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Hamiltonian problem: observables

by noblegas
Tags: hamiltonian, observables
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noblegas
#1
Oct13-09, 07:26 PM
P: 386
1. The problem statement, all variables and given/known data


A particle that moves in 3 dimensions has that Hamiltonian

[tex] H=p^2/2m+\alpha*(x^2+y^2+z^2)+\gamma*z[/tex] where [tex] \alpha[/tex] and [tex]\gamma[/tex] are real nonzero constant numbers.

a) For each of the following observables , state whether or why the observable is conserved: parity , [tex]\Pi[/tex]; energy [tex]H[/tex] ; the z component of orbital angular momentum , [tex] L_z[/tex] ; the x component of orbital angular momentum , [tex] L_x[/tex] , the z componetm of the linear momentum [tex]p_z[/tex]

b) Reduce the expression for the time rate of change of the expectation value of the y component of orbital angular momentum , [tex]d<L_y>/dt[/tex] , to the simplest possible form. Find the classical analog to the result.

2. Relevant equations



3. The attempt at a solution

a) parity: [tex] \Pi \phi(r)=\phi(-r); [/tex] Have to show that H(r)=H(-r)

x -> -x
y-> -y
z -> -z

therefore , [tex] H=p^2/2m+\alpha*(x^2+y^2+z^2)+\gamma*z, H(-r)=] H=p^2/2m+\alpha*(-x)^2+(-y)^2+(-z)^2)+\gamma*(-z)=] H=p^2/2m+\alpha*(x^2+y^2+z^2)-\gamma*z[/tex]

observable for parity is not conserved since H(r) and H(-r) are not equal to each other.

For energy, I don't know how to show that the observable is observed, other than stating the Law of energy conservation, which I know thats not what you do; Same goes for the rest of the observables Should I take the commutator of : [H, L_z] , [H,L_x], [H,p_z]?

b) [tex] <L_y>=\varphi^2*L_ydy[/tex]. What do I set [tex] \varphi[/tex] equal to?
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kuruman
#2
Oct13-09, 09:25 PM
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What is true when an operator commutes with the Hamiltonian?
noblegas
#3
Oct13-09, 10:24 PM
P: 386
Quote Quote by kuruman View Post
What is true when an operator commutes with the Hamiltonian?
it equals zero

kuruman
#4
Oct14-09, 07:47 AM
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Hamiltonian problem: observables

Yes, that is true, but there are other things that are also true when an operator commutes with the Hamiltonian. What are they? What does your textbook say?
noblegas
#5
Oct14-09, 08:17 AM
P: 386
Quote Quote by kuruman View Post
Yes, that is true, but there are other things that are also true when an operator commutes with the Hamiltonian. What are they? What does your textbook say?
[tex]\Pi*p^2[/tex]=[tex]-p*\Pi*p=p^2*\Pi[/tex]?

[tex]\Pi*r=-r*\Pi[/tex]
[tex]\Pi*\varphi(r)=\varphi(-r)=-\varphi(r)[/tex] if [tex]\pi=+1[/tex], [tex]\Pi*\varphi(r)=\varphi(-r)=\varphi(r)[/tex] if [tex]\pi=-1[/tex]
kuruman
#6
Oct14-09, 09:00 AM
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Quote Quote by noblegas View Post
[tex]\Pi*p^2[/tex]=[tex]-p*\Pi*p=p^2*\Pi[/tex]?

[tex]\Pi*r=-r*\Pi[/tex]
[tex]\Pi*\varphi(r)=\varphi(-r)=-\varphi(r)[/tex] if [tex]\pi=+1[/tex], [tex]\Pi*\varphi(r)=\varphi(-r)=\varphi(r)[/tex] if [tex]\pi=-1[/tex]
Is that what your textbook says?
noblegas
#7
Oct14-09, 10:17 AM
P: 386
Quote Quote by kuruman View Post
Is that what your textbook says?
yes. should I calculated the commutators I listed in my OP
kuruman
#8
Oct14-09, 12:44 PM
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Is there nothing in your textbook about what is true when an operator commutes with the Hamiltonian? This is a rather important result in Quantum Mechanics.
noblegas
#9
Oct14-09, 05:47 PM
P: 386
Quote Quote by kuruman View Post
Is there nothing in your textbook about what is true when an operator commutes with the Hamiltonian? This is a rather important result in Quantum Mechanics.
Sorry I don't know what else to say: When an operator commutes with a hamiltonian, [tex][H,\Pi]=0[/tex] My book says that [tex]H[/tex] and [tex]\Pi[/tex] are simultaneous eigenfunctions of [tex]H[/tex] and [tex]\Pi[/tex]
kuruman
#10
Oct14-09, 09:04 PM
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Yes, when an operator commutes with the Hamiltonian the energy eigenfunctions are also eigenfunctions of the operator. That's the first thing. The second thing is that the observable represented by the operator is a constant of the motion, i.e. is conserved. Do you see what you ought to do now?


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