf(bx), fiddle round until I get Rf(x)+Rf(a)=R[f(x)+f(a)]=f(-x)+f(-a) or something similar and Rf(bx)=bRf(x). Right?

b) Show that the wavefunction is an eigenfunction of R with eigenvalue -1
-Well Rf(x)=f(-x), if I put -x in sin(x) I have -sin(x) so that's pretty easy

c) Suppose you made a measurement of the parity on the state described by the wavefunction. What result would you get?
-Not so sure about this, my guess is -1 but that's almost completely a guess. This is the bit that I'm most unsure and concerned about, the rest is just maths.

physi_marc

For a), I'd suggest you change f(a) to f(y), though. You used b as a constant, so it seems weird to use a as a variable in the same problem. I know they can be whatever you want them to be, but I'm sure your teacher will prefer if you use conventions as x,y,z for variables and a,b,c for constants. Also, you want to show that R(b*f(x)) = b[Rf(x)]. The constant multiplies the function, not the argument.

For c), I also believe you are correct. According to one of the postulates of QM, the only precise result of a measurement of a quantity is one of the eigenvalues of the linear operator associated with the quantity. In your case, the only eigenvalue is -1, so that's the only measurement you can get.

It's been a while since I've done QM, but I think that's it.

L|ama

Yeah, you're completely right on both counts with A, having b inside there means it would depend on whether f is linear or not. C sounds reasonable and familiar too, thanks.