Quantum Mechanics => Operators

[jonnylane]

I'm doing some past papers for my QM finals and I've come across a question that is a bit strange. I'm not sure if it's as easy as it sounds.

X and P are one dimensional position and momentum operators, which take the explicit forms of x and -ihd/dx.

i) write down the explicit forms of X^2 and P^2

now then, is this just x^2 and hd^2/dx^2?

im ok on the next few bits, but:

iv) which, if any, of the two functions exp(ikx) and exp(-ax^2) are eigenfunctions of P^2?

my guess is that, since the eigenfunction of P is exp(ikx), its the other one (and the i has disapeared in the squaring process), but how can i prove this?


Im probably just being paranoid, but can someone verify these answers?

thanks

 

[ahrkron]

Originally posted by jonnylane
now then, is this just x^2 and hd^2/dx^2?

How did you obtain them? that's the important part. I think you're missing a sign on P^2.

iv) which, if any, of the two functions exp(ikx) and exp(-ax^2) are eigenfunctions of P^2?
how can i prove this?

Remember what an eigenfunction is. You need to plug in both functions on P^2, and see if you get a constant times the original function.

 

[M98Ranger]

Hi there,

Yes, you are correct in your guess that the eigenfunction of P is exp(ikx) and that the other function, exp(-ax^2), is not an eigenfunction of P^2.

To prove this, we can use the definition of an eigenfunction, which is a function that when operated on by an operator, results in a constant multiple of itself. In this case, we are looking for a function f(x) such that P^2f(x) = cf(x), where c is a constant.

For the function exp(ikx), we have P^2(exp(ikx)) = -k^2(exp(ikx)). Therefore, exp(ikx) is an eigenfunction of P^2 with eigenvalue -k^2.

However, for the function exp(-ax^2), we have P^2(exp(-ax^2)) = -2a(exp(-ax^2) + 2ax^2(exp(-ax^2)). This is not a constant multiple of exp(-ax^2), so it is not an eigenfunction of P^2.

I hope this helps! Good luck on your QM finals.