QM: Linear momentum of angular momentum eigenstate

Using the commutation relations and the given equations, we get -i\hbar\langle l' m' \lvert P_x \rvert l m \rangle = \hbar^2\sqrt{l(l+1)-m(m+1)}\langle l' m' \lvert P_x \rvert l (m+1) \rangle + \hbar^2\sqrt{l(l+1)-m(m-1)}\langle l' m' \lvert P_x \rvert l (m-1) \rangle.Since m' \neq m \pm 1, the only way for the right hand side to be equal to zero is if both terms are equal to zero.
  • #1
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Homework Statement


Find [Lz, Px] and [Lz,Py] and use this to show that [tex]\langle l'm'|P_x|lm\rangle = 0[/tex] for [tex]m' \neq m \pm 1.[/tex]


Homework Equations


[tex]L_z|lm\rangle = \hbar m |lm\rangle[/tex]
[tex]L^2|lm\rangle = \hbar^2 l(l+1)|lm\rangle [/tex]
[tex]L_{\pm}|lm\rangle = \hbar \sqrt{l(l+1)-m(m\pm 1)}|l,m\pm 1 \rangle[/tex]


The Attempt at a Solution


It was easy to show that [tex][L_z,P_x] = i \hbar P_y[/tex] and that [tex][L_z,P_y] = - i\hbar P_x[/tex] but how might I use this to show that [tex]\langle l'm'|P_x|lm\rangle = 0[/tex] for [tex]m' \neq m \pm 1?[/tex]
 
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  • #2
Start with [itex]-i\hbar\langle l' m' \lvert P_x \rvert l m \rangle = \langle l' m' \lvert [L_z,P_y] \rvert l m \rangle[/itex].
 

1. What is the difference between linear momentum and angular momentum?

Linear momentum is the product of an object's mass and its velocity in a straight line. Angular momentum, on the other hand, is the product of an object's moment of inertia and its angular velocity around a fixed axis.

2. What is a quantum mechanical eigenstate?

An eigenstate is a quantum state that represents a single, definite value for a particular observable, such as linear or angular momentum. In other words, the state is not a superposition of multiple possible values, but rather a single, specific value.

3. How is linear momentum of an eigenstate determined in quantum mechanics?

In quantum mechanics, the linear momentum of an eigenstate is determined by the operator p, which is defined as the rate of change of the wavefunction with respect to position. The expectation value of this operator gives the average linear momentum of the eigenstate.

4. What is the significance of the quantum mechanical eigenstate in the study of linear momentum?

The quantum mechanical eigenstate is significant because it allows us to determine the precise value of linear momentum for a given system. This is essential in understanding the behavior and properties of particles at the quantum level.

5. Can an eigenstate have both non-zero linear and angular momentum?

Yes, an eigenstate can have non-zero values for both linear and angular momentum. This is known as a coupled eigenstate and is often observed in systems with multiple particles interacting with one another.

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