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1. (15 points) Commutator exercise. Consider the angular momentum operator: L_i, where i is equal x, y, and z. For example: L_x is the x component of the angular momentum, etc... In the textbook (3rd edition), equation 6.33 gives the commutation relations between L_i and V_j: [ L_i, V_j ] = i ? ?_ijk V_k (eq. 6.33) V? is what is called a “vector operator”, which is any operator with 3 components (x, y, z for example), which transforms the same way as the position operator, r = x î + y ? + z k? in three dimensions. An example is the position operator, r?, and the momentum operator, p?. For now we don’t need the details of the vector operator V?. We just need to know that L_i and V?_i satisfy (eq. 6.33). In (eq. 6.33), i is the imaginary unit, i = ??1. ?_ijk is the Levi-Civita symbol. It is given by: 1, if i j k = x y z, y z x, or z x y (cyclic permutation). Case (A) ?1, if any two indices above are interchanged (anti-cyclic permutation). Case (B) 0, otherwise. Case (C) We will define the operator: V_± = V_x ± i V_y. Using the above information, show that the following is true: (a) [ L_z, V_z ] = 0. (b) [ L_z, V_± ] = ± ? V_±. (c) [ L_±, V_z ] = 0. (d) [ L_±, V_± ] = ? ? V_±. (e) [ L_+, V_- ] = ± 2 ? V_z.

          1. (15 points) Commutator exercise. Consider the angular momentum operator: L_i, where i is equal x, y, and z. For example: L_x is the x component of the angular momentum, etc...

In the textbook (3rd edition), equation 6.33 gives the commutation relations between L_i and V_j:

[ L_i, V_j ] = i ? ?_ijk V_k   (eq. 6.33)

V? is what is called a “vector operator”, which is any operator with 3 components (x, y, z for example), which transforms the same way as the position operator, r = x î + y ? + z k? in three dimensions. An example is the position operator, r?, and the momentum operator, p?. For now we don’t need the details of the vector operator V?. We just need to know that L_i and V?_i satisfy (eq. 6.33).

In (eq. 6.33), i is the imaginary unit, i = ??1. ?_ijk is the Levi-Civita symbol. It is given by:

1, if i j k = x y z, y z x, or z x y (cyclic permutation). Case (A)
?1, if any two indices above are interchanged (anti-cyclic permutation). Case (B)
0, otherwise. Case (C)

We will define the operator: V_± = V_x ± i V_y. Using the above information, show that the following is true:

(a) [ L_z, V_z ] = 0.
(b) [ L_z, V_± ] = ± ? V_±.
(c) [ L_±, V_z ] = 0.
(d) [ L_±, V_± ] = ? ? V_±.
(e) [ L_+, V_- ] = ± 2 ? V_z.

1. (15 points) Commutator exercise. Consider the angular momentum operator: Li, where i is equal x, y, and z. For example: Lx is the x component of the angular momentum, etc...

In the textbook (3rd edition), equation 6.33 gives the commutation relations between Li and Vj:

[ Li, Vj ] = i ? ?ijk Vk (eq. 6.33)

V? is what is called a “vector operator”, which is any operator with 3 components (x, y, z for example), which transforms the same way as the position operator, r = x î + y ? + z k? in three dimensions. An example is the position operator, r?, and the momentum operator, p?. For now we don’t need the details of the vector operator V?. We just need to know that Li and V?i satisfy (eq. 6.33).

In (eq. 6.33), i is the imaginary unit, i = ??1. ?ijk is the Levi-Civita symbol. It is given by:

1, if i j k = x y z, y z x, or z x y (cyclic permutation). Case (A)
?1, if any two indices above are interchanged (anti-cyclic permutation). Case (B)
0, otherwise. Case (C)

We will define the operator: V± = Vx ± i Vy. Using the above information, show that the following is true:

(a) [ Lz, Vz ] = 0.
(b) [ Lz, V± ] = ± ? V±.
(c) [ L±, Vz ] = 0.
(d) [ L±, V± ] = ? ? V±.
(e) [ L+, V- ] = ± 2 ? Vz.

Added by Jessica E.

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