Kinetic and Potential energy operators commutation

In summary, the conversation is about the Schrodinger equation and the concept of commuting operators. The sum of the kinetic and potential energy operators does not commute and therefore, their wave functions cannot be the same. However, this does not matter when finding eigenfunctions for the total Hamiltonian, as they are only concerned with the total energy, not the individual energies. This concept is also applicable in functional analysis, where two self-adjoint operators have a common set of eigenvectors if and only if they commute.
  • #1
Dinara
5
0
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.
 
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  • #2
Does the sum of them commute?
 
  • #3
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
Why do you say that if they occur together?
 
  • #5
Dinara said:
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!

Do you know what functional analysis is ?
 
  • #6
No - I don't. Is the answer to my question in functional analysis?
 
  • #8
Dinara said:
The sum doesn't matter.
Sorry, but it does matter.

The thing is (KE+PE) psi is a wrong expression
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.
 
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  • #9
dextercioby said:
Pretty much so.
strangerep said:
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.
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
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
Dinara said:
This does not make sense!

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.
 
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  • #12
Dinara said:
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.
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]
 
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  • #13
Nugatory said:
[It would be a good exercise to check my algebra here]
Heh, I was about to suggest the exact same exercise (so I'd already "checked" your algebra before your posted). :oldwink:

[Moderator's note: off topic comment deleted.]
 
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  • #14
Nugatory said:
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]
Nugatory said:
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]

Thank you so much! Great example!
 
  • #15
Everyone, please keep the discussion civil. I have deleted an off topic comment and an off topic post.
 

1. What is the difference between kinetic and potential energy?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object possesses due to its position or configuration.

2. How do kinetic and potential energy operators commute?

The commutation of kinetic and potential energy operators refers to the order in which they are applied to a wavefunction. In general, these operators do not commute, meaning the order in which they are applied can affect the outcome of the calculation.

3. What is the significance of commutation in quantum mechanics?

Commutation is an important concept in quantum mechanics because it determines the observables (such as energy) that can be measured simultaneously with a certain degree of accuracy. Operators that commute with each other can be measured simultaneously with high accuracy, while non-commuting operators cannot.

4. How do kinetic and potential energy operators affect the wavefunction of a quantum system?

Kinetic and potential energy operators are both used to describe the behavior of a quantum system. The kinetic energy operator determines how the wavefunction changes in space, while the potential energy operator determines how the wavefunction changes in time.

5. Can the commutation of kinetic and potential energy operators change over time?

Yes, the commutation of kinetic and potential energy operators can change over time as the potential energy of a system changes. This is because the potential energy operator is dependent on the position of the system, which can vary over time.

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