Trying to understand Hermitian adjoint proof

[Mary]

I'm currently reading the book Introductory Quantum Mechanics by Richard Liboff 4th edition.

I'm reading one of the proofs and I don't understand what is happening in one of the steps.

The problem is trying to find the Hermitian adjoint of the operator $\hat{D}$=$\partial$/$\partial$x defined in hilbert space

I have attached an image of the problem to this forum. The step I don't understand is circled. I understand where the last part (the - integral) comes in but the first part I do not understand. All I know is that it is equal to 0 and I don't know why. Thanks for any input.

I know this is probably just a simple math step I have probably forgotten about.

 

[micromass]

Integration by parts?

 

[WannabeNewton]

The first part is the boundary term that comes from integration by parts. It vanishes because the physical wave-functions are taken to decay quickly enough while approaching infinity so as to be square-integrable (and thus normalizable).

 

[micromass]

It vanishes because the physical wave-functions are taken to decay quickly enough while approaching infinity so as to be square-integrable (and thus normalizable).

Square integrability does not imply that the functions decay quickly enough approach infinity. You'll need some special conditions on the wave functions.

But ok, these kind of physics problems don't need to be complicated by a mathematician, so I'll leave...

 

[strangerep]

Square integrability does not imply that the functions decay quickly enough approach infinity. You'll need some special conditions on the wave functions.

But ok, these kind of physics problems don't need to be complicated by a mathematician, so I'll leave...

It doesn't hurt for physicists to learn a little more...

In this case, I would have said that for $\psi$ to be square-integrable over the real line, and be such that $\partial_x \psi$ is well-defined everywhere (which is physically important for the momentum operator), then it's reasonable to deduce that
$$\lim_{|x|\to\infty} \psi(x) = 0 ~.$$Anything else needed?

-> Mary: Is your question now answered?

 

[micromass]

It doesn't hurt for physicists to learn a little more...

In this case, I would have said that for $\psi$ to be square-integrable over the real line, and be such that $\partial_x \psi$ is well-defined everywhere (which is physically important for the momentum operator), then it's reasonable to deduce that
$$\lim_{|x|\to\infty} \psi(x) = 0 ~.$$Anything else needed?

It's still not true then. We've actually been going through this in this thread: https://www.physicsforums.com/showthread.php?t=745413

In practice, can't you always choose $\psi$ in the Schwarz class? In that case, it would be true. Or perhaps if you demand $xf(x)$ and other to be square-integrable or something.

 

[WannabeNewton]

Square integrability does not imply that the functions decay quickly enough approach infinity.

I never said it did.

 

[micromass]

I never said it did.

Not sure why the square-integrable part was necessary then. Ah well, never mind

 

[WannabeNewton]

Not sure why the square-integrable part was necessary then. Ah well, never mind

The usual justification for why we take physical wave-functions to decay fast enough at infinity is so that we can ensure square-integrability through physical boundary conditions (at infinity) on the wave-functions. There are physically relevant cases where we have to consider wave-functions that don't decay fast enough e.g. free particles (plane waves) but we can get around this by using box normalization. Of course if the wave-functions correspond to bound states as opposed to scattering then boundary conditions are imposed on finite regions and may involve periodicity instead but that's not what the OP is considering.

 

[micromass]

The usual justification for why we take physical wave-functions to decay fast enough at infinity is so that we can ensure square-integrability through physical boundary conditions (at infinity) on the wave-functions. There are physically relevant cases where we have to consider wave-functions that don't decay fast enough e.g. free particles (plane waves) but we can get around this by using box normalization. Of course if the wave-functions correspond to bound states as opposed to scattering then boundary conditions are imposed on finite regions and may involve periodicity instead but that's not what the OP is considering.

Of course, Liboff mentions none of that and just takes a generic $\psi$ in $L^2$. Sigh...

 

[strangerep]

It's still not true then. We've actually been going through this in this thread: https://www.physicsforums.com/showthread.php?t=745413

In practice, can't you always choose $\psi$ in the Schwarz class? In that case, it would be true. Or perhaps if you demand $xf(x)$ and other to be square-integrable or something.

I was trying to find a subset of the full Schwarz distribution theory framework that would answer the OP's question. But perhaps that's unwise. Physicists need to understand distribution theory.

So yes, I guess it's better (more direct) to answer the question in the way you suggest (Schwarz theory).

 

[micromass]

I was trying to find a subset of the full Schwarz distribution theory framework that would answer the OP's question. But perhaps that's unwise. Physicists need to understand distribution theory.

So yes, I guess it's better (more direct) to answer the question in the way you suggest (Schwarz theory).

Hmm, interesting. You could always see $\psi$ as a distribution and then use the definition of distributional derivative (which then of course does not agree with the usual notions). It's a bit of a stretch, but I need to think about this.

 

[strangerep]

Hmm, interesting. You could always see $\psi$ as a distribution and then use the definition of distributional derivative (which then of course does not agree with the usual notions). It's a bit of a stretch, but I need to think about this.

Indeed. (Remember the conversations we've had in the past about Rigged Hilbert Space and QM?) But QFT people use the distributional (aka weak) derivative quite a lot, often without realizing that it has a name.

 

[dextercioby]

I think for practical purposes the vector space of wavefunctions can always be taken as the Schwartz space, so that integration by parts with limits to infinity is reduced to 0.

 

[Mary]

This makes a lot more sense now. Thanks for all of your help. I just couldn't accept that it was 0 and I wanted to know why. This is my first thread and its been so helpful!

 

[bhobba]

In practice, can't you always choose $\psi$ in the Schwarz class?

I think for practical purposes the vector space of wavefunctions can always be taken as the Schwartz space, so that integration by parts with limits to infinity is reduced to 0.

That's the way I tend to look at it.

Simply take it as a test function or for all practical purposes approximated by a test function and Bob's your uncle.

I used to really worry about such things but got a hold of the following nifty book that helped a lot:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Thanks
Bill