Q38P The Hamiltonian for a certain th... [FREE SOLUTION]

Chapter 3: Q38P (page 129)

The Hamiltonian for a certain three-level system is represented by the matrix
$H = h 蝇 [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}]$ Two other observables, A and B, are represented by the matrices $A = 位 [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 2 \end{matrix}], B = 渭 [\begin{matrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$ ,where 蝇, , and 渭 are positive real numbers.
(a)Find the Eigen values and (normalized) eigenvectors of H, A and B.
(b) Suppose the system starts out in the generic staterole="math" localid="1656040462996" $| S (0) > = (\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \end{matrix})$
with ${| c_{1} |}^{2} + {| c_{2} |}^{2} + {| c_{3} |}^{2} = 1$ . Find the expectation values (at t=0) of H, A, and B.
(c) What is $| S (t) >$ ? If you measured the energy of this state (at time t), what values might you get, and what is the probability of each? Answer the same questions for A and for B.

Short Answer

Expert verified

(a) Eigen values and Eigen vectors of H, A and B

$| h_{1} > = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), | h_{2} > = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), | h_{3} > = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) . | a_{1} > = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}), | a_{2} > = \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}), | a_{3} > = \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}) . | b_{1} > = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), | b_{2} > = \frac{1}{\sqrt{2}} (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}), | b_{3} > = \frac{1}{\sqrt{2}} (\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix}) .$

(b) Expectation values of H, A and B

$< H > = h 蝇 ({|c_{1}|}^{2} + 2 {|c_{2}|}^{2} + 2 {|c_{3}|}^{2}) . < A > = 位 (c_{1}^{*} c_{2} + c_{2}^{*} c_{1} + 2 {|c_{3}|}^{2}) . < B > = 渭 ({|c_{1}|}^{2} + c_{2}^{*} c_{3} + c_{3}^{*} c_{2}) .$

Step by step solution

(a) Finding the Eigen values and Eigen vectors of H, A and B

$E_{1} = h 蝇, E_{2} = E_{3} = 2 h 蝇; | h_{1} > = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), | h_{2} > = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), | h_{3} > = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) .$

localid="1656041551772" $|\begin{matrix} - a & 位 & 0 \\ 位 & - a & 0 \\ 0 & 0 & (2 位 - a) \end{matrix}| = a^{2} (2 位 - a) - (2 位 - a) 位^{2} = 0 鈬� a_{1} = 2 位, a_{2} = 位, a_{3} = - 位位 (\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 2 \end{matrix}) (\begin{matrix} 伪 \\ 尾 \\ 纬 \end{matrix}) = a (\begin{matrix} 伪 \\ 尾 \\ 纬 \end{matrix}) 鈬� \{\begin{cases} 位尾 = a 伪 \\ 位伪 = a 尾 \\ 2 位纬 = a 纬 \end{cases} .$

(1)

$\begin{matrix} 位尾 \\ 位伪 \\ 2 位纬 \end{matrix} \begin{matrix} = 2 位伪鈬� 尾 = 2 伪 \\ 2 位尾 \\ 2 位纬 \end{matrix}\} 伪 = 尾 = 0; | a_{1} > = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) .$

(2)

localid="1656042243144" $\begin{array}{r} \begin{matrix} 位尾 = 位伪鈬� 尾 = 伪 \\ 位伪 = 位尾鈬� 伪 = 尾 \\ 2 位纬 = 位纬; 鈬� 纬 = 0 \end{matrix} \end{array}\}; | a_{2} > = \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) .$

(3)

$\begin{array}{r} \begin{matrix} 位尾 = - 位伪鈬� 尾 = - 伪 \\ 位伪 = - 位尾鈬� 伪 = - 尾 \\ 2 位纬 = - 位纬; 鈬� 纬 = 0 \end{matrix} \end{array}\}; | a_{3} > = \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}) .$

$|\begin{matrix} (2 渭 - b) & 0 & 0 \\ 0 & - b & 渭 \\ 0 & 渭 & - b \end{matrix}| = b^{2} (2 渭 - b) - (2 渭 - b) 渭^{2} = 0 鈬� b_{1} = 2 渭, b_{2} = 渭, b_{2} = - 渭 . 渭 (\begin{matrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}) (\begin{matrix} 伪 \\ 尾 \\ 纬 \end{matrix}) = b (\begin{matrix} 伪 \\ 尾 \\ 纬 \end{matrix}) 鈬� \{\begin{cases} 2 渭伪 = b 伪 \\ 渭纬 = b 尾 \\ 渭尾 = b 纬 \end{cases} .$

(1)

role="math" localid="1656043183873" $\begin{array}{r} 2 渭伪 = 2 渭伪 \\ 渭纬 = 2 渭尾鈬� 纬 = 2 尾 \\ 渭尾 = 2 渭纬鈬� 尾 = 2 纬 \end{array}\} 尾 = 纬 = 0; | b_{1} > = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) .$

(2)

$\begin{array}{r} 2 渭伪 = 渭伪鈬� 伪 = 0 \\ 渭纬 = 渭尾鈬� 纬 = 尾 \\ 渭尾 = 渭纬; 鈬� 尾 = 纬 \end{array}\}; | b_{1} > = \frac{1}{\sqrt{2}} (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}) .$

(3)

$\begin{array}{r} 2 渭伪 = - 渭伪鈬� 伪 = 0 \\ 渭纬 = - 渭尾鈬� 纬 = - 尾 \\ 渭尾 = - 渭纬; 鈬� 尾 = - 纬 \end{array}\}; | b_{1} > = \frac{1}{\sqrt{2}} (\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix}) .$

Step2: (b) Finding the expectation values of H, A and B

$<H> = <S (0) |H| S (0)> = h 蝇 (c_{1}^{*} c_{2}^{*} c_{3}^{*}) (\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}) (\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \end{matrix}) = h 蝇 ({|c_{1}|}^{2} + {|c_{2}|}^{2} + {|c_{3}|}^{2}) . <A> = <S (0) |A| S (0)> = 位 (c_{1}^{*} c_{2}^{*} c_{3}^{*}) (\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 2 \end{matrix}) (\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \end{matrix}) = 位 (c_{1}^{*} c_{2} + c_{2}^{*} c_{1} + 2 {|c_{3}|}^{2}) . <B> = <S (0) |B| S (0)> = 渭 (c_{1}^{*} c_{2}^{*} c_{3}^{*}) (\begin{matrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}) (\begin{matrix} c_{1} \\ c_{2} \\ c_{3} \end{matrix}) = 渭 (2 {|c_{1}|}^{2} + c_{2}^{*} c_{3} + c_{3}^{*} c_{2}) .$

(c) Finding and the probabilities of H, A and B

$| S (0) > = c_{1} | h_{1} > + c_{2} | h_{2} > + c_{3} | h_{3} > 鈬� | S (t) > = c_{1} e^{- i E_{1} t I h} | h_{1} > c_{2} e^{- i E_{2} t I h} + | h_{2} > c_{3} e^{- i E_{3} t I h} + | h_{3} > = c_{1} e^{- i 蝇 t} | h_{1} > c_{2} e^{- 2 i 蝇 t} + | h_{2} > c_{3} e^{- 2 i 蝇 t} + | h_{3} > .$

$= e^{- 2 i 蝇 t} [c_{1} e^{i 蝇 t} (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + c_{2} (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) + c_{3} (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})] = e^{- 2 i 蝇 t} (\begin{matrix} c_{1} e^{i 蝇 t} \\ c_{2} \\ c_{3} \end{matrix}) .$

$H : : h_{1} = h 蝇,$ probability ${|c_{1}|}^{2}; h_{2} = h_{3} = 2 h 蝇$ , probability $({|c_{2}|}^{2} + {|c_{3}|}^{2}) .$

$A : a_{1} = 2, <a_{1} | S (t)> = e^{- 2 i 蝇 t} (0 0 1) (\begin{matrix} c_{1} e^{i 蝇 t} \\ c_{2} \\ c_{3} \end{matrix}) = e^{- 2 i 蝇 t} c_{3} 鈬� probability {|c_{3}|}^{2} a_{2} = 位 / 2, <a_{1} | S (t)> = e^{- 2 i蝇t} \frac{1}{\sqrt{2}} (1 1 0) (\begin{matrix} c_{1} e^{i蝇t} \\ c_{2} \\ c_{3} \end{matrix}) = \frac{1}{\sqrt{2}} e^{- 2 i蝇t} (c_{1} e^{i蝇t} + c_{2}) 鈬�$

localid="1656045665650" $probability = \frac{1}{2} (c_{1}^{*} e^{- i 蝇 t} + c_{2}) (c_{1} e^{- i 蝇 t} + c_{2}) = \frac{1}{2} ({|c_{1}|}^{2} + {|c_{2}|}^{2} + c_{1}^{*} c_{2} e^{- i 蝇 t} + c_{2}^{*} c_{1} e^{i 蝇 t}) . a_{3} = - 位 / 3, <a_{3} | S (t)> = e^{- 2 i 蝇 t} \frac{1}{\sqrt{2}} (1 - 1 0) (\begin{matrix} c_{1} e^{i 蝇 t} \\ c_{2} \\ c_{3} \end{matrix}) = \frac{1}{\sqrt{2}} (c_{1} e^{i 蝇 t} - c_{2}) 鈬� probability = \frac{1}{2} (c_{1}^{*} e^{- i 蝇 t} - c_{2}^{*}) (c_{1} e^{- i 蝇 t} - c_{2}) = \frac{1}{2} ({|c_{1}|}^{2} + {|c_{2}|}^{2} - c_{1}^{*} c_{2} e^{- i 蝇 t} + c_{2}^{*} c_{1} e^{i 蝇 t}) .$

Note that the sum of the probabilities is 1.

$b_{1} = 渭 / 2, <b_{1} | S (t)> = e^{- 2 i 蝇 t} \frac{1}{\sqrt{2}} (1 0 0) (\begin{matrix} c_{1} e^{i 蝇 t} \\ c_{2} \\ c_{3} \end{matrix}) = e^{i 蝇 t} - c_{1} probability {|c_{1}|}^{2} b_{2} = 渭 / 2, <b_{2} | S (t)> = e^{- 2 i 蝇 t} \frac{1}{\sqrt{2}} (0 1 1) (\begin{matrix} c_{1} e^{i 蝇 t} \\ c_{2} \\ c_{3} \end{matrix}) = \frac{1}{\sqrt{2}} e^{i 蝇 t} (c_{2} + c_{3}) 鈬� probability = \frac{1}{2} (c_{2}^{*} - c_{3}^{*}) (c_{2} + c_{3}) = \frac{1}{2} ({|c_{2}|}^{2} + {|c_{3}|}^{2} + c_{2}^{*} c_{3} + c_{3}^{*} c_{2}) . b_{3} = - 渭 / 3, <b_{3} | S (t)> = e^{- 2 i 蝇 t} \frac{1}{\sqrt{2}} (0 1 - 1) (\begin{matrix} c_{1} e^{i 蝇 t} \\ c_{2} \\ c_{3} \end{matrix}) = \frac{1}{\sqrt{2}} e^{- 2 i 蝇 t} (c_{2} - c_{3}) 鈬� probability = \frac{1}{2} (c_{2}^{*} - c_{3}^{*}) (c_{2} - c_{3}) = \frac{1}{2} ({|c_{2}|}^{2} + {|c_{3}|}^{2} - c_{2}^{*} c_{3} - c_{3}^{*} c_{2}) .$

Again, the sum of the probabilities is 1.

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Short Answer

Step by step solution

(a) Finding the Eigen values and Eigen vectors of H, A and B

Step2: (b) Finding the expectation values of H, A and B

(c) Finding and the probabilities of H, A and B

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