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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).

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closed as off-topic by CR Drost, ZeroTheHero, Kyle Kanos, Cosmas Zachos, stafusa Nov 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.

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  • $\begingroup$ 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? $\endgroup$ – Mandeep Nov 12 '18 at 21:45
  • $\begingroup$ @Mandeep If you wanted to see if $\psi(x)$ is a momentum eigenfunction? $\endgroup$ – Aaron Stevens Nov 12 '18 at 21:50
  • $\begingroup$ Yes, how would I go about "operating" on the operator to test if the wavefunction is an eigenfunction of the momentum operator $\endgroup$ – Mandeep Nov 12 '18 at 21:51
  • $\begingroup$ @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 $\endgroup$ – Aaron Stevens Nov 12 '18 at 21:53

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