Timeline for Difference between real operators and Hermitian operators in quantum mechanics
Current License: CC BY-SA 4.0
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Feb 29 at 13:43 | history | edited | Albertus Magnus | CC BY-SA 4.0 |
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Feb 29 at 13:35 | history | edited | Albertus Magnus | CC BY-SA 4.0 |
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Feb 29 at 12:33 | comment | added | Albertus Magnus | @Hyperon I was thinking in terms of a specific eigen basis, as this highlights how talking about a "real" operator might not be definitive, i.e. an operator that looks good in a particular basis might not be Hermitian. Thank you for your sagacity! | |
Feb 29 at 8:00 | comment | added | Hyperon | @AlbertusMagnus Finally, defining a "real" operator $A$ by the property $\langle \psi | A \psi \rangle \in \mathbb{R}$ for all vectors $\psi$ in your Hilbert space (disregarding the well-known domain issues in the case of an unbounded linear operator in an infinite dimensional Hilbert space), would make more sense. However, this property is equivalent to hermiticity (take $\psi=f+g$ and $\psi=f+ig$) and in this case your remark "not all real operators are hermitean" would not make sense. | |
Feb 29 at 7:51 | comment | added | Hyperon | @AlbertusMagnus Your answer is unclear. What exactly do you mean by "a real operator is one whose matrix elements are real"? All matrix elements $\langle f |A g \rangle$ (in the sense used in QM)? Take the identity operator as an example: $\langle f | \mathbf{1} g \rangle =\langle f | g\rangle$ can take any complex value by choosing suitable vectors $f,g$ in your Hilbert space. Or do you mean the matrix elements $\langle u_m |A u_n\rangle$ with respect to a certain orthonormal basis $\{ u_1,u_2,\ldots \}$? In this case, the reality of the matrix elements is a basis dependent statement. | |
Feb 28 at 23:41 | history | edited | Albertus Magnus | CC BY-SA 4.0 |
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Feb 28 at 22:49 | history | edited | Albertus Magnus | CC BY-SA 4.0 |
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Feb 28 at 22:46 | comment | added | Albertus Magnus | @Andrea Ignore the stuff I said about real/complex and follow the advice above. I am terribly sorry if I have caused you any inconvenience. | |
Feb 28 at 22:45 | comment | added | Albertus Magnus | @Andrea The most important thing to take away from this stuff today is that operators in quantum mechanics should be Hermitian, which is the same as self-adjoint (whenever the underlying domain of the operator is a complex linear space). Personally, I would speak about Hermitian operators as opposed to real operators. I think that I have actually been mislead by a past professor and that I read that into what others were saying, however, it didn't do me any harm because I ultimately understood that it was the self-adjoint nature of the operator that was the heart of the issue. | |
Feb 28 at 20:36 | vote | accept | Andrea | ||
Feb 28 at 20:36 | comment | added | Andrea | Thank you for your enlightening answer. Can you please explain a bit more how the momentum operator is real in the momentum basis? And is the property of being hermitian dependant on the basis as well or is it intrinsic to the operator? | |
Feb 28 at 16:33 | comment | added | Albertus Magnus | @infinitezero Thanks! | |
Feb 28 at 16:32 | history | edited | Albertus Magnus | CC BY-SA 4.0 |
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Feb 28 at 16:17 | comment | added | Albertus Magnus | @TobiasFünke Yes, that was discussed in the above comments, perhaps you might take a look at the edited post and see how it holds up? Your acumen would be appreciated, thanks. | |
Feb 28 at 16:16 | history | edited | Albertus Magnus | CC BY-SA 4.0 |
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Feb 28 at 16:13 | comment | added | Tobias Fünke | The phrase "real operator" makes no sense in an abstract Hilbert space... | |
Feb 28 at 15:52 | history | edited | Albertus Magnus | CC BY-SA 4.0 |
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Feb 28 at 15:46 | history | edited | Albertus Magnus | CC BY-SA 4.0 |
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Feb 28 at 15:33 | history | answered | Albertus Magnus | CC BY-SA 4.0 |