0
$\begingroup$

I'm trying to understand why in quantum harmonic oscillator when finding ground state eigenfunction we don't use $a^\dagger$.

For a simple harmonic oscillator the Hamiltonian is given by $$H=\hbar\omega_0\left(\hat{a}\hat{a}^\dagger-1/2\right),$$ where the operators $\hat{a}$ and $\hat{a}^\dagger$ may be expressed as $$\hat{a}=\frac{1}{\sqrt{2}}\left(\xi+\frac{\partial}{\partial\xi}\right),$$ $$\hat{a}^\dagger=\frac{1}{\sqrt{2}}\left(\xi-\frac{\partial}{\partial\xi}\right),$$ where $\xi=\beta x$. You are required to find the expression for the normalized ground state eigenfunction.

This question is from my exam I just want to understand why when solution came we didn't use $a^\dagger$ its solution was given as:

solution enter image description here

$\endgroup$
4
  • 7
    $\begingroup$ You'll find that eigenstates of $a^\dagger$ can't be normalized; not a bad exercise to do. People on this site g generally refer that you typeset your equations as opposed to posting pictures of them $\endgroup$ Jan 8 at 16:57
  • 1
    $\begingroup$ I'm sry I'm new to this site my friend told me about it i will try to keep that in mind next time $\endgroup$ Jan 8 at 17:09
  • 4
    $\begingroup$ Welcome to PhysicsSE! Note that this site supports MathJax. You can click the link to learn the basics. I've taken the liberty of typesetting the first part of the post for you, as an example. $\endgroup$
    – Chris
    Jan 8 at 19:35
  • $\begingroup$ To be pedantic, the professor's solution misses a phase factor in the normalization constant... $\endgroup$
    – DanielC
    Jan 9 at 13:54

1 Answer 1

1
$\begingroup$

You could have equally well used $$ \hat{a}^\dagger= \frac{1}{\sqrt{2}}\int\!\!d\xi ~ |\xi\rangle \left(\xi-\frac{\partial}{\partial\xi}\right)\langle \xi|, $$ instead, since the hermitian conjugate of your starting expression is $$ \langle 0| \hat{a}^\dagger=0. $$ You then have $$ 0=\langle 0| \hat{a}^\dagger = \frac{1}{\sqrt{2}}\int\!\!d\xi ~ \langle 0|\xi\rangle \left(\xi-\frac{\partial}{\partial\xi}\right)\langle \xi| \\= \frac{1}{\sqrt{2}}\int\!\!d\xi ~ \left(\xi \psi^*_0(\xi) +\frac{\partial \psi^*_0(\xi) }{\partial\xi}\right)\langle \xi| ~~, $$ the last step involving integration by parts and use of the definition $\langle 0|\xi\rangle\equiv \psi^*_0(\xi) $.

Consequently, you get the same equation you had before, $$ \xi \psi^*_0(\xi) +\frac{\partial \psi^*_0(\xi) }{\partial\xi}=0, $$ with real solution, $$ \psi^*_0(\xi) \propto e^{-\xi^2/2} ~, $$ so you may complex conjugate and suitably normalize, etc.

It should be evident that the crucial step connecting states to functions is identical.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.