# Eigenvalues and functions in quantum mechanics [closed]

How do I determine if $$\psi(x)$$ is a eigenfunction of some operator and find the corresponding eigenvalues, where $$\psi(x)$$ is the wave function of free particle (potential = zero).

## closed as off-topic by CR Drost, ZeroTheHero, Kyle Kanos, Cosmas Zachos, stafusaNov 18 '18 at 22:39

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How do I determine if $$\psi(x)$$ is a eigenfunction of some operator and find the corresponding eigenvalues

In general, if a wavefunction $$\psi(x)$$ is an eigenfunction of some operator $$\hat A$$, then the following equation must be true: $$\hat A\psi(x)=a\psi(x)$$ Where $$a$$ is the eigenvalue of the corresponding eigenfunction.

Therefore, to determine if a wavefunction is an eigenfunction of the operator in question, all you have to do is operate on $$\psi(x)$$ by $$\hat A$$ and see if you get the function $$\psi(x)$$ multiplied by a constant back.

where $$\psi(x)$$ is the wave function of free particle

There is no single $$\psi(x)$$ of a free particle. You can have some initial wavefunction $$\psi(x,t_0)$$ that then evolves according to the free particle Hamiltonian $$\hat H=\frac{\hat{P}^2}{2m}$$, but there are many things $$\psi(x,t_0)$$ can be. This is kind of analogous to the classical physics question "what is x(t) for a free particle?", where we can say how $$x(t)$$ evolves due to its initial conditions ($$x(t_0)$$ and $$v(t_0)$$), but there is not a single $$x(t)$$ for a free particle.

• So for example, if the operator was the momentum operator: -i*h/2pi * d/dx, I would have to multiply the operator by the wavefunction, and then integrate with respect to x to remove the d/dx? Is this correct? – Mandeep Nov 12 '18 at 21:45
• @Mandeep If you wanted to see if $\psi(x)$ is a momentum eigenfunction? – Aaron Stevens Nov 12 '18 at 21:50
• Yes, how would I go about "operating" on the operator to test if the wavefunction is an eigenfunction of the momentum operator – Mandeep Nov 12 '18 at 21:51
• @Mandeep You perform the operation. So for momentum you just take the derivative and multiply by $-i\hbar$. This is what $-i\hbar\frac{\text d}{\text d x}$ means – Aaron Stevens Nov 12 '18 at 21:53