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Feb
6
comment Does a set of eigenvalues in QM correspond to a random variable in statistics?
The term "eigenvalue" does not appear in the answer of ValterMoretti you link to, so I think my question is different.
Feb
6
asked Does a set of eigenvalues in QM correspond to a random variable in statistics?
Feb
1
comment Is what statisticians call a “random variable” what physicists call an “observable” in QM?
Terence Tao seems to share my view "...and quantum mechanics (with physical observables taking the role of random variables, and their expected value on a given quantum state being the expectation)" (terrytao.wordpress.com/2010/02/10/245a-notes-5-free-probability)
Feb
1
comment Is what statisticians call a “random variable” what physicists call an “observable” in QM?
That is interesting. Can you put your statement about four tuples of binary random variables into a mathematical form?
Jan
31
asked Is what statisticians call a “random variable” what physicists call an “observable” in QM?
Jan
29
comment What is the point of complex fields in classical field theory?
@Numrok But isn't a set of two real scalars a vector? The two scalars also have the same units and it is often said, that complex numbers can be represented as vectors.
Jan
29
comment What is the point of complex fields in classical field theory?
@Numrok I'm not sure about that point, too. The idea is that a complex number is by definition a en.wikipedia.org/wiki/Scalar_(physics) quantity, so invariant under coordinate transformation. But a vector of two real quantities is not a scalar. So it is not invariant under coordinate transformations.
Jan
29
asked Is the polarization of light changed by gravity?
Jan
29
accepted Why is the Fourier transform more useful than the Hartley transform in physics?
Jan
29
asked What is polarisation, spin, helicity, chirality and parity?
Jan
29
asked What is the point of complex fields in classical field theory?
Jan
28
awarded  Yearling
Jan
22
comment Why don't antiparticles have antispin?
@ACuriousMind: Wouldn't a change in P (=parity?) imply spinning backwards?
Jan
22
asked Why don't antiparticles have antispin?
Jan
14
comment Why is the Fourier transform more useful than the Hartley transform in physics?
@Daniel Sank: Thank you. Actually I was expecting somebody to say something along the line that a optical lens performs a Fourier transformation and not(?) a Hartley transformation, and that is why Fourier is more "physical". But your edit is very interesting, too.
Jan
12
comment Why is the Fourier transform more useful than the Hartley transform in physics?
I would love if someone could address this part of the question: "Are their any properties that make the Fourier transformation more "physical"?" .
Jan
12
comment Why is the Fourier transform more useful than the Hartley transform in physics?
@DanielSank I know that hermitian means, that the operator has real eigenvalues. What confuses me is that an operator involving the imaginary unit like $i (d/dt)$ can in fact have real eigenvalues. But maybe this leads to far away from the original question. Maybe I ask another question about this. Thank you.
Jan
11
comment Why is the Fourier transform more useful than the Hartley transform in physics?
@DanielSank Isn't the eigenvector of $(d/dt)$ simply the real exponential function $e^{x}$ ? Why is the exponential function with a complex argument be used for the Fourier transformation?
Jan
9
comment Why is the Fourier transform more useful than the Hartley transform in physics?
I think it is interesting to note that when looking at second derivatives both kernels show again the same behaviour: $(d^2/dt^2) \text{cas}(\omega t) = -\omega^2 \text{cas}(\omega t)$ and $(d^2/dt^2)\exp(i\omega t) = - \omega^2 \exp(i \omega t)$
Jan
9
asked Why is the Fourier transform more useful than the Hartley transform in physics?