Commutation of (L^2)op and (Lz)op

In summary, the conversation discusses the non-commutativity of the operators (Lx)op and (Ly)op, as well as the equation they satisfy: (Lx)op(Ly)op - (Ly)op(Lx)op = i(hbar)(Lz)op. The question then asks for a proof that (L2)op(Lz)op = (Lz)op(L2)op, using the fact that (AB)C = A(BC). The solution provided by the person asking the question contains an error in the first three lines, where they set L_z = L_x L_y - L_y L_x, instead of the correct expression of L_z = \frac{1
  • #1
metgt4
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0

Homework Statement



It has been shown that the operators (Lx)op and (Ly)op do not commute but satisfy the following equation:

(Lx)op(Ly)op - (Ly)op(Lx)op = i(hbar)(Lz)op

(a) Use this relation and the two similar equations obtained by cycling the coordinate labels to show that (L2)op(Lz)op = (Lz)op(L2)op, that is, these two operators commute. [Hint: You do not need to introduce the differential formulas for the operators. Use the fact that (AB)C = A(BC) where A, B, and C are operators]


The question continues, but this is the part I am having trouble with.



Homework Equations



Relevant equations are included in the question

The Attempt at a Solution



My attempted solution is attached (Scan0004.jpg). I work it out, but I'm going wrong somewhere as I'm finding that they do not commute as they should.



Thank you in advance for any help!

Andrew
 

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  • #2
This is a rather convoluted way to do this, so I'm not going to try to pick apart all of it. But I can tell you that you have a significant error in the very first three lines. You have attempted to set

[tex]L_z = L_x L_y - L_y L_x[/tex]

when the correct expression is

[tex]L_z = \frac{1}{i \hbar} (L_x L_y - L_y L_x)[/tex]

Now, if you like, you can set [itex]\hbar = 1[/itex], as this amounts to just choosing a special system of units. However, you cannot set [itex]i = 1[/itex]. The imaginary unit i is crucial to getting the cancellation you need.
 

Related to Commutation of (L^2)op and (Lz)op

What is commutation of (L^2)op and (Lz)op?

Commutation of (L^2)op and (Lz)op refers to the mathematical operation of finding the commutator between the operator for the total angular momentum squared (L^2) and the operator for the z-component of the angular momentum (Lz). This operation is often used in quantum mechanics to determine the simultaneous eigenstates of these two operators.

Why is the commutation of (L^2)op and (Lz)op important?

The commutation of (L^2)op and (Lz)op is important because it allows us to determine the simultaneous eigenstates of these two operators, which are fundamental properties in quantum mechanics. These eigenstates represent the quantized values of the total angular momentum and its z-component, which are essential for understanding the behavior of particles at the atomic and subatomic levels.

How is the commutator of (L^2)op and (Lz)op calculated?

The commutator between (L^2)op and (Lz)op is calculated by taking the product of these two operators and subtracting the product in the reverse order. Mathematically, this can be written as [L^2, Lz] = L^2Lz - LzL^2. The resulting commutator will depend on the specific form of the operators and can be solved using mathematical techniques such as angular momentum algebra.

What is the physical significance of the commutator of (L^2)op and (Lz)op?

The commutator of (L^2)op and (Lz)op has physical significance because it represents the uncertainty in the simultaneous measurement of the total angular momentum and its z-component. In quantum mechanics, the commutator of two operators is related to their uncertainty through the Heisenberg uncertainty principle. Therefore, the commutator of (L^2)op and (Lz)op provides important information about the uncertainty in the angular momentum of a particle.

How does the commutation of (L^2)op and (Lz)op relate to the Schrödinger equation?

The commutation of (L^2)op and (Lz)op is related to the Schrödinger equation through the eigenvalue problem. The simultaneous eigenstates of these two operators satisfy the eigenvalue equation (L^2)op|lm> = l(l+1)|lm> and (Lz)op|lm> = m|lm>, where l and m are the quantum numbers for the total angular momentum and its z-component, respectively. These eigenvalue equations can then be used to solve the time-independent Schrödinger equation for a particle with a given potential.

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