Physics Help Forum Expectation Value

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 May 6th 2008, 12:25 PM #1 Junior Member   Join Date: May 2008 Posts: 6 Expectation Value .......... Last edited by nairbdm; May 8th 2008 at 02:07 PM.
May 6th 2008, 01:24 PM   #2

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 Originally Posted by nairbdm Say you have an arbitraty vector: /w> = {a b c} where a, b, c are complex numbers. Calculate the "expectation value" w = and show that w is always a real number, no matter what a,b,c are. Hint - Dont forget how to get the dual vector,
I'm sorry, but I'm just drawing a complete blank here. What is the operator M? (Or is it simply an arbitrary Hermitian operator?)

-Dan
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May 6th 2008, 08:02 PM   #3

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 Originally Posted by nairbdm Say you have an arbitraty vector: /w> = {a b c} where a, b, c are complex numbers. Calculate the "expectation value" w = and show that w is always a real number, no matter what a,b,c are. Hint - Dont forget how to get the dual vector,
 Originally Posted by nairbdm I'm sorry, M is the 3X3 matrix: M= {1,-i(2^(1/2)),0} {i(2^(1/2)),0,0} {0,0,2} Thanks.
$\displaystyle <M> = <w|M|w> = < a^* b^* c^* | \left ( \begin{matrix} 1 & -i\sqrt{2} & 0 \\ i\sqrt{2} & 0 & 0 \\ 0 & 0 & 2 \end{matrix} \right ) \left | \begin{matrix} a \\ b \\ c \end{matrix} \right >$

$\displaystyle = < a^* b^* c^* | \left | \begin{matrix} a - ib\sqrt{2} \\ ia\sqrt{2} \\ 2c \end{matrix} \right >$

$\displaystyle = a^*(a - ib\sqrt{2}) + b^*(ia\sqrt{2}) + c^*2c$

$\displaystyle = a^*a + 2c^*c + i(ab^* - a^*b)\sqrt{2}$

Now, given any two complex numbers a and b, the number $\displaystyle ab^* - a^*b$ will be pure imaginary. Thus the expectation value is a real number.

-Dan
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