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Mar
26
comment Perturbative Quantum Mechanics
@linkhyrule5 well, you can successively calculate first-order corrections, second-order corrections, and so forth and check the convergence. If perturbative series do converge, you can safely use the theory, but if not, then you are in a trouble.
Mar
26
comment Perturbative Quantum Mechanics
@linkhyrule5 hmmm... the higher-order terms are not less significant. Where did you get that? Perturbative series are not even guaranteed to converge.
Mar
26
comment Perturbative Quantum Mechanics
First-order approximation to energy (state vector) is the coefficient of the first power of $\lambda$ in the expansions of energy (state vector) in powers of $\lambda$, i.e. $E^{(1)}$ ($n^{(1)}$) in the notation you used.
Mar
26
answered Perturbative Quantum Mechanics
Mar
5
comment Quantum State Function $\psi$
@Ruslan, true. I meant that $\cos(6 \pi x)$ is not square-integrable on the whole real line, so it can't be the wave function for a particle not confined to some region of space a priori.
Mar
5
comment Questions on electron orbits
The well known fact that electrons do not move in orbits makes all your questions meaningless.
Mar
5
comment Quantum State Function $\psi$
@innisfree, yes, that is true. I was wrong.
Mar
5
revised Quantum State Function $\psi$
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Mar
5
comment Quantum State Function $\psi$
I still think $\cos(6 \pi x)$ is just an example in the question, though a bad one, since this function is not square-integrable.
Mar
5
comment Quantum State Function $\psi$
Just to note: you do not "write" a wave function, you write the Schrödinger equation and solve it. The solution is the wave function. A function is not a wave function unless it is a solution of the Schrödinger equation for some physical system.
Mar
5
comment Quantum State Function $\psi$
The wave function is postulated to be square-integrable (normalizable). If a function is not square-integrable, then it can not be a wave function. Nothing to discuss in this case.
Mar
5
awarded  Yearling
Mar
5
comment Quantum State Function $\psi$
@innisfree, I meant that a wave function is normalized to one to conform to the ("standard") probability theory, in which 1 indicates certainty. But in principle you are not required to adopt this standard. Yes, it is desirable to do so, but it is not required. You can work with probabilities in the range from 0 to 100 (thinking in terms of %), then you normalize your wave function to 10. Math would be a little different, physics will be the same.
Mar
5
comment Quantum State Function $\psi$
@lota You apply a Fourier transform to get the position wave function out of the momentum one, not the Born rule.
Mar
5
revised Quantum State Function $\psi$
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Mar
5
revised Quantum State Function $\psi$
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Mar
5
answered Quantum State Function $\psi$
Mar
5
comment Is the uncertainty principle axiomatic or derived?
@SideshowBob, second, the situation, when the energy of a system is given, rather than its state, is totally different. In this case you don't have some uncertainty in $x$, rather, you do not even know $x$, since you do not know the state of a system. So you can not say anything about $x$ in this case, and not just specify it exactly. And, again, it has nothing to do with uncertainty principle. You simply do not know the state of a system, thus you do not know values of physical observables. Except energy, of course, since it was given to you by someone for some reason.
Mar
5
comment Is the uncertainty principle axiomatic or derived?
@SideshowBob, first, given a state of a system it is impossible to know both $x$ and $p$ of each particle in the system precisely. That is what uncertainty principle tells you. And, of course, it works for the system in question: if I give you the state of such system, you won't be able to specify both $x$ and $p$ of any particle precisely.
Mar
1
comment How can the product of two real linear operators be not real?
@asmaier yes, in physics the word "Hermitian" is often used as a synonym for "self-adjoint".