X^4 perturbative energy eigenvalues for harmonic oscillator

[PBTR3]

Homework Statement

Compute first order correction matrix elements $\langle \n | x^4 |m \rangle$ for a one dimensional harmonic oscillator.

Relevant Equations

$\langle \n |x^4 |m \rangle = 2 \left( \frac hbar 2m\omega \right) \left( 3n^2 +2n +1 \right)$
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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?

 

[TSny]

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.

 

[PBTR3]

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.

 

[PBTR3]

(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]

(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.

 

[TSny]

(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.

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.

$\langle n | a^2a*^2 | m \rangle=\langle n | (m+1) (m+2) | m \rangle$
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 $\large \langle n | x^4| n \rangle = \frac{3}{4} \left( \frac{\hbar}{m \omega} \right)^2 (2n^2+2n+1)$?

 

[PBTR3]

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.