Commutation of observables

In summary, the commutator [Q1, Q2] is also a constant of the motion if the observables (operators) Q1 and Q2 are both constant of the motion for some Hamiltonian H.
  • #1
syang9
61
0
[SOLVED] commutation of observables

Homework Statement



Prove: If the observables (operators) Q1 and Q2 are both constant of the motion for some Hamiltonian H, then the commutator [Q1, Q2] is also a constant of the motion.

okay, first question.. am i being asked to prove that [[Q1, Q2], H] = 0? If so, then consider the following.. if not.. well, that sucks.

Given that they are 'constant of the motion', that means they commute with the Hamiltonian, right..? So that gives us

[tex]
\[[H,Q_1 ] = 0\]
\[[H,Q_2 ] = 0\]
[/tex]

Then maybe I can say..

[tex]
\[HQ_1 - Q_1 H = HQ_2 - Q_2 H\]
[/tex]

I can multiply through the left side by Q2 to get

[tex]
\[Q_2 HQ_1 - Q_2 Q_1 H = 0\]
[/tex]

Since Q2 is an observable, it's a hermitan operator, which allows me to move Q2 to the other side in that first term, like this..

[tex]
\[HQ_1 Q_2 - Q_2 Q_1 H = 0\]
[/tex]

which is equivalent to saying

[tex]
\[[H,Q_1 Q_2 ] = 0\]
[/tex]

Going back to what I was asked to prove (hopefully), which was:
[tex]
\[[[Q_1 Q_2 - Q_2 Q_1 ],H] = 0\]
[/tex]

Expanding that out, we have
[tex]
\[Q_1 Q_2 H - Q_2 Q_1 H - HQ_1 Q_2 + HQ_2 Q_1 = 0\]
[/tex]

but I've already proved that [H, Q1Q2] = 0, so we can rewrite this as:

[tex]

\[
({Q_1 Q_2 H - HQ_1 Q_2 }) - ({Q_2 Q_1 H - HQ_2 Q_1 }) = 0
\]

[/tex]

which we know is equivalent to
[tex]
\[[H,Q_1 Q_2 ] - [H,Q_1 Q_2 ] = 0\]
[/tex]

Since [H, Q1Q2] = 0, both sides are equal to zero, and the proof is... complete? Have I done anything wrong?
 
Last edited:
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  • #2
syang9 said:

Homework Statement


Since Q2 is an observable, it's a hermitan operator, which allows me to move Q2 to the other side in that first term, like this..

[tex]
\[HQ_1 Q_2 - Q_2 Q_1 H = 0\]
[/tex]

There might be a breakdown of logic here. It seems to me that you are also assuming that Q1 and Q2 commute, which is an unnecessary assumption. I think the proof should be easier than you are making it.
 
  • #3
Why is it unnecessary? Is what I've said incorrect?
 
  • #4
It is only incorrect if Q1 and Q2 don't commute. In general, they don't have to for the statement that you need to prove to be correct.
You can only move the Q2 through the Q1 (toward the right) if they commute. But nothing in the problem says that you should assume they commute. In fact, you can prove what you need to without this assumption.
 
  • #5
syang9 said:
Why is it unnecessary? Is what I've said incorrect?

Like Pacopag said, you are assuming that Q1 Q2 = Q2 Q1 which is not necessarily the case.

It's easy. Simply expand [[Q1,Q2],H] completely. Then use that fact that you can move H though Q1 and Q2.
 

What is the commutation of observables?

The commutation of observables refers to the mathematical operation of calculating the difference between two observables in a quantum system. This operation is essential in determining the fundamental properties of a system and predicting its behavior.

Why is the commutation of observables important?

The commutation of observables is important because it allows us to determine the uncertainty in the measurement of two observables. It also helps us understand the relationship between different properties of a system and how they affect each other.

What is the commutator and how is it related to the commutation of observables?

The commutator is a mathematical operation that represents the commutation of two operators. It is used to calculate the commutation of observables and is essential in quantum mechanics for understanding the behavior of particles.

What is the Heisenberg uncertainty principle and how is it related to the commutation of observables?

The Heisenberg uncertainty principle states that it is impossible to know both the position and momentum of a particle with absolute certainty. This principle is related to the commutation of observables because the commutator of position and momentum operators is equal to the imaginary unit multiplied by Planck's constant, which is a fundamental constant in quantum mechanics.

What are some examples of observables that commute and do not commute?

Some examples of observables that commute are position and energy, momentum and spin, and angular momentum components along different axes. Some examples of observables that do not commute are position and momentum, energy and time, and angular momentum components along the same axis.

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