hamiltonian problem: observables


by noblegas
Tags: hamiltonian, observables
noblegas
noblegas is offline
#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?
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
kuruman
kuruman is offline
#2
Oct13-09, 09:25 PM
HW Helper
PF Gold
kuruman's Avatar
P: 3,444
What is true when an operator commutes with the Hamiltonian?
noblegas
noblegas is offline
#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
kuruman is offline
#4
Oct14-09, 07:47 AM
HW Helper
PF Gold
kuruman's Avatar
P: 3,444

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
noblegas is offline
#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
kuruman is offline
#6
Oct14-09, 09:00 AM
HW Helper
PF Gold
kuruman's Avatar
P: 3,444
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
noblegas is offline
#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
kuruman is offline
#8
Oct14-09, 12:44 PM
HW Helper
PF Gold
kuruman's Avatar
P: 3,444
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
noblegas is offline
#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
kuruman is offline
#10
Oct14-09, 09:04 PM
HW Helper
PF Gold
kuruman's Avatar
P: 3,444
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?


Register to reply

Related Discussions
electron phonon interaction hamiltonian problem Atomic, Solid State, Comp. Physics 2
hamiltonian problem concerning the simple harmonic oscillator Advanced Physics Homework 14
Hamiltonian problem Advanced Physics Homework 4
Hamiltonian matrix problem Quantum Physics 4
[Q]Hamiltonian of many body problem. Quantum Physics 1