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13:05
Why does the mode expansion of the Schrodinger field contain only one term when all other equations(like KG equation,Dirac equation) have two terms in their mode expansion?
I can reason it as "cz if I include the other term I get antiparticles which shouldn't arise in NRQM", but I am wondering if there is reasoning without using the above so that I can say-"This is why we don't get antiparticles after quantizing Schrodinger field"
I feel this has something to do with relativity...
@ManasDogra the mode expansion isn't something we impose, it's something we find when we look for nice variables in which the Hamiltonian is diagonal
I think the right way to think about why we only get a single mode for a Schrödinger field is because of the combination of a single time derivative and the squared spatial derivatives - while both the Dirac and the KG-field obey a second order equation (Dirac fields still obey Klein-Gordon in every component) which gives you two "independent" modes in the solutions, this is a true first order equation
The way I saw mode expansion of the KG field was...if we Fourier transform, we see that the Fourier modes satisfy SHO eqn which means we should expand the field in terms of the ladder operators which were used in determining the spectrum of the harmonic oscillator, cz we are interested in the spectrum of the field now...
But the Schrodinger equation's fourier transform doesn't satisfy the SHO eqn...so how to *find* the mode expansion in this case?
@ACuriousMind You said "... why we only 'get' a single mode..." How to get, can you sketch a proof or give a reference?
@ACuriousMind Btw I was thinking that somehow one could impose some condition on the already obtained mode expansion of a complex scalar, to eliminate one term...not impose the mode expansion directly from nowhere...
13:39
@ManasDogra the mode expansion is not a Fourier transform
Yeah I understand that, when I said about the Fourier transform, I just spoke about the way we arrive at the picture involving harmonic oscillators...
In case the conversation is over. I had a question of my own I wanted to ask?
I'll wait 5 mins as a safety measure I guess
@ACuriousMind But in the case of Schrodinger field, Fourier transforming works because there is a single mode, the question was why there is a single-mode, and why such a Fourier decomposition wouldn't diagonalize the KG field hamiltonian for example?
I mean, you can just check that it doesn't
guess I wont have to wait 5 mins
13:47
what sort of "why" are you looking for here?
@ACuriousMind More along this line...you said that this is related to Schrodinger field involving space and time derivatives on unequal footing, how do you see that this fact is related to a simple Fourier transform working in this case?
for the KG equation isnt it it's own anti-particle?
@ManasDogra what diagonalizes the Hamiltonian is when you express the field as a supoerposition of eigenfunctions
the second-order KG equation has both $\mathrm{e}^{\mathrm{i}px}$ and $\mathrm{e}^{-\mathrm{i}px}$ as such eigenfunctions
the first order non-rel version has only $\mathrm{e}^{\mathrm{i}px}$ (or maybe with a minus, it's too hot here for me to think :P), there's no choice of sign there
I thought the reason the dirac equation had to have antiparticles was cause it was a first order differential equation?
@ACuriousMind Oh okay okay, this is the same thing which is why we get negative energy solutions for KG eqn in RQM but no such thing for the Schrodinger eqn...
@MoreAnonymous Complex scalar field theory's equation of motion being 2nd order in time also have antiparticles...
13:55
@ManasDogra and that is consistent because negative energy solutions and antiparticles are somewhat related!
@ACuriousMind Hmm...I know it from this part...Thank you godfather :)
As far as I know there is no RQM from the Klien Gordan equation but there is for the Dirac
@ManasDogra Yes for example the KG equation
My point was you cannot expect to 2 unknowns (second order differential equation) reduce to a single unknown (1st order differential equation) unless there are 2 of them
@MoreAnonymous sure, but so what the first-order does is not "allow antiparticle" but "allow an interpretation as probability density/unitary evolution"
@ACuriousMind Griener begs to differ
Page 76 relativistic quantum mechanics
Okay time for my question
Is there a way I can maximize a function (like done for entropy) and get this?
A is a normalization constant
and $ \tilde P $ is a probability distribution
any help or hints would be great
?
 
6 hours later…
20:31
@ManasDogra From $E = \mathbf{p}^2 + m^2$ we have $E = \pm \varepsilon$ for $\varepsilon = \sqrt{\mathbf{p}^2+m^2}$, therefore the eigenfunctions $e^{i xp}$ of KG are $e^{i( \mathbf{p} \cdot \mathbf{r} - \varepsilon t)}$ and $e^{i( \mathbf{p} \cdot \mathbf{r} + \varepsilon t)}$, so the general solution is a linear combination of such solutions
$$\psi(x,t) = \sum_{\mathbf{p}} N_{\mathbf{p}}[ a_{\mathbf{p}}^{(+)} e^{i( \mathbf{p} \cdot \mathbf{r} - \varepsilon t)} + a_{\mathbf{p}}^{(-)} e^{i( \mathbf{p} \cdot \mathbf{r} + \varepsilon t)}]$$
 
2 hours later…
Rlz
Rlz
22:26
Hey all, found two questions that seem to ask the same thing, yet have different answers. (physics.stackexchange.com/questions/172127/…) (physics.stackexchange.com/questions/171996/…)
Could someone please let me know what one is incorrect, or if they are both correct and I'm mistaken.
@Rlz you mean because one answer has $eh$ and the other $e^2h$?
Can you see something wrong with the derivation for $e^2 h$?
I'm not sure why you'd just believe someone here when they told you which is correct - but one of these answers provides a derivation and the other doesn't, so why would you believe the one without one if the derivation is correct?

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