1. Aug 18, 2007

### nklohit

Why must the ladder operators be
$$\sqrt{\dfrac{m\omega}{2\hbar}}(x+\dfrac{ip}{m\omega})$$ and
$$\sqrt{\dfrac{m\omega}{2\hbar}}(x-\dfrac{ip}{m\omega})$$?
What is the method that obtain them from schrodinger Equation?
And why we know that they are creation and anihilation operator?

2. Aug 18, 2007

### malawi_glenn

Maybe this can be useful:

Last edited by a moderator: May 3, 2017
3. Aug 18, 2007

### olgranpappy

Don't worry about all the constants out front, they don't really matter too much, since they are just a convenient normalization.

The important point is that one of those operators you wrote down gives zero when it acts on a Gaussian (which is the ground state of the simple harmonic oscillator). That operator is that "annihilation operator" or "lowering operator."

The other operator is the annihilation operator's Hermitian conjugate and is called the "creation operator" or "raising operator."

4. Aug 18, 2007

### quetzalcoatl9

why would you say that? they follow from the true hamiltonian

5. Aug 18, 2007

### olgranpappy

I said that because I believe it is important to realize the difference between important aspects of this problem and trivial aspects or this problem.

Of course, in the end, you want to get the trivial aspects correct as well.

Perhaps what I should have said was: "chose your units in such a way that you can set all the messy crap out front equal to one." Or, better yet, choose your units such that the unit of mass is $$m$$ and the unit of time is $$1/\omega$$ and the unit of angular momentum is $$2\hbar$$, in which case:

$$a=x+\frac{d}{dx}$$

and
$$a^{\dagger}= x - \frac{d}{dx}$$

and the ground state is
$$\psi_0(x)=\sqrt{\frac{1}{\sqrt{\pi}}}e^{-x^2/2}$$

My point was that one of those operators has a relative minus sign which is important because that one doesn't annihilate the ground state and the other has a relative plus sign which then does annihilate the ground state.