This is from Laine and Vuorinen’s Basics of Thermal Field Theory. I do not understand why the fact that the integral over $x(\tau)$ implies the following regularization scheme. That is, I don’t understand why Eq. 1.52 (the fact that the average of $x(\tau)$ is $Ta_0$) implies Eq. 1.53 (that $\int_{-\infty}^{\infty} \mathrm{d}a_0 \to \int_{\Delta x/T} \mathrm{d}a_0 ..read more

In the book QFT by Schwartz, in section 4.1 "Lippmann-Schwinger equation", he says that: If we write Hamiltonian as $H=H_0+V$ and the energies are continuous, and we have eigenstate of $H_0$: $$H_0|\phi\rangle =E|\phi\rangle\tag{4.2}$$ then we are able to find an eigenstate of $H$ with same energy:$$H|\psi\rangle =E|\psi\rangle.\tag{4.3}$$ My question is, why we can find such an eigenstate? And what has it to do with the continuity of energies ..read more

This is from Laine and Vuorinen’s Basics of Thermal Field Theory. I do not understand why the fact that the integral over $x(\tau)$ implies the following regularization scheme. That is, I don’t understand why Eq. 1.52 (the fact that the average of $dx(\tau)$ is $Ta_0$) implies Eq. 1.53 (that $\int da_0 = \int_{\Delta x/t} da_0 ..read more

If photon doesn't have probability to be in dark (destructive interference) area, what will be the effect of adding obstacles (walls) in the dark (destructive interference) area for the double slit experiment ..read more

I am looking at fusion reactions in stars and came across how particles will bypass the Coulomb barrier through quantum tunneling. I was wondering if there is an equation for the probability of a particle to tunnel when it encounters an energy barrier ..read more

If I have a wave function $ \psi (x) = \langle x|\psi \rangle$, now I want to use kinetic energy representation $$ T = \frac{p^2}{2\mu} ,$$ where $ \mu $ is the mass of particle. I try to \begin{align*} \langle T|\psi \rangle = \int {\rm d} x \langle T|x \rangle \langle x|\psi \rangle \end{align*} But how to calculate $ \langle T|x \rangle $? May be I can write \begin{align*} \langle T|x \rangle = \int {\rm d} p' \langle T|p' \rangle \langle p'|x \rangle = \frac{1}{\sqrt{2\pi\hbar}} \int {\rm d} p' \delta_{\frac{p^2}{2\mu},p'} e^{-\frac{ip'x}{\hbar}} = \frac{1}{\sqrt{2\pi\hbar}} e^{-\frac{ip^2 ..read more

If I have a wave function $ \psi (x) = <x|\psi>$, now I want to use kinetic energy representation $ T = \frac{p^2}{2\mu} $, where $ \mu $ is the mass of particle. I try to \begin{align*} <T|\psi> = \int {\rm d} x <T|x><x|\psi> \end{align*} But how to calculate $ <T|x> $. May be I can write \begin{align*} <T|x> = \int {\rm d} p' <T|p'><p'|x> = \frac{1}{\sqrt{2\pi\hbar}} \int {\rm d} p' \delta_{\frac{p^2}{2\mu},p'} e^{-\frac{ip'x}{\hbar}} = \frac{1}{\sqrt{2\pi\hbar}} e^{-\frac{ip^2x}{2\mu}} \end{align*} I don't know something wrong above. And I try ..read more

In quantum mechanics, the radial equation of the SHO takes the form \begin{align} \frac{d^2 u}{dx^2}+\left(\epsilon-x^2-\frac{l(l+1)}{x^2}\right)u=0, \end{align} where $x=\sqrt{\frac{m\omega}{\hbar}}r$, $\epsilon=\frac{2mE}{\hbar^2}$. In the far field ($x\to\infty$) limit, the equation becomes \begin{align} \frac{d^2 u}{dx^2}-x^2 u=0. \end{align} Every literature says that the asymptotic solution of this equation is $u(x)\sim e^{\pm\frac{1}{2}x^2}$ without detailed explanation. I suspect that the asymptotic solution is not unique. If we take the ansatz $u=e^{-ax^2}$, then \begin{align} (-2a+(4 ..read more

I learned from this article that when a quantum state passes through an optical amplifier based on stimulated emission (e.g., EDFA), the variance of its quadrature becomes: $(\Delta X)^2 = G(\Delta X_0)^2 + (2 n_\text{sp} - 1)(G - 1) \frac{1}{4}$ where $X_0$ is the quadrature of the input state, $X$ is the quadrature of the output state, $G$ is the gain, and $n_\text{sp} = \frac{N_2}{N_2 - N_1}$ is a factor depending on the popular inversion. The first term $G(\Delta X_0)^2$ is because the noise of the input is amplified by $G$ times. The second term is due to the amplified spontaneous emissio ..read more

This question might be confusing so let me try to clarify this carefully. The wavefunction is a tool that allows us to calculate probability distributions that model uncertainties. Thus makes sense. My question is, are the wavefunctions themselves modeled with some uncertainty? If so, how is this done? If not, why is this given special treatment over other quantitative objects in experimental physics? By the PBR theorem, the quantum state of a quantum object/system is unique. So any modeling of the quantum state in an experiment will have to be with some uncertainty of the form $|\psi\rangle ..read more