Quantum harmonic oscillator wave function

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SUMMARY

The discussion focuses on finding the wave function Φα for a quantum harmonic oscillator (QHO) using the Hamiltonian (p²/2m) + (mω²x²)/2 and the lowering operator a defined as a = (1/√2)((x/(ħ/mω)^(1/2)) + i(p/(mħω)^(1/2))). The key equation aΦα(x) = αΦα(x) leads to a recurrence relation for the coefficients c_n in the expansion of Φα in terms of the eigenstates ψ_n of the Hamiltonian. The participants clarify that the eigenfunctions ψ_n are linearly independent and emphasize the importance of understanding how ladder operators act on these functions.

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  • Quantum mechanics fundamentals, specifically the Schrödinger equation.
  • Understanding of quantum harmonic oscillator concepts and Hamiltonians.
  • Familiarity with raising and lowering operators in quantum mechanics.
  • Knowledge of Hermite polynomials and their role in QHO solutions.
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  • Study the derivation and properties of Hermite polynomials in quantum mechanics.
  • Learn how to apply ladder operators to quantum states in detail.
  • Explore the normalization of wave functions in quantum mechanics.
  • Investigate the implications of complex parameters in quantum harmonic oscillator wave functions.
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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!
 
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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.
 
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.
 
Last edited:
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.
 
  • #10
Would I be correct in saying ∑(n-1)^½ Cn-1 Ψn(x)
 
  • #11
The eigenfunctions ##\psi_n## are linearly independent, therefore ##\alpha c_n = \sqrt{n+1}c_{n+1}## (why?). So what are the first few terms in the expansion of ##\Phi_\alpha (x)## ?
 
  • #12
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.
 
  • #13
^ He's not looking for the eigenfunctions of the hamiltonian, the problem is about the eigenfunctions of the lowering operator.
 

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