# Testing for hermiticity

amolv06
Science 22 Feb '10 19:54
1. 22 Feb '10 19:54
I'm having a hard time proving that A is hermitian where A acting on f gives the integral from a to x of f(x'ðŸ˜‰dx'. Could anyone help me out? I can't seem to force hermiticity after doing <g|Af> by parts, but I'm not sure if I've gone as far as I can. It's kind of hard to post the math I have tried here, but if anyone could give me some advice, I'd appreciate it.
2. AThousandYoung
West Coast Rioter
22 Feb '10 22:14
Originally posted by amolv06
I'm having a hard time proving that A is hermitian where A acting on f gives the integral from a to x of f(x'ðŸ˜‰dx'. Could anyone help me out? I can't seem to force hermiticity after doing <g|Af> by parts, but I'm not sure if I've gone as far as I can. It's kind of hard to post the math I have tried here, but if anyone could give me some advice, I'd appreciate it.
I barely passed linear algebra and differential equations...sorry!
3. 22 Feb '10 22:37
How do you define your inner product? If you use the L2 definition of an inner product, then A is not hermitian at all if I'm not mistaken... (take e.g. g(x) = x, f(x) = x^2, a = 0)

I could be wrong though, I'm a bit rusty on this subject.
4. 23 Feb '10 00:08
Originally posted by KazetNagorra
How do you define your inner product? If you use the L2 definition of an inner product, then A is not hermitian at all if I'm not mistaken... (take e.g. g(x) = x, f(x) = x^2, a = 0)

I could be wrong though, I'm a bit rusty on this subject.
"How do you define your inner product?"

Correct, using the L2 definition. Both f and g lie in Hilbert space, so they are complex. I must show that A is either hermitian or not hermitian more generally. I do not think the professor would allow us to select specific vectors. I think I've got it, though. A is non-hermitian, as <f|Ag>=-<Af|g>.
5. 23 Feb '10 07:16
Originally posted by amolv06
"How do you define your inner product?"

Correct, using the L2 definition. Both f and g lie in Hilbert space, so they are complex. I must show that A is either hermitian or not hermitian more generally. I do not think the professor would allow us to select specific vectors. I think I've got it, though. A is non-hermitian, as <f|Ag>=-<Af|g>.
If you want to show A is not hermitian, then showing a single example where it does not apply suffices.
6. 23 Feb '10 08:121 edit
Good point.

That said, our professor has instructed us to consider the wavefunctions localized, which leads certain operators (such as the derivative) to be hermitian in these cases, whereas in general (from a mathematical, and not a physical perspective) they would not be. Such results seem to be most readily apparent by actually performing the calculus on abstract vectors, at least to me. If you have a more efficient method, please do share.

Thanks,

Amol

Edit: Just some musing on my part -- please let me know if it sounds reasonable --

Localizing the wavefunction must put all the work we're doing in some subspace of the complex function space. Using your point above, if I could find two vectors for which A was not hermitian in this subspace, I would have proved non-hermiticity for A. That said, I'm having a hard time convincing myself that I would be able to accurately pick out vectors which lie in the appropriate subspace.

I believe I've solved the problem from a more abstract perspective, but what you said never dawned on me for some reason. Does everything above sound reasonable to you?
7. 23 Feb '10 10:59
I'm not sure what your professor means by "considering the wavefunctions localized", but if certain wavefunctions f and g do not overlap, <f|g> = 0, and for these wavefunctions, A is probably hermitian (not sure though). It's an important concept in quantum physics because in principle, all e.g. electrons in the universe are correlated with each other due to the symmetrization requirement - but if the individual wavefunctions do not overlap, you can describe the (clusters of) electrons seperately.
8. 23 Feb '10 15:14
Originally posted by KazetNagorra
I'm not sure what your professor means by "considering the wavefunctions localized",
The wavefunction goes to 0 at infinity.

9. 23 Feb '10 16:59
Originally posted by amolv06
The wavefunction goes to 0 at infinity.