Midterm 5 Solutions

2 and 3 will only contribute to the diagonal elements

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Unformatted text preview: m of (1) this case. (2) and (3) will only contribute to the diagonal elements. Thus, our matrix representation will have the structure ` H Ø ` ` Sz Äȩ Sz basis 1 2 3 4 1 0 0 0 2 0 0 3 0 0 4 0 0 0 - 1 2 3 4 for Ø Ø Ø Ø + z, +z , - z, - z, + z\ - z\ + z\ - z\ ` We calculate the diagonal elements by substituting each of the Si z in (2) and (3) with their corresponding eigenvalues and ignoring (1) to obtain ` H Ø ` ` Sz Äȩ Sz basis Aê2 0 — w0 0 0 0 - HA ê 2L _ -— w0 0 0 0 _ 0 - HA ê 2L 0 0 Aê2 Finally, we calculate the off-diagonal elements by noting that (1) maps states like a, -a\ to —2 -a, a] (only one of the terms in (1) actually contributes as the other term will annihilate the product state on both degrees of freedom). Therefore ` H Ø ` ` Sz Äȩ Sz basis Aê2 0 — w0 0 0 0 - HA ê 2L A -— w0 0 0 0 A 0 - HA ê 2L 0 0 Aê2 Method 2 ` We may also use the tensor product to calculate the matrix representation of H directly. Recall that for two matrices A and B A Äȩ B = a00 B a01 B a10 B a11 B = a00 b00 a00 b10 a10 b00 a10 b10 a00 b01 a00 b11 a10 b01 a10 b11 a01 b00 a01 b10 a11 b00 a11 b10 a01 b01 a01 b11 a11 b01 a11 b11 A=K where B= Using ` Sx — Ø2 ` Sz basis ` Sy 01 10 — Ø2 ` Sz basis 0 -Â Â0 ` Sz — Ø2 ` Sz basis gives ` ` S1 x Äȩ S2 x ` ` S1 y Äȩ S2 y ` ` S1 z Äȩ S2 z 2 — Ø ` ` 4 Sz Äȩ Sz basis —2 Ø ` ` 4 Sz Äȩ Sz basis —2 Ø ` ` 4 Sz Äȩ Sz basis 0001 0010 0100 1000 0 0 0 -1 0 01 0 0 10 0 -1 0 0 0 10 00 0 -1 0 0 0 0 -1 0 Printed by Mathematica for Students 10 0 -1 a00 a10 b00 b10 a01 O a11 b01 b11 gives ` ` S1 x Äȩ S2 x ` ` S1 y Äȩ S2 y ` ` S1 z Äȩ S2 z 2 Ø ` ` Sz Äȩ Sz basis — 4 2 — Ø ` ` 4 Sz Äȩ Sz basis —2 Ø ` ` 4 Sz Äȩ Sz basis 0 0 0 1 0 0 1 0 0 1 0 0 00 00 01 -1 0 10 0 -1 00 00 1 0 0 0 MidtermExam5Solutions.nb 0 -1 10 00 00 00 00 -1 0 01 Thus `` ` ` ` ` ` ` S1 ÿ S2 = S1 x Äȩ S2 x + S1 y Äȩ S2 y + S1 z Äȩ S2 z —2 Ø ` ` 4 Sz Äȩ Sz basis 10 0 0 -1 2 0 2 -1 00 0 Similarly ` ` S1 z Äȩ 1ཽ ` ` 1ཽ Äȩ S2 z — Ø ` ` 2 Sz Äȩ Sz basis — Ø ` ` 2 Sz Äȩ Sz basis 1 0 0 0 1 0 0 0 00 0 10 0 0 -1 0 0 0 -1 000 -1 0 0 010 0 0 -1 and ` ` ` `` ` S1 z - S2 z = S1 z Äȩ 1ཽ - 1ཽ Äȩ S2 z — Ø ` ` 2 Sz Äȩ Sz basis 0 0 0 0 00 20 0 -2 00 0 0 0 0 so we have ` H ` H Ø ` ` Sz Äȩ Sz basis Ø ` ` Sz Äȩ Sz basis 1 A0 2...
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