Why Must Ladder Operators Have Specific Forms in Quantum Mechanics?

[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?

 

[malawi_glenn]

Maybe this can be useful:

http://www.oru.se/oru-upload/Institutioner/Naturvetenskap/Dokument/Fysik/PJ/Kvantmekanik/stegop.pdf

 

[olgranpappy]

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?

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."

 

[quetzalcoatl9]

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

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

 

[olgranpappy]

why would you say that?

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.