if $\psi(x)$ is a solution to the schrodinger equation and $\psi(-x)$ is also a solution, then a combination of these two is a solution. however, doesnt this mean that any combination of them should be a solution? i am looking at an exercise that uses this ideaology to prove that the solution to the schrodinger equation with an even potential can always be taken to be even or odd, but even if the lin comb of these two [...]
[...] has constants that satisfy this (i.e. using 1 and -1 to make the sum even or odd), surely there are many constants that make the combination neither even or odd
@Relativisticcucumber Consider e.g. the harmonic oscillator. The potential is even and $\lvert n \rangle$ is odd if $n$ is odd and even if $n$ is even. A combination (i.e. a generic state) is not a solution to the TDSE as the solutions are by definition eigenstates of the Hamiltonian
@Relativisticcucumber I think @Feynman_00 might have meant the Time Independent SE not the Time Dependent SE above.
If you combine two solutions to the time independent SE the result is time dependent, so combinations of the eigenfunctions are not solutions to the TISE.
But any combination of solutions to the time dependent SE is also a solution to the time dependent SE.
@JohnRennie Yes, sorry. I always mess with that acronym :P
@Relativisticcucumber What we really mean is that (if the solutions you combine belong to different eigenspaces), the combination undergoes non-trivial time evolution
Eigenstates of the Hamiltonian are stationary states: it's not like they don't evolve over time but the evolution is given by attaching a phase factor i.e. it is trivial as this won't affect the observables. Different eigenstates get different phase factors, which means that the combination won't have a trivial evolution, because the phase factors have different time dependence
@SillyGoose The TISE is the eigenvalue problem of the Hamiltonian
@SillyGoose Yes, it depends on the energy. When you want to deal with time dependent SE, you attach the $\exp(iEt/\hbar)$ factors to each eigenstate of the combination. As long as the state is just an eigenstate this phase factor is irrelevant, while combinations have different oscillation frequency in each term of the sum thus they evolve differently yielding overall non trivial time evolution
Does there exist a closed-form analytical solution to Lorentz force law for general EM fields...(like we have Jeffimenko equations for Maxwell equations)?
I have to find how the einbein transforms given that the action should be reparametrization invariant
But im not entirely sure what reparametrization ivariant means.
we have $\lambda \mapsto \lambda '(\lambda)$
What I did was to vary the action, for $x^{\mu} \mapsto x^{\mu} + \delta x^{\mu}$ and $g_{\mu \nu} \mapsto g_{\mu \nu} + \partial_{\sigma} g_{\mu \nu} \delta x^{\sigma}$
as well as vary the einbein, but all I get is a huge action with two main terms, one with $\delta e$ and one with $\delta x^{\sigma}$
I set both terms to zero and I get what it seems like a constraint equation for e, as well as a modified geodesic equation
I feel like the answer should be much easier that what I'm doing
I'm not sure what you mean by "what I did was to vary the action"
Just plug the transformation $\lambda\mapsto\lambda'$ into everything whose transformation behavior you know, and then stare long enough at the result to figure out how $e$ must transform
e.g. for the terms involving $x^\mu$ you have $\frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda} \mapsto \frac{\mathrm{d}{x^\mu}}{\mathrm{d}\lambda'}\frac{\mathrm{d}\lambda'}{\mathrm{d}\lambda}$
writing $\lambda' = \lambda - \eta(\lambda)$ is just doing the usual physics thing of looking only at an infinitesimal transformation, but that's not really necessary here
If I have a scalar function $f(x)$ and I use another parametrization $x'(x)$ to get an $f(x') := f(x(x'))$, then nothing else happens to $f$, it doesn't "change", even though of course $f(x')$ has a different value at $x'=2$ than $f(x)$ did at $x=2$.
Alright alright. I meant change in the sense that if $\lambda$ transforms, then so should the metric. If the metric was given explicitly, they would have different formulas.
contrast this with the derivative $f'(x) = \frac{\mathrm{d}f}{\mathrm{d}x}$ - if I transform to $f'(x') = \frac{\mathrm{d}f}{\mathrm{d}x'} = \frac{\mathrm{d}x}{\mathrm{d}x'}f'(x(x'))$, you get a factor of $\frac{\mathrm{d}x'}{\mathrm{d}x}$
@ShikiRyougi "transform" and changing formulas don't exactly mean the same thing
a scalar function will "look different" in two different coordinate systems in terms of explicit formulae
that doesn't mean it transforms non-trivially under coordinate transformations
most physics texts don't really explain this very clearly and I've also done a rather bad job above because the notation doesn't really hold up, this is one of the points where I find the mathematicians' clear separation between abstract entities and coordinate expression really valuable to understand what's going on
So the transformed action should be $\int d \lambda ' [ g_{\mu \nu}(\lambda(\lambda ') \frac{d x^{\mu}}{d \lambda ') \frac{d x^{\nu}}{d \lambda ')\frac{d \lambda '}{d \lambda )\frac{d \lambda '}{d \lambda ) - m^2 e^2(\lambda ' )$. But how does the metric transform? I mean, it transforms as a tensor between coordinate systems but what about reparametrizations :P
@ACuriousMind I think I still don't understand what reparametrization invariant means. What I would do is to equate the old and new action (I ask this before I try this)
"equate" is a bit difficult a word to use here. What you want to show is that the action after the reparametrization has the same form as before that - in practical terms that all the $\frac{\mathrm{d}\lambda'}{\mathrm{d}\lambda}$ terms and the $k$ cancel each other
im still very confused about the discussion with johnrennie, Feynman, and silly goose. so 1) i dont get why a linear combination of solutions isnt a solution bc in the free particle case, we use this fact to get normalizable solutions i think? 2) this exercise is saying exactly that we can take any solution with even potential to be even or odd because we can show that psi(x) and psi(-x) are solutions, so we choose out constants such that the sum of them can be positive or negative
but that doesnt make sense to me because we can also choose them to not be that way
there is exactly one TDSE, $\partial_t \psi = -\mathrm{i}H\psi$, but there is a family of TISEs $E\psi = -\mathrm{i}H\psi$, one equation for each constant $E\in\mathbb{R}$
@Relativisticcucumber okay, so there the difference indeed doesn't matter - the claim is that if $\psi(x)$ fulfills one of these equations (and the TISE for one particular $E$), then $\psi(-x)$ fulfills the same equation (with the same $E$ in the case of the TISE)
so i used this result in a double delta potential well problem to say "okay lets look for only even and odd solutions and this should do it", so do you mean this is fine and this represents a possible basis but there are other bases also ?
and so the point is to say wecan always look for such a basis ?
quick question. I now want to vary the action with respect to $e(\lamba)$. That would mean that I only vary $e(\lamba)$ in the action and not $x^{\mu}$ and $g_{\mu \nu}$, correct?
Wait a minute, if I vary with respect to $x^{\mu}$, $g_{\mu \nu}$ varies but $e(\lambda)$ does not, so I get the geodesic equation but with a factor of $\frac{1}{e}$, which is basically the geodesic equation. How is that possible?
