The time-independent Shrödinger equation (TISE) is   $$\dfrac{-\hbar^2}{2m} \nabla^2 \psi + V \psi = E \psi,$$  where $m$ is the mass of the particle.  I just had a thought: If $m$ is the particle's mass, then the TISE is invalid for photons, since they're massless! But I know that the Shrödinger equation is used to model photons. So how can this be?
123184
In basic quantum mechanics courses, one describes the evolution of quantum mechanics chronologically. Interference experiments with particles showed that particles should have a wave character; on the other hand, the photo-electric effect showed that electromagnetic waves should have a particle character. So there's a particle-wave duality.  The course continues with postulating the Schrödinger equation, and solves the problems for the wave character of classical particles (e.g. with respect to harmonic oscillators to solve the UV-catastrophe in the description of blackbody radiation). On the other hand, one describes the photo-electric effect extensively with perturbation theory, as the electromagnetic wave acts as a periodical force on an electron. But the electromagnetic wave is still described with the Maxwell equations, and not by Schrödinger.  Suppose history went a little differently: the corpuscular character would still be the generally accepted behaviour of light (as Newton preferred), and one discovered the wave character of photons and electrons simultaneously by doing interference experiments. The laws of Maxwell, Gauss, Faraday and Ampère weren't discovered yet. Schrödinger comes up with his equation, and one describes the wave character of electrons as well as photons with this equation.   Would this equation give a exhaustive description of the electromagnetic (and vector like) behaviour of photons? In other words: is the Schrödinger theory, in some way, equivalent to the Maxwell theory for the description of photons? Or are the Maxwell equations some kind of limit for greater dimensions (like the Newton equations for mechanics)? What is the link between these two "wave character" descriptions of photons?
45265
I want to ask if it is reasonable that I use the Dirac-Delta function as an initial 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? If I use this initial state, can I also use it to predict how the wavefunction should evolve in time? That is, if $\Psi (x,0) = \delta(x) $, then, $$ \phi(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \delta(x)e^{-ikx} dx=\frac{1}{\sqrt{2 \pi}} $$ then, $$ \Psi(x,t)= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \phi(k)e^{i(kx-\omega t)} dk $$ or, $$ \Psi(x,t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty}e^{i(kx-\omega t)} dk. $$  Is the above wavefunction a valid expression to see how a particle that is initially localized in the origin evolves in time?
100859
Motivation: In odd dimensions, solutions to the wave equation: $$u_{tt}(x,t)=\nabla u(x,t), \qquad u_t(x,0)=0, \qquad u(x,0)=f(x)$$ where $t \geq 0$ and $x \in \mathbb{R}^n$, have the nice property that the value of $u(x,t)$ only depends on the values $f(y)$ with $|y-x|=t$. For even dimensions, the value $u(x,t)$ depend on all the values $f(y)$ with $|y-x|\leq t$. A consequence of this is that when you switch a light bulb on and then off (in 3D), there will be a light wave traveling with the speed on light and behind the wave, there will be total darkness. But when you throw a rock into a pond (with a 2D surface), there will be lots of waves traveling outwards from where the rock hit the water and, in theory, the water will never a still again.  Question: Can anyone give an intuitive explanation of this difference between odd and even dimensions?
95226
If a particle is totally localized at $x=0$, its wave function $\Psi(x,t)$ should be a Dirac delta function $\delta(x)$. Accordingly, its Fourier transform $\Phi(p,t)$ would be a constant for all $p$, thus the particle's momentum is totally uncertain. I guess this is what the Uncertainty Principle told us.  But in another hand, since the particle is totally localized at $x=0$, it is not moving and therefore static. Its velocity should be 0, so is its momentum, which contradicts the conclusion above.  So, what's wrong with my thinking?  Is it that there shouldn't be a TOTALLY STATIC particle, or it's in a non-normalizable state, so the uncertainty principle is not applicable?
87891
I am able to understand how light can be modeled to have wave characteristics from Young's double slit experiment.  But I am unable to comprehend how we can understand light to have particle characteristics from the photoelectric experiment. How is it wave character not able to explain the phenomena observed in this experiment? And how is it that particle nature defeats the wave theory?
115416
Disclaimer: this is a homework question, so I am happy with just a hint or the expressions needed to proceed with my understanding.  I am working on the momentum conservation of a particle/anti-particle annihilation process, and I have been asked to show that the annihilation of a particle with a finite mass and its anti-particle cannot lead to the emission of only one photon.   I understand why this happens: the conservation of momentum. However, I would like show this in a more sophisticated 4-momentum proof...how would I go about showing that momentum is conserved for two photons but it is not conserved when the annihilation process creates just one photon...?  This may be a duplicate of:
109438
Let's denote by $|\psi(t)\rangle$ some wavefunction at time $t$. Then let's define the time evolution operator $U(t_1,t_2)$ through $$ U(t_2,t_1) |\psi(t_1)\rangle = |\psi(t_2)\rangle \tag{1}$$ and Hamiltonian $H$ through $$ H |\psi(t)\rangle = i\frac{\partial}{\partial t} |\psi(t)\rangle. \tag{2}$$ (We have set $\hbar = 1$.) Question. What is the relation between $U$ and $H$, given that $H$ doesn't depend explicitly on $t$? Attempts. It is well-known that the answer is $U(t_2,t_1) = c \cdot e^{-i(t_2-t_1)H}$ for some scalar $c$. Some sources postulate it follows from operator equation $HU = i \frac{\partial}{\partial t} U$ but we cannot integrate it like we could an ordinary differential equation $f(x)g(x) = \frac{\partial}{\partial x} g(x)$ (at least I don't know how to integrate operators!). Then we could also &quot;guess&quot; the solution $U(t_2,t_1) = c \cdot e^{-i(t_2-t_1)H}$ and verify by inserting into $(1)$, but then we still need to show that this is the general solution, i.e., $e^{-i(t_2-t_1)H}$ spans the set of all solutions. Also this approach doesn't enlighten as to why we chose that particular guess.
92904
I'm an undergraduate mathematics student trying to understand some quantum mechanics, but I'm having a hard time understanding what is the status of the Schrödinger equation.  In some places I've read that it's just a postulate. At least, that's how I interpret e.g. the following quote:     Where did we get that (equation) from? Nowhere. It is not possible to derive it from anything you know. It came out of the mind of Schrödinger.    -- Richard Feynman   (from the Wikipedia entry on the Schrödinger equation)  However, some places seem to derive the Schrödinger equation: just search for "derivation of Schrödinger equation" in google.  This motivates the question in the title: Is the Schrödinger equation derived or postulated? If it is derived, then just how is it derived, and from what principles? If it is postulated, then it surely came out of somewhere. Something like "in these special cases it can be derived, and then we postulate it works in general". Or maybe not?  Thanks in advance, and please bear with my physical ignorance.
83616
My book on quantum mechanics suggests a derivation of the wave equation  $$\left(\Delta - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \right) \psi(\bar{r},t) = 0$$  from the photon energy-impulse relation  $$E^2 = c^2p^2$$  using the substitutions  $$\bar{p} \to -i\hbar \nabla$$ and $$E \to i \hbar \frac{\partial}{\partial t}$$  I suppose the $\hbar$s are there for the non-photon generalisation, but is there a reason for the minus sign in the $p$ substitution? Removing it would yield the same result.
36657
I am going through Shankar's Principles of Quantum Mechanics and am having trouble finding the free particle propagator $U(t)$ that satisfies $$\lvert\psi (t)\rangle = U(t)\lvert\psi (0)\rangle$$ due to the degeneracy of the $E$ eigenkets. Shankar says that if the Hamiltonian has degenerate eigenvalues, we change the propagator equation from $$U(t) = \sum_{E} \lvert E\rangle\langle E\rvert e^{-iEt/{\hbar}}\tag{4.3.13}$$ to $$U(t) = \sum_{E} \sum_\alpha \lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar} $$ where $\lvert E, \alpha \rangle $ are the orthonormal eigenkets for the $E$ eigenspace. This makes sense to me, because it seems like whenever we deal with degeneracy to use a projection. He also says for a Hamiltonian with no degeneracy to change the sum into an integral, so I imagine that for a Hamiltonian with a continuous, degenerate spectrum we would have this propagator: $$ U(t) = \sum_\alpha\int_\Re dE\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}$$  I tried applying this to the free particle and had some trouble. In the book, after solving for the eigenkets and eigenvalues, $\lvert E\rangle = \lvert p\rangle$ and $p = \pm\sqrt{2mE}$, he chooses the propagator as a function of the momentum eigenvalues $p$ so that $$U(t) = \int_\Re dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m).\tag{5.1.9}$$ I understand how he got to this, but I'm having some trouble because this doesn't appear to be the same expression that I have above. Splitting the integral up, we have $$U(t) = \int_{-\infty}^0dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ The left integral can changed to an integral over $-p$, and the bounds switch from $0$ to $\infty$ such that: $$U(t) =  \int_{0}^\infty d(-p)\lvert -p\rangle\langle -p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ Now we can substitute and interchange $\lvert p\rangle = \lvert E, + \rangle$,  $\lvert -p\rangle = \lvert E, - \rangle$, $E = p^2/2m$, and $dp = \pm mdE/\sqrt{2mE}$ to get: $$U(t) = \sum_{\alpha = \pm} \int_0^\infty dEm/\sqrt{2mE}\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}. $$ This is the result that we are asked to prove in Exercise 5.1.1. Why is this not the same as the first expression for the propagator? Where did the extra $ m/\sqrt{2mE} $ come from? Am I just wrong in assuming that that is the general expression for a Hamiltonian with continuous and degenerate eigenvalues?
80084
This is inspired by Griffiths QM section 2.2, on the infinite square well, which is about how far I've gotten (so, sorry if this is addressed later in the book).  For any given starting wavefunction, you can express it as a sum over the solutions of the time-independent Schrödinger equation. The coefficients in the sum are constant in time, therefore the expectation value of energy is constant in time. The chapter says "this is a manifestation of conservation of energy in quantum mechanics." Ok.  Now in problem 2.5, we have a wavefunction that is an even mixture of the first two stationary states. Part D has us compute the expectation value of momentum, and the solution is sinusoidal in time. So momentum is not conserved.  How can momentum be not conserved? Why is energy conserved but not momentum?
62384
I have been studying the semi-classical light matter interaction from the book, &quot;Light matter interaction&quot; by Weiner and Ho. They have defined the total Hamiltonian of a two level atom placed in an EM radiation as $$\hat{H}=\hat{H_{0}}+\hat{V}(t)$$ where $H_0$ is the unperturbed Hamiltonian of the two level atom and $\hat{V}(t)$ is the dipole interaction term given by $\hat{V}(t)=\hat{\vec{d}}\cdot\vec{E}$. $\vec{d}$ is the dipole moment of the atom given by $\vec{d}=-e\vec{r}$. The eigenstates of $H_0$ are $|1\rangle$ and $|2\rangle$. Now, it has been argued that since $V(t)$ has an odd parity with respect to $\vec{r}$, the diagonal terms $V_{ii}=\langle i|\hat{V}|i \rangle=0$. However, I am not able to understand why this should be so. Some thoughts Without loss of generality, we can take the dipole moment to be $\hat{\vec{d}}=-e\hat{x} \mathbf{e_x}$ and the driving field $\vec{E}=E_0 cos(\omega t) \mathbf{e_n}$, so that $\hat{V}(t)=-e\hat{x} E_0 cos(\omega t) cos(\phi)$ where $\phi$ is the angle between $\mathbf{e_n}$ and $\mathbf{e_x}$. In such a case, we have $$V_{ii}=\langle i|\hat{V}|i \rangle=e E_0 cos(\omega t)cos(\phi)\int_{-\infty}^{\infty}\phi^{*}_i(\vec{r}) x \phi_i(\vec{r}) d\vec{r}$$ If the wavefunction $\phi^{*}_i(\vec{r})$ has a definite parity(assumption 1), then indeed this integral is $0$. In case of a spherically symmetric potential with no interaction between electrons in the atom, assumption 1 indeed holds. However, in the presence of interaction between electrons, I am not so sure if it will hold true! Am I thinking about it the right way? Is there a way to prove this without assumption 1? Or is the assumption 1 always true?
63304
Consider the time-dependent  (or some equation in Schrödinger form) written down as $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{ H}~ \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0). $$ However, this form for the solution of $(1)$ is actually built by the method of successive approximations which actually returns a solution of the form $$ \tag 3 \Psi (t) ~=~ \hat{\mathrm T} \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime})~\mathrm  dt^{\prime}\right]\Psi (0), \qquad t&gt;0, $$ where $\hat{\mathrm T}$ is the time-ordering operator.  It seems that $(3)$ doesn't coincide with $(2)$, but formally $(2)$ seems to be perfectly fine: it satisfies $(1)$ and the initial conditions. So where is the mistake?
82440
Earlier questions under a similar title referred to the short range of the strong force. My question is completely different. I'd like to know why gluons are considered massless in the Standard Model (regardless of the range of the strong force). For example, if mass is a result of the Higgs interactions, then why do gluons not interact with the Higgs boson? In addition to the theoretical reasons, is there any experimental evidence of gluons being actually massless? (I do realize that such experiments are not easy, because of the confinement.)  In the past, neutrinos were also considered massless, but not anymore. Is there a similar possibility for gluons to have perhaps a small invariant mass or is there an overriding reason for them to be definitely massless?  EDIT: The answer I've been looking for is deep in the comments below and not immediately apparent. To make it clear, I am repeating it here: To interact with a boson, a particle (the Higgs) must have a charge mediated by this boson. The Higgs has a weak charge and therefore interacts with the W and Z bosons thus giving them mass. The Higgs does not have the electric or color charge and therefore does not interact with the photon or gluons thus leaving them massless.
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The $\delta(x)$ Dirac delta is not the only "point-supported" potential that we can integrate; in principle all their derivatives $\delta', \delta'', ...$ exist also, do they?  If yes, can we look for bound states in any of these $\delta'^{(n)}(x)$ potentials? Are there explicit formulae for them (and for the scattering states)?  To be more precise, I am asking for explicit solutions of the 1D Schroedinger equation with point potential,  $$- {\hbar^2 \over 2m} \Psi_n''(x) + a \  \delta'^{(n)}(x) \Psi(x) \ = E_n \Psi_n(x) $$  I should add that I have read at least of three set of boundary conditions that are said to be particular solutions:   $\Psi'(0^+)-\Psi'(0^-)= A \Psi(0)$ with $\Psi(0)$ continuous, is the zero-th derivative case, the "delta potential". $\Psi(0^+)-\Psi(0^-)= B \Psi'(0)$ with $\Psi'(0)$ continuous, was called "the delta prime potential" by Holden. $\lambda \Psi'(0^+)=\Psi'(0^-)$ and $\Psi(0^+)=\lambda\Psi(0^-)$ simultaneusly, was called "the delta prime potential" by Kurasov   The zero-th derivative case, $V(x)=a \delta(x)$ is a typical textbook example, pretty nice because it only has a bound state, for negative $a$, and it acts as a kind of barrier for positive $a$. So it is interesting to ask for other values of $n$ and of course for the general case and if it offers more bound states or other properties. Is it even possible to consider $n$ beyond the first derivative?  Related questions  (If you recall a related question, feel free to suggest it in the comments, or even to edit directly if you have privileges for it)  For the delta prime, including velocity-dependent potentials, the question has been asked in   In the halfline $r&gt;0$, the delta is called "Fermi Pseudopotential". As of today I can not see questions about it, but  seems to be the same potential.  A general way of taking with boundaring conditions is via the theory of self-adjoint extensions of hermitian operators. This case is not very different of the "particle in 1D box", question  A general take was the question  A related but very exotic question is  because obviosly the point-supported interacctions have a peculiar scaling  Comments  Of course upgrading distributions to look as operators in $L^2$ is delicate, and it goes worse for derivatives of distributions when you consider its evaluation $&lt;\phi | \rho(x) \psi&gt;$. Consider the case $\rho(x) = \delta'(x) = \delta(x) {d\over dx}$. Should the derivative apply to $\psi$ only, or to the product $\phi^*\psi$?
23488
In the particle in a box, harmonic oscillator and in Hydrogen Atom, we can safely assume $$\Psi(x,t) = \psi(x)e^{-i\omega t}.$$ So why not make it a postulate to consider the wave function to be always in the form $$\psi(x)e^{-i\omega t}$$ We can still explain all the foundational experiments, so my question is why not make this amendment to the theory so that we can avoid a lot of nightmares and some peace of mind. Why unnecessarily consider so general thing when we can do away with simple things most of the time?  Oh my actual question is where do we run into problems (practical physical situations/experiments) if we consider such a thing? If we run into problem anywhere, I hope to make any other simple changes to fix them but without going back on this one.
22934
Let $u$ solve the initial-value problem or the wave equation in one dimension: $$\begin{cases}u_{tt}-u_{xx}=0 &amp; \text{in } \mathbb{R} \times (0,\infty) \\ u = g, u_t = h &amp; \text{on } \mathbb{R} \times \{t=0\}. \end{cases}$$ Suppose $g,h$ have compact support. The kinetic energy is $k(t) := \frac 12 \int_{-\infty}^\infty u_t^2(x,t) \, dx$ and the potential energy is $p(t) := \frac 12 \int_{-\infty}^\infty u_x^2(x,t) \, dx$. Prove (a) $k(t)+p(t)$ is constant in $t$, (b) $k(t)=p(t)$ for all large enough times $t$.  This is Chapter 2, Exercise 24 of PDE Evans, 2nd edition. I am only doing part (a) right now; my work is shown below: Define $$e(t):=k(t)+p(t)=\int_{-\infty}^\infty u_t^2+u_x^2 \, dx.$$ Then $$e'(t)=\frac 12 \int_{-\infty}^\infty 2u_tu_{tt}+2u_{x}u_{xt} \, dx= \int_{-\infty}^\infty u_tu_{tt}+u_{x}u_{xt} \, dx.$$ Now, I want to get $e'(t)=0$ so that $e(t)$ is constant. How can I go about doing this? I do know that I can use $u_{tt}-u_{xx}=0$, if $t &gt; 0$. With $t&gt;0$ into mind, do I have to consider the $t=0$ case separately? Or can I treat both cases together as $t \ge 0$, since $g$ and $h$ have compact support?