Understanding Ansatz in Quantum Mechanics: Types and Applications

  • Context: Graduate 
  • Thread starter Thread starter AxiomOfChoice
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 10K views
AxiomOfChoice
Messages
531
Reaction score
1
What is an "ansatz?"

I am reading a quantum mechanics textbook, and they keep talking about different kinds of ansatz's - most commonly, separation ansatz's. But I thought that an ansatz was nothing more than an educated guess! If you show that your ansatz is *a* solution to the Schrödinger equation, do you then know that it is *the* solution?

For example, in the two-body problem in quantum mechanics, how can you assume that [itex]\psi(\mathbf{x}_{cm}) = e^{i \mathbf{k}_{cm} \cdot \mathbf{x}_{cm}}[/itex] is the solution to [itex]H_{cm} \psi_{cm} = E_{cm} \psi_{cm}[/itex]?
 
Physics news on Phys.org


Yes, basically Ansatz is a fancy word for (educated) guess. So at least it can help you find a solution. Then usually we have one of two options:
* The Ansatz gives the general solution. For example, if we have a second order differential equation and plugging in [itex]\psi = e^{\lambda x}[/itex] gives a quadratic equation in [itex]\lambda[/itex] with solutions [itex]\lambda_\pm[/itex] we have two linearly indpendent solutions [itex]e^{\lambda_\pm x}[/itex]. From the theory of differential equations we know that the most general solution is a linear combination of those two.

or

* The solutions of that form are all we are interested in. For example, if you know that any wavefunction can be written as a linear combination of stationary states times an evolution factor (exp(...t)) we can safely suppose that the wavefunction is of the form X(x) T(t) because we basically only want to know what X(x) can be.