is $p=\frac{h}{\lambda}$ only true for massless particles? because generally $E=\sqrt{p^2c^2+m^2c^4}$, then if we equate it to $h\nu$ we get $$p=\sqrt{\frac{h^2}{\lambda^2}-m^2c^2}\neq\frac{h}{\lambda}$$
97245
I'm curious where the  $p=\frac{h}{\lambda}$ comes from?   I know that for light (which has no rest mass), the following is true:  $E=pc$ and $E=hf$  so,  $$pc=hf \Rightarrow p=\frac{hf}{c}=\frac{h}{\lambda}$$  But how is the expression $p=\frac{h}{\lambda}$ obtained for a massive particle where $E\neq pc$? I've read some people claim that the expression can be derived, and others saying it's an experimentally verified relationship.
70903
Given the identically distributed and independant random variables $X_1,X_2,\ldots\sim\operatorname{Po}(\lambda)$ and $S_n=X_1+\ldots+X_n$ show with induction that   $$\Pr[S_n=k]=\frac{(n\lambda)^k}{k!}e^{-n\lambda}.$$   So far for $n=1$ via the definition of the density function for Poisson:$$\Pr[S_1=k]=\Pr[X_1=k]=\frac{e^{-\lambda}\lambda^k}{k!}$$ With $n=2$ and independency: $$\Pr[X_1+X_2=k]=\sum\limits_{n=0}^k\Pr[X_1=n]\Pr[X_2=k-n]=\sum\limits_{n=0}^k\frac{\lambda^n}{n!}e^{-\lambda}\frac{\lambda^{k-n}}{(k-n)!}e^{-\lambda}$$$$=\frac{1}{k!}e^{-2\lambda}\sum\limits_{n=0}^k\binom{k}{n}\lambda^n\lambda^{k-n}=\frac{(2\lambda)^k}{k!}e^{-2\lambda}$$ I assume now that with $S_n\sim\operatorname{Po}(\lambda)$ every $X_i\sim\operatorname{Po}(\lambda/n)$ and every Poisson variable can obviously split up, but how can I prove this?
88883
Consider a  $\phi(x)$, and let its  be $$ \frac{1}{p^2-m^2-\Pi(p^2)}=\int_0^\infty\mathrm d\mu^2\ \rho(\mu^2)\ \frac{1}{p^2-\mu^2} $$  We usually have $\Pi(m^2)=0$, which implies that there is a one-particle state with mass $m^2$. In terms of the , this reads $$ \rho(\mu^2)=\delta(\mu^2-m^2)+\sigma(\mu^2) $$ where the support of $\sigma(\mu^2)$ is disconnected from $m^2$.  On the other hand, the continuum contribution usually begins at $(2m)^2$, meaning that $\Pi(k^2)$ has a branch cut from $(2m)^2$ to $\infty$. Again, in terms of the spectral function, this is written as $\text{supp}(\sigma)=[(2m)^2,\infty)$.  My question  I would expect that the first bound state to lie close to $(2m)^2$, but slightly below, something around $(2m)^2-\mathcal O(\alpha^n)$, where $\alpha$ is the coupling constant and $n\in\mathbb N$. For example, in a simplified QED model, the bound state of an  has binding energy $\sim 7\mathrm{eV}\sim m\alpha^2$, and so I would expect the branch point to be somewhere around $(2m)^2-m^2\alpha^4$.  I know the value of the self-energy to one loop, and the branch point is indeed exactly at $(2m)^2$. I don't know where to find higher loop corrections, so I cannot conclude whether   1) at higher loops, the branch point moves a bit closer to $m^2$, or 2) it stays at $(2m)^2$.   I would expect that the correct behaviour is 1), but this contradicts the fact that all books I've read always depict the threshold of pair production as an hyperboloid with its bottom sitting exactly at $(2m)^2$. On the other hand, if it turns out that the correct option is 2), I am lead to ask why is perturbation theory unable to correctly predict the position of the branch point (I know this is a subtle issue because bound states are, to some extent, non-perturbative entities; but AFAIK, perturbative calculations do contain a lot of information about bound states).
2651
What is the explanation between equality of proton and electron charges (up to a sign)? This is connected to the gauge invariance and renormalization of charge is connected to the renormalization of photon field, but is this explanation enough? Do we have some experimental evidence that quarks have 2/3 and -1/3 charges? By the way I think about bare charge of electron and proton. And I am also wondering if this can be explained by Standard Model.
16822
In section 134 of Vol. 3 (Quantum Mechanics), Landau and Lifshitz make the energy complex in order to describe a particle that can decay:  $$ E = E_0 - \frac{1}{2}i \Gamma. $$  The propagator $U(t) = \exp(-i H t)$ then makes the wavefunction die exponentially with time.  But also, $H$ is non-Hermitian.  My question: Do we have to modify the basic postulates of quantum mechanics (as described by Shankar, say, or the earlier sections of Landau &amp; Lifshitz) to describe unstable particles?
120491
I am currently studying for an exam in Quantum Mechanics and came across a solution to a problem I have trouble with understanding.  The Problem:  A Particle sits in an infinite potential well described by  \begin{align} V(x) &amp;= 0, &amp; 0 \leq x \leq L \\ V(x) &amp;= \infty, &amp; \text{otherwise} \end{align}  We know that the energies are given by $E_n = \dfrac{n^2 \pi^2 \hbar^2}{2 m L^2}$ and $\Psi(x) = A_n \sin(n \pi x /L)$.  At time $t_0$ the potential well is suddenly doubled in size, such that the potential is now  \begin{align} V(x) &amp;= 0, &amp; 0 \leq x \leq 2L \\ V(x) &amp;= \infty, &amp; \text{otherwise} \end{align}  So the energies are now given by $\tilde{E}_n = \dfrac{n^2 \pi^2 \hbar^2}{2 \cdot 4 m L^2}$ and $\tilde{\Psi}(x) = \tilde{A}_n \sin(n \pi x /2L)$.   If the particle is in the ground state of the potential well before the change, what is the probability to find the particle in the ground state of the new potential after the change?   This is absolutely clear to me. We find a non vanishing probability as a result. But now it gets tricky:   What is the expectation value of the energy of the particle directly after the change? How does the expectation value of the energy evolve in time?   The solution suggests that the expectation value of the energy does not evolve in time, which is clear to me, since the Hamiltonian is time independent and thus energy is conserved. But it also suggests that the expectation value does not change after we double the width of the potential wall which I understand from the argument of energy conservation but not in terms of quantum mechanics. If the probability that the particle is in the state $\tilde{\Psi}$ does not vanish the particle could have the energy $\tilde{E}_n$ which is lower than $E_n$ and this would mean that the expectation value of energy could change (with a given probability).  What am I missing here, where is my mistake? Any help is appreciated!
81415
In a quantum mechanics script I'm reading, the Schrödinger equation is "derived" (more precisely, motivated) by the De Broglie hypothesis. It starts at  $$ \lambda = \frac{2\pi h}{p} $$ $$ \omega = \frac{E}{h} $$  then takes the partial derivatives of the wave $\Psi(x,t) = \Psi(0,0)e^{\frac{2\pi ix}{\lambda}-it\omega}$  $$ \tag{1} \frac{\partial}{\partial x}\Psi(x,t) = \frac{ip}{h}\Psi(x,t) $$ $$ \frac{\partial}{\partial t}\Psi(x,t) = -i\frac{E}{h}\Psi(x,t) $$  With the non-relativistic free particle $E=\frac{1}{2m}p^2$ one gets  $$ \tag{2} ih\frac{\partial}{\partial t}\Psi(x,t) = E\Psi(x,t) = \frac{p^2}{2m}\Psi(x,t) $$  From there, they miraculously get the time-dependent Schrödinger equation. I cannot understand this step. If I insert formula (1) for $p^2$ in (2), I get something with $(\frac{\partial}{\partial x}\Psi)^2$, which is not the second partial derivative of $\Psi$ with respect to $x$.  Any hints?
28403
Photons are massless, but if $m = 0$ and , then $E = 0c^2 = 0$. This would say that photons have no energy, which is not true.  However, given the formula $E = ℎf$, a photon does have energy as $ℎ$ is a non-zero constant and photons with a zero frequency do not exist.  What am I missing here, or does the famous formula not apply to photons?
100195
Question solve in $C$ :   $P(z)=z^4+2z^3+5z^2+4z+1=0$  where $P(i+-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$   My attempt :   Let $\lambda=i-\frac{1}{2}+i\frac{\sqrt 3}{2}$  $(2\lambda+1)^2=(i(2+\sqrt 3))^2$  $(2\lambda+1)^2+7)^2=(-4\sqrt 3)^2$  $({\lambda}^2+\lambda+2)^2=3$  So we find :   $\lambda^2++2\lambda^3+5\lambda^2+4\lambda+1=0$  But which step !?  can be find  all root of P(z) !!
50221
Suppose we have a particle of mass $m$ that is in an eigenstate $|\psi\rangle$ of the Hamiltonian $\hat{H}=\hat{T}+\hat{V}$, where $\hat{T}$ is the kinetic energy operator and $\hat{V}=V(\hat{r})$ is the potential energy operator. If the potential has lower bound $V_0$, then is it necessary for the energy eigenvalue $E$ of $|\psi\rangle$ to be greater than $V_0$? Classically this is true, since we regard negative kinetic energy as physically unrealizable/meaningless. However, I don't know if I can necessarily say the same in the quantum case. For example, what I'm tempted to do is write  $$\langle\psi | \hat{T}|\psi \rangle = \langle\psi | E-\hat{V}|\psi\rangle$$  and say "If $E&lt;V_0$, then the RHS is necessarily negative, implying the LHS is as well, which we will regard as physically meaningless. If $E=V_0$, then $|\psi\rangle$ is trivial.", but I'm not sure if that's right.   Confused by that, I then wanted to show that, if $E&lt; V_0$, then a non-trivial $|\psi\rangle$ is non-normalizable. However, I'm not entirely sure how to do this.
47199
In the theory of free scalar bosons (KG field) it is a basic result that the propagator $\Delta(p)$ has poles at $p^2=m^2$, with residue $1$ (or any other constant, depending on conventions). Thinking of $p^2$ as a variable is acceptable because it can be shown that the propagator is a function of $p^2$, even in the interacting case.  On the other hand, in the theory of free spinor fields (Dirac field), we can show that the propagator is not a function of $p^2$, but of $\not p$, so we can't use the $p^2$ trick. Conventions aside, the propagator is $$ S(p)=\frac{\not p+m}{p^2-m^2+i \epsilon} $$  It is clear that we cannot think of poles as $p^2\to m^2$, because $S(p)$ depends on $p$ in a non-so-trivial way. The usual solution is to think of $S$ as a function of $\not p$, as if $\not p$ was a number instead of a matrix, thus writing $$ S(\not p)=\frac{1}{\not p-m+i\epsilon} $$ and, again, we find a pole at $\not p=m$, with residue $1$. I think this is not mathematics. This is just nonsensical to me (may be I'm too skeptic, and we can give a meaning to this last formula).  In an interacting theory, we define the field strength normalisation $Z$ as the residue of the propagator at the poles: $$ \Delta(p^2)=\frac{Z}{p^2-m^2+i\epsilon}+\int_{M_\text{thresh}^2}^\infty \mathrm d\mu^2\rho(\mu^2)\frac{1}{p^2-\mu^2+i\epsilon} $$ $$ S(\not p)=\frac{Z}{\not p-m+i\epsilon}+\int_{M_\text{thresh}^2}^\infty \mathrm d\mu^2\frac{\not p\rho_1(\mu^2)+\mu \rho_2(\mu^2)}{\not p^2-\mu^2+i\epsilon} $$  Now, I don't understand what's the proper definition of $Z$ as a residue. In what sense is it a residue? What's the variable? I just can't accept it is a residue as $\not p\to m$. Do we really take this "think of $\not p$ as a variable'' seriously? Is it possible to formalise these matters, by thinking of a residue for an actual variable (as $p_0$)?  Perhaps the $\not p$ trick is just that: a trick, which may simplify calculations, but such that it can be shown that the actual result, found by a more standard procedure, is the same. Is this the case? If so, what is the correct procedure?  (I would really appreciate if the answers don't assume that we can boost to the rest frame of the particle, as I'd like it to be as general as posible. I want to take into account the possibility of $m=0$, so please don't boost into $k=(m,\boldsymbol 0)$ if it's not really necessary)
61359
I believe that the MLE of parameter $p$ in the geometric distribution, $\hat p = 1/(\bar x +1)$, is an unbiased estimator for $p$ and would like to prove it. So far, I have:  $E[\bar x + 1] = E[\bar x] + 1 = \frac{1}{p} - 1 + 1 = \frac{1}{p}$  I can see that the relationship is likely there, but I don't know how to work with the $(\bar x + 1)$ being in the denominator.
352
Up to which precision has the coulomb law proven to be true? I.e.  if you have two electrons in a vacuum chamber, 5 meters appart, have the third order terms been ruled out? Are there any theoretical limits to measure the precision ( Planck's constant?). Obviously there are practical limitations ( imperfect vacuum, cosmic rays, vacuum fluctuation). Still, does anyone know what was the smallest amount ever correctly predicted by that law?    Edit : Summary  On the high end of the energy spectrum a precision of 10^-16 has been shown ( 42 years ago )  For electron point charges at large distances the law might brake down due to practical reasons.  For moving particles QED gives a correction to the law:
70683
If $E(X^2)=1$ and $E(|X|)\ge a &gt;0$, then $P(|X|\ge\lambda a)\ge (1-\lambda)^2a^2$ for $0\le \lambda \le 1$.  I can see from the well known inequality $E(|X|) \le E(|X|^2)^{1/2}$ that it must be the case that $a\le 1$. But what to do next I'm not sure.
34694
Two fundamental equations regarding  are: $$ \lambda = \frac{h}{p}, \\ \nu = E/h .$$  We talk about , is it meaningful to talk about de Broglie frequency ($\nu$ above) and de Broglie velocity ($\nu \lambda$)? Are these two equations independent or can one derive one from the other? Or mid-way, does one impose constraint on other? In case of light or photons we can relate frequency and wave, is there similar interpretation in case frequency and wavelengths in above equations? Comparing with that of light, if we multiple $\lambda$ and $\nu$ we get velocity, what does this velocity mean here?  If we do the above calculation for an average human, what would be the meaning of $\nu$ and $\nu \lambda$? Are we jiggling with that $\nu$?
1740
The simplest equation expressing mass–energy equivalence is the famous $E=mc^2$ where $c$ represents the speed of light. Compare this with $E_K = \frac{1}{2}mv^2$.   Since $E=mc^2$ can be applied to rest mass ($m_0$) and rest energy ($E_0$) to show their proportionality as  $E_0=m_0c^2$, I ask whether this resemblance is just a coincidence created by the need of any equation to be homogeneous for units, or are the two equations fundamentally related?  I know about  but I could not find the answer I'm looking for there.
25543
states that:    $$\sigma(x)\sigma(  p_x )\ge \frac {\hbar}{2}.$$  What is the scientific proof of this principle?