- #1
Somali_Physicist
- 117
- 13
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 I am a bit Iffy about a step i took:
<x>n = <Ψn|x|Ψn> = L
for a given Ψn = (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|x3|L>+<C|x|C>
as ∫x2n+1dx for limits [-∞,∞] and n =0,1,2,3...
=> 0
we find B=0
therefore
<x>n = 0
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 I am a bit Iffy about a step i took:
<x>n = <Ψn|x|Ψn> = L
for a given Ψn = (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|x3|L>+<C|x|C>
as ∫x2n+1dx for limits [-∞,∞] and n =0,1,2,3...
=> 0
we find B=0
therefore
<x>n = 0
For odd ns.
Last edited: