# KE operator and eigenfunctions

• dyn
In summary, the conversation discusses the question of whether the function u = x - iy is an eigenfunction of the kinetic energy operator in 3-D. The question asks if applying the operator to u results in a zero output, indicating an eigenvalue of zero. While the solution agrees with this interpretation, it also points out that u is not a physically realizable function due to its non-zero value at infinity. This leads to a discussion on the definition of an eigenfunction in quantum mechanics and the requirement for square-integrability. The conversation concludes with the suggestion for the question to be addressed by a teacher or further study into distribution theory.
dyn
I have just done a question and then looked at the solution which I don't get. The question gives a wavefunction as u = x - iy. It then asks if this function is an eigenfunction of the kinetic energy operator in 3-D. Applying this operator to u gives zero. I took this to mean that u is an eigenfunction with eigenvalue zero but the solution says that u is not an eigenfunction. Am I right or what is wrong with my argument ? Thanks

It is zero - so is an eigenfunction with eigenvalue 0.

Its not physically realisable of course - but formally it is true.

Thanks
Bill

Last edited:
The kinetic energy operator contains a Laplacian which involves 2nd derivatives in x , y , z so the function u becomes zero. The solution agreed with me that the KE operator acting on the function gives zero but I took that to mean u is an eigenfunction with eigenvalue zero but the solution says it is not an eigenfunction.

Well obviously it is - there is an error.

What it may be alluding to is its not physically realisable since its non zero at infinity. But then again so are the normal solutions.

Thanks
Bill

dyn
Just to expand a bit further what's going on is you have run into a pathological case with degeneracy issues due to an eigenvalue of zero.

If you solve the free SE via separation of variables you end up with the multiple of wave solutions. But things get tricky when those waves have infinite wavelength as in your example. Work through the detail - I just did and its interesting.

Thanks
Bill

Last edited:
I suppose the most straightforward argument that that is not an eigenvector is simply that it is not orthogonal to any of the eigenvectors with nonzero eigenvalue. Since the operator is Hermitian, it is not a proper eigenfunction (pun intended!)

fzero said:
I suppose the most straightforward argument that that is not an eigenvector is simply that it is not orthogonal to any of the eigenvectors with nonzero eigenvalue. Since the operator is Hermitian, it is not a proper eigenfunction (pun intended!)

In QM, the usual meaning of "eigenfunction" is a square-integrable function that has a definite value for some operator. But $\psi = x - iy$ isn't square-integrable, so it's not a legitimate wave function. In many instances in quantum mechanics, it's the constraint that the wavefunction be square-integrable that forces its eigenvalues to be discrete. So usually when people talk about the eigenvalues of an operator $O$, they mean those values $\lambda$ such that there is a square-integrable function $\psi$ with $O \psi = \lambda \psi$.

stevendaryl said:
In QM, the usual meaning of "eigenfunction" is a square-integrable function that has a definite value for some operator. But $\psi = x - iy$ isn't square-integrable, so it's not a legitimate wave function. In many instances in quantum mechanics, it's the constraint that the wavefunction be square-integrable that forces its eigenvalues to be discrete. So usually when people talk about the eigenvalues of an operator $O$, they mean those values $\lambda$ such that there is a square-integrable function $\psi$ with $O \psi = \lambda \psi$.

I am pretty sure that's the reason they wanted it excluded as an eigenfunction - it blows up to infinity at infinity. My concern is I have zero doubt wherever the OP got it from would have the usual wave solution to the free particle SE. That, while not blowing up to infinity, is nonetheless bounded there so is not valid either. This is the type of thing that can easily confuse a beginning student that actually thinks.

In fact it can confuse more advanced students until they come to grips with Rigged Hilbert Spaces.

Thanks
Bill

the question was from an introductory QM course. At that level Rigged Hilbert Spaces are unheard of. So at the introductory level is there a definite answer of eigenfunction or not ?

dyn said:
the question was from an introductory QM course. At that level Rigged Hilbert Spaces are unheard of. So at the introductory level is there a definite answer of eigenfunction or not ?

No it isn't because it blows up at infinity so isn't normalisable - but like I said its inconsistent.

Thanks
Bill

Sorry to be a pain here but the mathematical definition of eigenvectors doesn't put any requirements on whether they can be normailised or not. So if it is given as a wavefunction surely it is an eigenfunction with eigenvalue zero. Another part of the question asked whether Aexp(ikx) + Bexp(-ikx) was an eigenfunction of the kinetic energy operator and gave the answer that it was but this function is also not square-integrable

dyn said:
Sorry to be a pain here but the mathematical definition of eigenvectors doesn't put any requirements on whether they can be normailised or not.

They are part of a normed vector space - that vector has infinite norm.

That function you gave has exactly the same issue but is considered valid - but it really isn't - that's the part that makes thinking students pull their hair out.

I could explain what's going on but it will involve a long sojourn into distribution theory:
https://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/#more-2072

Really your teacher needs to explain it - post here with what he/she says - it should prove interesting.

Thanks
Bill

Last edited:
I don't have a teacher. Just self-studying. Obviously the QM case is never straight forward - no surprise there but I have a last question. If it was just a maths question regarding eigenvalues and I had a function similar to the KE operator and the function x-iy would it be an eigenfunction with eigenvalue zero ? Whenever I have done eigenfunction questions I have just had to check if it satisfies the eigenvalue equation. Is this not enough ? Do I now need to check normed vector spaces ?

dyn said:
I don't have a teacher. Just self-studying. Obviously the QM case is never straight forward - no surprise there but I have a last question. If it was just a maths question regarding eigenvalues and I had a function similar to the KE operator and the function x-iy would it be an eigenfunction with eigenvalue zero ? Whenever I have done eigenfunction questions I have just had to check if it satisfies the eigenvalue equation. Is this not enough ? Do I now need to check normed vector spaces ?

You need to specify a space. To really understand this stuff you need to study linear algebra first, then Hilbert Spaces. For now my advice is simply to persevere and move on.

Thanks
Bill

One way to deal with this kind of issue, at least in part, and without going to the full machinery of rigged Hilbert spaces, is to put the particle in a box with some boundary conditions, like the wave function must vanish on the walls, or periodic boundary conditions. Then require wave functions to be square integrable. Then x-iy is not an allowed wave function, because it doesn't satisfy the boundary conditions.

Whatever source the OP had that asked the original question is badly written and should be dropped.

Avodyne said:
Whatever source the OP had that asked the original question is badly written and should be dropped.

Agreed.

Students should not be confronted with tricky issues like this to start with.

Thanks
Bill

Thanks guys. I'm going to forget I ever saw that question.

## 1. What is the KE operator?

The KE operator, also known as the kinetic energy operator, is a mathematical representation of the energy associated with the motion of a particle. It is used in quantum mechanics to calculate the kinetic energy of a particle in a given system.

## 2. How is the KE operator represented mathematically?

The KE operator is represented by the symbol "T" and is defined as T = -ħ²/2m (∂²/∂x² + ∂²/∂y² + ∂²/∂z²), where ħ is the reduced Planck's constant, m is the mass of the particle, and ∂/∂x, ∂/∂y, and ∂/∂z represent partial derivatives in the x, y, and z directions, respectively.

## 3. What are eigenfunctions in relation to the KE operator?

Eigenfunctions of the KE operator are mathematical functions that, when acted on by the KE operator, result in a multiple of the original function. In other words, they are functions that satisfy the eigenvalue equation Tψ = Eψ, where ψ is the eigenfunction and E is the corresponding eigenvalue.

## 4. How are eigenfunctions related to the energy of a particle?

The eigenfunctions of the KE operator are directly related to the energy of a particle. The eigenvalue E represents the energy of the particle in the system, and the corresponding eigenfunction ψ describes the spatial distribution of the particle's wavefunction.

## 5. What is the significance of eigenfunctions in quantum mechanics?

Eigenfunctions play a crucial role in quantum mechanics as they provide a way to describe the energy and spatial distribution of a particle in a given system. They also form a complete set of basis functions, which allows for the expansion of any wavefunction in terms of these eigenfunctions, making calculations and predictions in quantum mechanics possible.

Replies
31
Views
2K
Replies
22
Views
1K
Replies
1
Views
999
Replies
14
Views
1K
Replies
16
Views
1K
Replies
1
Views
1K
Replies
60
Views
6K
Replies
24
Views
2K
Replies
14
Views
2K
Replies
16
Views
2K