X^4 perturbative energy eigenvalues for harmonic oscillator

AI Thread Summary
The discussion revolves around the calculation of perturbative energy eigenvalues for a harmonic oscillator, specifically addressing discrepancies in the textbook solution for the expectation value of the operator x^4. Participants identify that the correct expression should yield 2n^2 + 2n + 1 instead of the textbook's 3n^2 + 2n + 1. There are also typographical errors noted in the relevant equations, particularly regarding the treatment of terms involving creation and annihilation operators. The importance of discarding odd power terms due to their integral yielding zero is emphasized. Overall, the conversation highlights the challenges faced when published solutions contain inaccuracies, impacting the learning process.
PBTR3
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Homework Statement
Compute first order correction matrix elements ##\langle \n | x^4 |m \rangle## for a one dimensional harmonic oscillator.
Relevant Equations
##\langle \n |x^4 |m \rangle = 2 \left( \frac hbar 2m\omega \right) \left( 3n^2 +2n +1 \right)##
\
The book(Schaum) says the above is the solution but after two hours of tedious checking and rechecking I get 2n^2 in place or the 3n^2. Am I missing something or is this just a typo?
 
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PBTR3 said:
Relevant Equations:: ##\langle \n |x^4 |m \rangle = 2 \left( \frac hbar 2m\omega \right) \left( 3n^2 +2n +1 \right)##

Please correct the typographical errors on both sides of this equation.

I do find that ##\langle n |x^4 |n \rangle## is equal to a constant factor times ##\left( 2n^2 +2n +1 \right)## instead of ##\left(3n^2 +2n +1 \right)##. So, the textbook's result does not appear to be correct.

You haven't shown any work, so we cannot tell if your method of solution is correct.
 
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TSny said:
Please correct the typographical errors on both sides of this equation.

I do find that ##\langle n |x^4 |n \rangle## is equal to a constant factor times ##\left( 2n^2 +2n +1 \right)## instead of ##\left(3n^2 +2n +1 \right)##. So, the textbook's result does not appear to be correct.

You haven't shown any work, so we cannot tell if your method of solution is correct.
 
(a + a*)^4= a^4+a^3a*+a^2a*a+a^2a*2+aa*a^2+aa*aa*+aa*^2a+aa*^3+a*a^3+a*a^2a*+
a*aa*a+a*aa^2+a*^2a^2+a*^2a^2+a*^2aa*+a*^3a+a*^4
then throw out all terms with unequal powers of a and a* which are 0.
##\langle n | a^2a*^2 | m \rangle=\langle n | (m+1) (m+2) | m \rangle##
Results for the rest are similar calculations to:
(m+1)(m+1), 2 m(m+1), m^2, and m(m-1)
Adding them all up = 6m^2 + 6m +3 or 3(2m^2 + 2m + 1) or since n=m
3(2n^2 + 2n + 1)
 
PBTR3 said:
(a + a*)^4= a^4+a^3a*+a^2a*a+a^2a*2+aa*a^2+aa*aa*+aa*^2a+aa*^3+a*a^3+a*a^2a*+
a*aa*a+a*aa^2+a*^2a^2+a*^2a^2+a*^2aa*+a*^3a+a*^4
then throw out all terms with unequal powers of a and a* which are 0.
##\langle n | a^2a*^2 | m \rangle=\langle n | (m+1) (m+2) | m \rangle##
Results for the rest are similar calculations to:
(m+1)(m+1), 2 m(m+1), m^2, and m(m-1)
Adding them all up = 6m^2 + 6m +3 or 3(2m^2 + 2m + 1) or since n=m
3(2n^2 + 2n + 1)
I am still trying to figure out how to fix that relevant equation post.
 
PBTR3 said:
(a + a*)^4= a^4+a^3a*+a^2a*a+a^2a*2+aa*a^2+aa*aa*+aa*^2a+aa*^3+a*a^3+a*a^2a*+
a*aa*a+a*aa^2+a*^2a^2+a*^2a^2+a*^2aa*+a*^3a+a*^4
OK. I think the term in boldface should be a*aa*^2, but that's probably just a typographical error.

PBTR3 said:
then throw out all terms with unequal powers of a and a* which are 0.
I assume the reason why you are throwing out these terms is that you are only considering the case where m = n.

PBTR3 said:
##\langle n | a^2a*^2 | m \rangle=\langle n | (m+1) (m+2) | m \rangle##
Results for the rest are similar calculations to:
(m+1)(m+1), 2 m(m+1), m^2, and m(m-1)
Adding them all up = 6m^2 + 6m +3 or 3(2m^2 + 2m + 1) or since n=m
3(2n^2 + 2n + 1)
Looks good.

Do you get ##\large \langle n | x^4| n \rangle = \frac{3}{4} \left( \frac{\hbar}{m \omega} \right)^2 (2n^2+2n+1)##?
 
TSny said:
OK. I think the term in boldface should be a*aa*^2, but that's probably just a typographical error.I assume the reason why you are throwing out these terms is that you are only considering the case where m = n.Looks good.

Do you get ##\large \langle n | x^4| n \rangle = \frac{3}{4} \left( \frac{\hbar}{m \omega} \right)^2 (2n^2+2n+1)##?
Yes! I am very much a novice at this and when books give an answer it usually helps greatly BUT when the answer published is wrong it causes much difficulty. Yes that is my typo. I am not perfect either. Also I throw out the odd powers because the integral they represent is zero from -infinity to plus infinity. Finally, all the *'s should be daggers but typing dagger every time seems wasteful. I have more confidence in my normalizations of a and a* now. Thank you very very much.
 
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