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Midterm 5 Solutions

# Midterm 5 Solutions - Intermediate QMI Chapter 5 Midterm...

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Intermediate QMI Chapter 5 Midterm Solutions Problem 1 Given the Hamiltonian for two spin- 1 2 particles H ` = 2 A 2 S ` 1 ÿ S ` 2 + w 0 J S ` 1 z - S ` 2 z N where A > 0, w 0 ¥ 0 . ü Part a ü Determining the matrix representation of H ` in the S ` z ƒ S ` z basis There are two ways of doing this, both of which are outlined below Method 1 See lecture note 5. Recall that using S ` i + = S ` i x + Â S ` i y S ` i - = S ` i x - Â S ` i y ñ S ` i x = 1 2 J S ` i + + S ` i - N S ` i y = 1 2 Â J S ` i + - S ` i - N gives S ` 1 x S ` 2 x = 1 4 J S ` 1 + + S ` 1 - N J S ` 2 + + S ` 2 - N = 1 4 J S ` 1 + S ` 2 + + S ` 1 + S ` 2 - + S ` 1 - S ` 2 + + S ` 1 - S ` 2 - N S ` 1 y S ` 2 y = - 1 4 J S ` 1 + - S ` 1 - N J S ` 2 + - S ` 2 - N = - 1 4 J S ` 1 + S ` 2 + - S ` 1 + S ` 2 - - S ` 1 - S ` 2 + + S ` 1 - S ` 2 - N where terms have been lined up so that we can see cancellations. Thus S ` 1 ÿ S ` 2 = S ` 1 x S ` 2 x + S ` 1 y S ` 2 y + S ` 1 z S ` 2 z = 1 2 J S ` 1 + S ` 2 - + S ` 1 - S ` 2 + N + S ` 1 z S ` 2 z Because we know the action of the rasing and lowering operators on the S ` z eigenstates S ` + + z ^ = 0 S ` + - z ^ = + z ^ S ` - + z ^ = - z ^ S ` - - z ^ = 0 we may evaluate all matrix elements as Z a , b H ` c , d ^ = A 2 X a , b BJ S ` 1 + S ` 2 - + S ` 1 - S ` 2 + N H 1 L + 2 S ` 1 z S ` 2 z F H 2 L c , d \ + w 0 X a , b J S ` 1 z - S ` 2 z N H 3 L c , d \ Printed by Mathematica for Students

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where a , b , c , d oe ± z . We see that (1) will only contribute to off-diagonal elements in the matrix and will map states like a , a \ to zero, as a + z \ ( - z \ ) will get hit with a S ` + ( S ` - ) in each term of (1) this case. (2) and (3) will only contribute to the diagonal elements. Thus, our matrix representation will have the structure H ` Ø S ` z Ä⊗ S ` z basis 1 2 3 4 1 2 3 4 - 0 0 0 0 - - 0 0 - - 0 0 0 0 - for 1 Ø + z , + z \ 2 Ø + z , - z \ 3 Ø - z , + z \ 4 Ø - z , - z \ We calculate the diagonal elements by substituting each of the S ` i z in (2) and (3) with their corresponding eigenvalues and ignoring (1) to obtain H ` Ø S ` z Ä⊗ S ` z basis A ê 2 0 0 0 0 w 0 - H A ê 2 L _ 0 0 _ - w 0 - H A ê 2 L 0 0 0 0 A ê 2 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 S ` z Ä⊗ S ` z basis A ê 2 0 0 0 0 w 0 - H A ê 2 L A 0 0 A - w 0 - H A ê 2 L 0 0 0 0 A ê 2 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 Ä⊗ B = a 00 B a 01 B a 10 B a 11 B = a 00 b 00 a 00 b 01 a 01 b 00 a 01 b 01 a 00 b 10 a 00 b 11 a 01 b 10 a 01 b 11 a 10 b 00 a 10 b 01 a 11 b 00 a 11 b 01 a 10 b 10 a 10 b 11 a 11 b 10 a 11 b 11 where A = K a 00 a 01 a 10 a 11 O B = b 00 b 01 b 10 b 11 Using S ` x Ø S ` z basis 2 0 1 1 0 S ` y Ø S ` z basis 2 0 Â 0 S ` z Ø S ` z basis 2 1 0 0 - 1 gives S ` 1 x Ä⊗ S ` 2 x Ø S ` z Ä⊗ S ` z basis 2 4 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 2 MidtermExam5Solutions.nb Printed by Mathematica for Students
S ` 1 y Ä⊗ S ` 2 y Ø S ` z Ä⊗ S ` z basis 2 4 0 0 0 - 1 0 0 1 0 0 1 0 0 - 1 0 0 0 S ` 1 z Ä⊗ S ` 2 z Ø S ` z Ä⊗ S ` z basis 2 4 1 0 0 0 0 - 1 0 0 0 0 - 1 0 0 0 0 1 Thus S ` 1 ÿ S ` 2 = S ` 1 x Ä⊗ S ` 2 x + S ` 1 y Ä⊗ S ` 2 y + S ` 1 z Ä⊗ S ` 2 z Ø S ` z Ä⊗ S ` z basis 2 4 1 0 0 0 0 - 1 2 0 0 2 - 1 0 0 0 0 1 Similarly S ` 1 z Ä⊗ 1 ` Ø S ` z Ä⊗ S ` z basis 2 1 0 0 0 0 1 0 0 0 0 - 1 0 0 0 0 - 1 1 ` Ä⊗ S ` 2 z Ø S ` z Ä⊗ S ` z basis 2 1 0 0 0 0 - 1 0 0 0 0 1 0 0 0 0 - 1 and S ` 1 z - S ` 2 z = S ` 1 z Ä⊗ 1 ` - 1 ` Ä⊗ S ` 2 z Ø S ` z Ä⊗ S ` z basis 2 0 0 0 0 0 2 0 0 0 0 - 2 0 0 0 0 0 so we have H ` Ø S ` z Ä⊗ S ` z basis A 2

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• Fall '13
• AkimasaMiyake
• Eigenvalue, eigenvector and eigenspace, Alice and Bob, QR algorithm, Quantum cryptography, stot, Sz Äȩ Sz

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