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1

It seems there is a mistake in your $\pi (x)$ expression: there must be one minus sign near $\hat{a}^{\dagger}$. The relation between classical and field is obvious since lagrangian (hamiltonian) of free $\varphi $ (it's not hard to see that $\varphi$ satisfies Klein-Gordon equation) field may be rewritten as lagrangian of free ossilator in terms of ...


2

This is essentially the statement that the Fourier transform is injective and therefore invertible. In looser language, it states that the functions $\mathbf x\mapsto\exp(i\mathbf p\cdot\mathbf x)$ are linearly independent, so any sum of them (i.e. $\int\mathrm d^3\mathbf p$) that gives zero must have identically zero coefficients. Thus, it applies to any ...


2

Your first equation can be compared to (in one dimension) $$ g(x)=\int dp\exp(ipx)f(p)=0 \,\,\, (*) $$ Note that this is a statement applies to all x-values. The inverse fourier transform of $g(x)$ should give $f(p)$, $$ f(p)=\frac{1}{2\pi}\int dx\exp(-ipx)g(x) $$ As $g(x)=0$ according to to $(*)$ then $f(p)=0$.


0

If you refer to a layered structure along growth direction (e.g. quantum well, superlattice), you can calculate your mass for different states and positions from the 2nd derivative of the wavefunctions. Perpendicular to it you would have the dispersion (and therefore mass) of a free particle, since you would usually assume that your layer planes are ...


2

Perhaps the best way to understand this is to start simply: Consider a function $f(x)$. Now, let's try to take the Fourier transform of its derivative $f'(x)$. Just use the definition of the Fourier transform: $$\mathscr{F}(f'(x))(k)=\frac{1}{\sqrt{2\pi}}\int dx\ e^{-ikx}f'(x) $$ and now use integration by parts (assuming $f(x) \to 0$ as $|x|\to \infty$, a ...


0

I already wrote elsewhere that Gaussian beams are just some approximations to solutions of the Maxwell equations. For this reason, I derived some exact solutions of the Maxwell equations that are approximated by Gaussian beams extremely well ("asymptotically accurately") when the beam waist is much larger than the wavelength. Please see ...


2

The problem with using the actual free space Fourier propagator is aliasing. I learned this through trial and error as well, after a few wavelengths the numerical model really begins to behave poorly, probably due to aliasing and roundoff error. The Fresnel approximation actually does a better job if you are anywhere further than the very near field ...


0

I want to ask if it is reasonable that I use the Dirac-Delta function as an intial state ($\psi(x,0)$) for the free particle wavefunction and interpret it such that I say that the particle is exactly at $x=0$ during time $t=0$? No, because the delta function is not compliant with the Born interpretation of the function $\psi$. Evolving function that is ...


1

That is indeed how you would go about it. Note, however, that there is nothing to guarantee that the solution is going to be reasonable, or that the integral even exists. In fact, because the Schrödinger equation is time reversible to a large extent, you are essentially guaranteed to not end up in physical states. One thing to note is that the frequency ...


5

Consider evolution of gaussian wave packet. Its wave function in position representation looks like: $$\Psi(\vec r,t)=\left(\frac a{a+i\hbar t/m}\right)^{3/2}\exp\left(-\frac{\vec r\cdot \vec r}{2(a+i\hbar t/m)}\right).\tag1$$ Corresponding relative probability density is $$P(r)=|\Psi|^2=\left(\frac a{\sqrt{a^2+(\hbar ...



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