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Old May 6th 2008, 12:25 PM   #1
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Expectation Value

..........

Last edited by nairbdm; May 8th 2008 at 02:07 PM.
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Old May 6th 2008, 01:24 PM   #2
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Originally Posted by nairbdm View Post
Say you have an arbitraty vector:
/w> = {a
b
c}
where a, b, c are complex numbers. Calculate the "expectation value"
<M>w = <w/M/w>

and show that <M>w is always a real number, no matter what a,b,c are. Hint - Dont forget how to get the dual vector, <w/.

Thanks.
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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Old May 6th 2008, 08:02 PM   #3
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Originally Posted by nairbdm View Post
Say you have an arbitraty vector:
/w> = {a
b
c}
where a, b, c are complex numbers. Calculate the "expectation value"
<M>w = <w/M/w>

and show that <M>w is always a real number, no matter what a,b,c are. Hint - Dont forget how to get the dual vector, <w/.

Thanks.
Originally Posted by nairbdm View Post
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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