Quantum harmonic oscillator wave function

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Dean Navels
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How do you find the wave function Φα when given the Hamiltonian, and the equation:

α(x) = αΦα(x)

Where I know the operator

a = 1/21/2((x/(ħ/mω)1/2) + i(p/(mħω)1/2))

And the Hamiltonian,

(p2/2m) + (mω2x2)/2

And α is a complex parameter.

I obviously don't want someone to do this question for me, just a point in the right direction. Thanks in advance!
 
on Phys.org
hilbert2 said:
Do a Google search with terms "harmonic oscillator raising and lowering operators" and you will probably find many web sites where this is discussed in detail.
I've read a lot of them, what I don't understand is that when you apply a lowering or raising operator Φα becomes Φα+1 or Φα-1 yet here it doesn't change.
 
Suppose the function ##\Phi_\alpha## can be expanded in terms of ##\psi_n##, the eigenstates of the harmonic oscillator hamiltonian:

##\Phi_\alpha (x) = c_0 \psi_0 (x) + c_1 \psi_1 (x) + c_2 \psi_2 (x) + \dots##.

Now form a recursion equation for the numbers ##c_n## from the condition ##a\Phi_\alpha = \alpha\Phi_\alpha##

EDIT: Sorry, I meant "recurrence relation", which is typical English-language term for that.
 
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hilbert2 said:
Suppose the function ##\Phi_\alpha## can be expanded in terms of ##\psi_n##, the eigenstates of the harmonic oscillator hamiltonian:

##\Phi_\alpha (x) = c_0 \psi_0 (x) + c_1 \psi_1 (x) + c_2 \psi_2 (x) + \dots##.

Now form a recursion equation for the numbers ##c_n## from the condition ##a\Phi_\alpha = \alpha\Phi_\alpha##

EDIT: Sorry, I meant "recurrence relation", which is typical English-language term for that.
Is this required to solve it via ladder operators?
 
^ Yes, you have to use the fact that any quantum state can be expanded in the basis of the eigenfunctions of the hamiltonian, and also the knowledge of how the ladder operators act on those functions.
 
hilbert2 said:
^ Yes, you have to use the fact that any quantum state can be expanded in the basis of the eigenfunctions of the hamiltonian, and also the knowledge of how the ladder operators act on those functions.
After doing extensive reading on and around it, I know how to find the formula for different energies corresponding to different eigenfunctions, but now I need to put Φα(x) in the form of a normalisation constant C multiplied by an exponential.
 
The action of a lowering operator ##a## on a QHO eigenstate ##\psi_n## is:

##a\psi_n (x) = \sqrt{n}\psi_{n-1}(x)## .

Now form a general linear combination from the functions ##\psi_n## and act on it with ##a##:

##a\sum_{n=0}^{\infty} c_n \psi_n (x) = \sum_{n=1}^{\infty} \sqrt{n} c_n \psi_{n-1} (x) = \sum_{n=0}^{\infty} \sqrt{n+1} c_{n+1} \psi_{n} (x)## .

Now you should be able to write an equation for ##c_n## in terms of ##c_{n-1}## and ##\alpha##.
 
hilbert2 said:
The action of a lowering operator ##a## on a QHO eigenstate ##\psi_n## is:

##a\psi_n (x) = \sqrt{n}\psi_{n-1}(x)## .

Now form a general linear combination from the functions ##\psi_n## and act on it with ##a##:

##a\sum_{n=0}^{\infty} c_n \psi_n (x) = \sum_{n=1}^{\infty} \sqrt{n} c_n \psi_{n-1} (x) = \sum_{n=0}^{\infty} \sqrt{n+1} c_{n+1} \psi_{n} (x)## .

Now you should be able to write an equation for ##c_n## in terms of ##c_{n-1}## and ##\alpha##.
I don't understand how to involve alpha. Thanks so much for all your help by the way.
 
Would I be correct in saying ∑(n-1)^½ Cn-1 Ψn(x)
 
Dean Navels said:
How do you find the wave function Φα when given the Hamiltonian, and the equation:

α(x) = αΦα(x)

Where I know the operator

a = 1/21/2((x/(ħ/mω)1/2) + i(p/(mħω)1/2))

And the Hamiltonian,

(p2/2m) + (mω2x2)/2

And α is a complex parameter.
See, for example http://quantummechanics.ucsd.edu/ph130a/130_notes/node153.html
The solutions of the Schrödinger equation can be written as a product of a Gaussian with a polynomial. You get recurrence relation among the coeffiients of the polynomials. These polynomials have the name Hermite polynomials.