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I've been trying to form a proof using , using majorly dirac notation.There has been claims that its much better to use in QM.

The question i wanted to generally show that the expected value is Zero for all odd energy levels.I believe i have solved the question but im a bit Iffy about a step i took:

<x>

for a given Ψ

Energy eigen functions have definite parity, assume for all odd n's if one is zero the rest should follow.

Take n = 1

=> L = <(A

= (n!)

B = <(A

Define A

=> <Lx+iC|x|Lx+iC>

(Bit iffy after these steps)

= <Lx|x|Lx|> + <iC|x|iC>

= <L|x

as ∫x

=> 0

we find B=0

therefore

<x>

For odd ns.

The question i wanted to generally show that the expected value is Zero for all odd energy levels.I believe i have solved the question but im a bit Iffy about a step i took:

<x>

_{n}= <Ψ_{n}|x|Ψ_{n}> = Lfor a given Ψ

_{n}= (A^{+})^{n}(n!)^{-2}Energy eigen functions have definite parity, assume for all odd n's if one is zero the rest should follow.

Take n = 1

=> L = <(A

^{+})(n!)^{-2}|x|(A^{+})(n!)^{-2}>= (n!)

^{-1}<(A^{+})|x|(A^{+})>B = <(A

^{+})|x|(A^{+})>Define A

^{+}= Lx + iC : B,C are Real=> <Lx+iC|x|Lx+iC>

(Bit iffy after these steps)

= <Lx|x|Lx|> + <iC|x|iC>

= <L|x

^{3}|L>+<C|x|C>as ∫x

^{2n+1}dx for limits [-∞,∞] and n =0,1,2,3....=> 0

we find B=0

therefore

<x>

_{n}= 0For odd ns.

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