Quantum mechanics, commutators...

[Murka]

Hey guys,

 

I have trouble grasping Quantum Mechanics...not such a rare condition, I guess I just need a nudge in the right direction, I seem to know the basics but somehow when I try and solve the homework we are given I get stuck and unsure on where to begin.

 

Let P be an operator that commutates with the Hamiltonian HP=PH show that if |phi> is an eigenvector of H, P|phi> is also an eigenvector of H with the same eigenvalue.

 

Thanks in advance!

 

[timo]

That one's actually so straightforward that it's hard to give a hint without telling you the solution. How about this: There is not so much you can do with HP=PH and the definition of an eigenvector f of H. Try the few things you can do. Maybe you'll see something.

[Murka]

That's the thing, I'm not sure I know how to proceed, maybe my definitions are wrong?

 

From the definition of eigenvectors and eigenvalues: If Av = av, v is the eigenvecor, a is the eigenvalue of T.

So if |phi> is the eigenvector of H with the same value as P|phi> as the eigenvector then...

H|phi>=a|phi>

HP|phi>=aP|phi>

?

 

sigh...I realize those are really basic firsttimer questions, but I'm at my 3rd year at the university, and the last time I saw the word "operator" even mentioned was at my linear algebra course, which was at the 2nd semester of the 1st year...I've done all my math courses a long time ago

[Murka]

Anyone?

[elfmotat]

H|phi>=a|phi>

HP|phi>=aP|phi>

?

 

What's the question mark for? You just proved it!

[timo]

I don't really see a proof, yet.

 

That's the thing, I'm not sure I know how to proceed, maybe my definitions are wrong?

 

From the definition of eigenvectors and eigenvalues: If Av = av, v is the eigenvecor, a is the eigenvalue of T.

So if |phi> is the eigenvector of H with the same value as P|phi> as the eigenvector then...

H|phi>=a|phi>

HP|phi>=aP|phi>

?

 

sigh...I realize those are really basic firsttimer questions, but I'm at my 3rd year at the university, and the last time I saw the word "operator" even mentioned was at my linear algebra course, which was at the 2nd semester of the 1st year...I've done all my math courses a long time ago

First of all: Don't worry if the question is a problem for you. Such "proof that ..." questions are, in my opinion, all about experience and perhaps a bit of endurance. They have little to do with physics skill.

 

To your question: What you want to show is [math]HP | \phi \rangle = a P | \phi \rangle[/math] knowing that [math] H | \phi \rangle = a | \phi \rangle[/math] and HP=PH. Perhaps an extra hint: Numbers commute with operators.

 

Finally, as you may have noted already this forum is not exactly about getting answers very quickly. Members here live on different continents and many, especially those who can answer questions about operators, have jobs: I've been sleeping for the last eight hours and now go to my office for the next twelve.

Edited April 23, 2013 by timo

[Murka]

Thank you everyone, timo I actually realized that cobstants commute with operators just before you posted it, thank you very much, as with many things, I have a feeling that the question is too simple thus I have overthought and overcomplicated things myself...

 

Thus I present my proof

H|psi>=a|psi> /*P()

PH|psi>=Pa|psi>

Use PH=HP as per given data

Use Pa=aP, can be moved because a is a constant

H(P|psi>)=a(P|psi>)

Q.E.D.

[elfmotat]

Thank you everyone, timo I actually realized that cobstants commute with operators just before you posted it, thank you very much, as with many things, I have a feeling that the question is too simple thus I have overthought and overcomplicated things myself...

 

Thus I present my proof

H|psi>=a|psi> /*P()

PH|psi>=Pa|psi>

Use PH=HP as per given data

Use Pa=aP, can be moved because a is a constant

H(P|psi>)=a(P|psi>)

Q.E.D.

 

There you go!