# I Kinetic and Potential energy operators commutation

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1. Jun 16, 2017

### Dinara

Hi All,
Perhaps I am missing something. Schrodinger equation is HPsi=EPsi, where H is hamiltonian = sum of kinetic energy operator and potential energy operator. Kinetic energy operator does not commute with potential energy operator, then how come they share the same wave function Psi???? The operators that do not commute cannot have the same wave function. Please help me to figure this out. Thanks.

2. Jun 16, 2017

### Jilang

Does the sum of them commute?

3. Jun 16, 2017

### Dinara

The sum doesn't matter. The thing is (KE+PE) psi is a wrong expression - since KE and PE operators have to have different wave functions. Yet, in every single quantum chemistry text it is this (KE+PE) psi that is solved for the hydrogen atom!

4. Jun 16, 2017

### Jilang

Why do you say that if they occur together?

5. Jun 16, 2017

### dextercioby

Do you know what functional analysis is ?

6. Jun 16, 2017

### Dinara

No - I don't. Is the answer to my question in functional analysis?

7. Jun 16, 2017

### dextercioby

Pretty much so.

8. Jun 16, 2017

### strangerep

Sorry, but it does matter.

No, it's a correct expression. One finds eigenfunctions for the total Hamiltonian. It doesn't matter that such an eigenfunction might not also be an eigenfunction for KE and PE separately.

9. Jun 17, 2017

### Dinara

This does not make sense! (KE+PE)(psi) = KE (psi) + PE (psi), It does matter that eigenfunction has to be an eigenfunction of both KE and PE.

10. Jun 17, 2017

### vanhees71

No! You obviously misunderstood the concept of eigenvectors and eigenvalues completely. Which book are you using to study QT? I'm pretty sure, this utmost important topic is treated there. So please, have a look again and then come back with concrete questions.

Here you should really understand that if $\hat{H}=\hat{H}_1+\hat{H}_2$ then the eigenvectors of $\hat{H}$ need not be eigenvectors of $\hat{H}_1$ or $\hat{H}_2$. That's only the case if $\hat{H}_1$ and $\hat{H}_2$ commute. That's also a very important concept: Two self-adjoint operators have a common complete set of orthonormalized eigenvectors if and only if they commute, and only then the two observables represented by these two operators are compatible, i.e., you can define states (the common eigenstates of the two operators), for which both observables take determined values, namely the corresponding eigenvalues.

11. Jun 17, 2017

### Staff: Mentor

Maybe not to you, but it's easy to find examples. Here's one: the ground state wave function of the electron in the idealized hydrogen atom (i.e., a spinless "electron" in the Coulomb potential of the nucleus). This wave function is not an eigenstate of either KE or PE separately, but it is an eigenstate of the total Hamiltonian H.

12. Jun 18, 2017

### Staff: Mentor

Consider the following operators:$$A=\begin{bmatrix}0&1\\1&0\end{bmatrix}~~B=\begin{bmatrix}1&0\\0&-1\end{bmatrix}~~C=A+B=\begin{bmatrix}1&1\\1&-1\end{bmatrix}$$
The eigenvectors of A are $\begin{bmatrix}1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\-1\end{bmatrix}$
The eigenvectors of B are $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$
The eigenvectors of C are $\begin{bmatrix}1\\ \sqrt{2}-1\end{bmatrix}$ and $\begin{bmatrix}-1\\ \sqrt{2}+1\end{bmatrix}$
A and B do not commute and have no common eigenvectors. Nonetheless, their sum C has eigenvectors and they are not eigenvectors of either A or B. It's the same thing with the Hamiltonian, which is the sum of the non-commuting kinetic and potential energy operators.

[It would be a good exercise to check my algebra here]

13. Jun 18, 2017

### strangerep

[Moderator's note: off topic comment deleted.]

Last edited by a moderator: Jun 18, 2017
14. Jun 18, 2017

### Dinara

Thank you so much! Great example!

15. Jun 18, 2017

### Staff: Mentor

Everyone, please keep the discussion civil. I have deleted an off topic comment and an off topic post.