Imaginary Part of QHO Solutions?

[LarryS]

I have seen a few online lectures on solving the Schrödinger equation for the Quantum Harmonic Oscillator. The various solutions are products of the real-valued Gaussian function and the real-valued Hermite Polynomials. But I have never seen a mathematical expression for the imaginary part of those solutions. The Wikipedia entry for the QHO shows a graph for the imaginary part but no expression for it. Please help.

Thanks in advance.

 

[DrClaude]

As always in QM, two wave functions differing only by a complex phase correspond to the same state. By convention, the eigenfunctions of the time-independent Schrödinger equation are chosen to be purely real. It is only when considering the time evolution of an eigenfunction, or when considering a wave function that is a superposition of eigenstates, that the complex part is not zero.

 

[jtbell]

The Wikipedia entry for the QHO shows a graph for the imaginary part but no expression for it.

Are you referring to the animated graphs which have the real and imaginary parts in different colors? I’m pretty sure they’re just the real solutions of the time-independent SE, multiplied by the complex time-dependent phase factor: $$e^{-iEt/\hbar} = \cos (Et/\hbar) - i \sin(Et/\hbar)$$

 

[George Jones]

I have seen a few online lectures on solving the Schrödinger equation for the Quantum Harmonic Oscillator. The various solutions are products of the real-valued Gaussian function and the real-valued Hermite Polynomials. But I have never seen a mathematical expression for the imaginary part of those solutions. The Wikipedia entry for the QHO shows a graph for the imaginary part but no expression for it.

As always in QM, two wave functions differing only by a complex phase correspond to the same state. By convention, the eigenfunctions of the time-independent Schrödinger equation are chosen to be purely real. It is only when considering the time evolution of an eigenfunction, or when considering a wave function that is a superposition of eigenstates, that the complex part is not zero.

Are you referring to the animated graphs which have the real and imaginary parts in different colors? I’m pretty sure they’re just the real solutions of the time-independent SE, multiplied by the complex time-dependent phase factor: $$e^{-iEt/\hbar} = \cos (Et/\hbar) - i \sin(Et/\hbar)$$

The first four real/imaginary animations are examples of this, but the last two animations are superpositions of time-dependent stationary states, i.e., as my first quantum instructor would say, "There is (spatial) sloshing."

 

[vanhees71]

Are you referring to the animated graphs which have the real and imaginary parts in different colors? I’m pretty sure they’re just the real solutions of the time-independent SE, multiplied by the complex time-dependent phase factor: $$e^{-iEt/\hbar} = \cos (Et/\hbar) - i \sin(Et/\hbar)$$

Well, that may well be true, but it doesn't have to do anything with the classical counterparts plotted as A and B. The only state referring to this classical case is a coherent state (depicted as plot H). It's of course misleading to show real and imaginary parts, which both are not observable. What's observable in a statistical sense, being position-probability distributions, are the modulus squares. Than it should be immediately clear that the energy eigenstates are stationary states, i.e., nothing is moving at all!

 

[LarryS]

Are you referring to the animated graphs which have the real and imaginary parts in different colors? I’m pretty sure they’re just the real solutions of the time-independent SE, multiplied by the complex time-dependent phase factor: $$e^{-iEt/\hbar} = \cos (Et/\hbar) - i \sin(Et/\hbar)$$

Yes, I was referring to those animated graphs. Thanks for the clarification.

 

[hilbert2]

The graphing of real and imaginary parts is useful with time-dependent states where the initial state is a displaced version of the gaussian ground state of SHO: $\Psi (x,0) = A\exp\left[-b(x-\Delta x)^2\right]$. In that case the wavepacket at later times, $\Psi (x,t)$, is also gaussian but the center of it will oscillate about the equilibrium point like a classical oscillator.

Here's a video about this: