I Imaginary Residuals in Solving the Schrodinger Equation

[Bob3141592]

TL;DR Summary

An amateur asks: Do any solutions of quantum systems have non-zero imaginary components?

I guess the summary says it all, if the question is clear enough. The last time I took physics courses was 45 years ago, and the QM course blew my mind, meaning I was mostly baffled. I could not wrap my head around it, and without a conceptual framework I couldn't remember the details. So I basically washed out into computers, which at the time was a good option. Anyway, I kinda know some stuff, but mostly I know nothing.

I sort of remember in solving the Schrodinger equation (probably time independent but I doubt it matters), there were leftover imaginary parts that were just crossed off the blackboard. I was under the impression at the time that these imaginary residuals were just ignored. I don't recall the professor ever arguing that they must be zero. He may have, but if he did I missed the point. I missed a lot, for example, I have never worked with Dirac notation. But that was then, and I'd like to know now. It's a fairly general question that I don't know how to research, so I'll just ask here.

Any comments will be very appreciated. TIA

 

[PeterDonis]

Do any solutions of quantum systems have non-zero imaginary components?

It depends on what you mean by "solutions". Wave functions, i.e., mathematical solutions of the Schrodinger Equation, can certainly take values with non-zero imaginary components. But any actual observed result of a measurement must be a real number.

 

[PeterDonis]

I sort of remember in solving the Schrodinger equation (probably time independent but I doubt it matters), there were leftover imaginary parts that were just crossed off the blackboard.

This sounds somewhat fishy. The nonzero imaginary parts of wave functions produce quantum interference effects, which are well established experimentally (and indeed were among the main results that convinced physicists that classical, pre-quantum physics would not work), so you can't just cross them out and expect to make valid predictions.

 

[Bob3141592]

It depends on what you mean by "solutions". Wave functions, i.e., mathematical solutions of the Schrodinger Equation, can certainly take values with non-zero imaginary components. But any actual observed result of a measurement must be a real number.

Yes, thanks, I get that. Let me ask this way. When the measurement is made, at that point is the coefficients of the imaginary components zero? Is the measurement only sampling the real part while the imaginary part could be anything, or is the measurement sampling the real part at it's "greatest extent" while the imaginary component is zero?

 

[PeterDonis]

When the measurement is made, at that point is the coefficients of the imaginary components zero?

The imaginary components of what? Of the wave function? Of course not. The result you get from the measurement has to be a real number, but that number is not the same as the value of the wave function.

Is the measurement only sampling the real part while the imaginary part could be anything, or is the measurement sampling the real part at it's "greatest extent" while the imaginary component is zero?

No to both. The wave function gives you probabilities for various possible measurement outcomes; you obtain those probabilities by taking the squared modulus of the complex amplitudes that appear in the wave function.

 

[weirdoguy]

you obtain those probabilities by taking the squared modulus of the complex amplitudes

And adding a little bit: the modulus of a complex number is real, but it certainly depends on both the real and imaginary parts of the number we start with. $z=1+2i$ has non-zero imgainary part, but $|z|^2=1^2+2^2=5$ is real and depends on the $2$.