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I understand now how I can derive the lowest energy state $W_0 = \tfrac{1}{2}\hbar \omega$ of the quantum harmonic oscillator (HO) using the ladder operators. What is the easiest way to now derive possible wavefunctions - the ones with Hermite polynomials?

I need some guidance first and then I will come up with a bit more detailed questions.

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it's a standard textbook derivation, see for example amazon.com/Quantum-Mechanics-2-vol-set/dp/0471569526 first volume chapter 5 –  Ikiperu May 28 '13 at 11:55
There is no "easiest way", there is just a standard way. But see Griffiths, Introduction to Quantum Mechanics for an intuitive, explanatory approach. –  Bogo May 28 '13 at 12:26
I am not sure that it is the easiest one, but the most common method is to solve corresponding Schrodinger equation in the coordinate representation. –  freude May 28 '13 at 15:27
I ll check Griffith for start –  71GA May 28 '13 at 17:18

1 Answer 1

up vote 2 down vote accepted

I think the easiest way to do this is to avoid solving differential equations to the greatest extent possible. There is, in fact, a way to use ladder operators and only requires you to solve one, fairly easy differential equation;

First, we note that the ladder operator technique can be used to derive the entire spectrum of one-dimensional harmonic oscillator. $$ E_n = (n+\tfrac{1}{2})\hbar\omega $$ The technique can also be used to show that the corresponding, properly normalized eigenvectors satisfy the following properties $$ a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle, \qquad a|0\rangle = 0 $$ The left-hand property shows that, once one has one of the eigenstates, every other eigenstate corresponding to a higher eigenvalue can be obtained by applying the raising operator. In particular, if one knows the position basis representation of the ground state, then one can obtain the position basis representation of every other eigenstate by applying the position basis representation of the raising operator.

The right-hand property shows that the ground state is annihilated by the lowering operator. Writing this condition in the position basis, one obtains a simple differential equation for the ground state wavefunction, and then, per the left-hand property, one generates all other wavefunctions.

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How is this helpful? He knows how to do it using ladder operators.. –  Bogo May 28 '13 at 17:16
@AmanAbhishek Huh? OP's exact words "I understand now how I can derive the lowest energy state...using the ladder operators. What is the easiest way to now derive possible wavefunctions..." Where does he indicate that he knows how to generate all other position basis wavefunctions in the way I have outlined? For the record, I think you'll find that you'll get much better responses from people without a contentious tone. –  joshphysics May 28 '13 at 17:25
I am sorry I didn't mean to be rude! Sorry :/ I thought he was asking for the non-operator approach. –  Bogo May 28 '13 at 17:31
@AmanAbhishek No worries man. You might actually have the right interpretation; maybe the OP will clarify. –  joshphysics May 28 '13 at 17:38

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