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**[SOLVED] commutation of observables**

**1. 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?

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