I have always been confused by the relationship between the  and the .  $$ i\hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m} \nabla^2+ U \psi  \hspace{0.25in}\text{-vs-}\hspace{0.25in}\nabla^2 E = \frac{1}{c^2}\frac{\partial^2 E}{\partial^2 t} $$  Because of the first derivative, the Schrödinger equation looks more like the .    Some derivations of the Schrodinger equation start from  for light and argue that matter should also exhibit this phenomenon.    In some notes by , it was derived by comparing the  $\delta \int n \;ds = 0 $ and Maupertuis  $\delta \int 2T(t) \; dt = 0 $.  Was this ever clarified?  How can we see the idea of a matter-wave more quantitatively?    To summarize, I am trying to understand why the Electromagnetic wave equation is hyperbolic while the Schrodinger equation is parabolic.
47934
Wave equations take the form:  $$\frac{ \partial^2 f}  {\partial t^2} = c^2 \nabla ^2f$$  But the Schroedinger equation takes the form:  $$i \hbar \frac{ \partial f}  {\partial t} = - \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$  The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation?  And why are interference patterns (e.g in the double-slit experiment) so similar for water waves and quantum wavefunctions?
17341
Is there anything in the physics that enforces the wave function to be $C^2$? Are weak solutions to the Schroedinger equation physical? I am reading the beginning chapters of Griffiths and he doesn't mention anything.
74734
I'm trying to teach myself separation of variables and have been following some notes for the wave equation, but there's one part which really confuses me and I'm not exactly sure how it makes the step.  For the wave equation   $$u_{tt} - c^2 u_{xx} = 0$$   with length $l$ and fixed ends, $u(0,t)=u(l,t)=0$ we seek a solution in the form $$u(x,t)=v(x)q(t)$$  and substituting this into the equation gives $$\frac{1}{q(t)}\frac{d^2 q}{dt^2}=c^2 \frac{1}{v(x)}\frac{d^2 v}{dx^2}=-\omega ^2$$   and I understand all of this so far. But when solving the equation  $$q''+\omega ^2 q = 0$$ the general solution is $$q(t)=A \cos(\omega t +\alpha)$$ where $A$ and $\alpha$ are constants - which is the part which I'm not exactly sure on how it gets to.   Could someone explain this step to me? Thanks!
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?
17972
I came across a problem in Griffiths where the derivative of the wave function (with respect to position in one dimension) evaluated at $\pm\infty$ is zero. Why is this? Is it true for any function that evaluates to zero at $\pm\infty$ or is there a special constraint on the wave function that I am forgetting?
45792
I read that it makes a difference whether you calculate $\frac{dE(\lambda) }{d \lambda}=0$ or $\frac{dE(\omega)}{d \omega}=0$ in the sense that the maximum energy density with respect to the wavelength does not coincide with the frequency maximum that one would assume to be at $\omega_{max} = \frac{c}{\lambda_{max}}$. Actually, I do not understand why this is so ( Now, I am only interested in a pure physical explanation, mathematically this is clear and due to the chain rule). Does anybody know how to explain this odd thing?  Somehow I feel that the core of the question has not been completely answered. Although I can easily look up wikipedia, where the need for the chain rule is explained, I am rather interested in understanding where the following argument breaks down, which seems to be why this question has caused some confusion in the past(as you can see by googling this question):  So we have $E(\omega)$ the energy radiated at a given frequency. Now this function has a maximum somewhere, so there is a frequency where a maximal amount of energy is emitted. In other words: If you add the sum of the photon's energies at each frequency that are emitted, you will notice that the maximum is reached at this frquency. Now $E(\lambda)$ tells you basically the same for the wavelength, but again: We know where at which frequency the maximal amount of energy is radiated, so we know the corresponding wavelength.   I think there is a need to explain this.   If anything is unclear, please let me know.
2666
In QED, according to Schwinger-Dyson equation $^{[1]}$, $$\left(\eta^{\mu\nu}(\partial ^2)-(1-\frac{1}{\xi})\partial^{\mu}\partial^{\nu}\right)\langle 0|\mathcal{T}A_{\nu}(x)...|0\rangle = e\,\langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle + \text{contact terms}.$$ And the term $\left(\eta^{\mu\nu}(\partial ^2)-(1-\frac{1}{\xi})\partial^{\mu}\partial^{\nu}\right)$ is just the inverse bare photon propagator$^{[5]}$, so if we put the photon on shell, then the l.h.s will yield the complete $n$-point Green function with the complete photon propagator removed and also multiplied by a factor $Z_3$, the vector field renormalization constant. But the r.h.s gives, according to the Ward Identity associated with global symmetry of the Lagrangian, $$\partial_{\mu}\, \langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle = \text{contact terms}^{[2]}$$ which is the sum of complete $(n-1)$-point complete Green functions each multiplied by a certain $\delta$ function.  So if we truncate all the $n-1$ external complete propagators, then we are left with the proper vertex Ward identity.    The problem is, now the constant $Z_3$ appeared.    For, example, if we apply this to $j^\mu=\bar\psi\gamma^\mu\psi$, $$\partial_{\mu}\, \langle 0|\mathcal{T}j^{\mu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle = -ie[\delta(x-x_1)\langle 0|\psi(x_1)\bar\psi(x_2)|0\rangle-\delta(x-x_2)\langle 0|\psi(x_1)\bar\psi(x_2)|0\rangle]^{[3]},$$ we will get an identity containing $Z_3$ RATHER THAN the well known proper vertex Ward identity (which is derived from the gauge or local symmetry of the Lagrangian), e.g.   $$q_\mu\Gamma^\mu_P(p,q,p+q)=S^{-1}(p+q)-S^{-1}(p)^{[4]}$$ which doesn't contain $Z_3$.    Where went wrong? Please help.  [1]. Peskin&amp;Schroeder, An Introduction to Quantum Field Theory, page308, eq.(9.88) [2]. Peskin&amp;Schroeder, An Introduction to Quantum Field Theory, page310, eq.(9.97) [3]. Peskin&amp;Schroeder, An Introduction to Quantum Field Theory, page311, eq.(9.103) [4]. Lewis H.Ryder, Quantum Field Theory, page 263-266, eq.(7.112) [5]. Peskin&amp;Schroeder, An Introduction to Quantum Field Theory, page297, eq.(9.58)
45781
In the book of Ashok Das, Field theory-path integral approach, he begin the demonstration of the Schwinger-Dyson equation using the fact that the $\delta Z[J]=0$, so  \begin{equation} \delta Z[J]=\int \mathcal{D} \phi \frac{\delta S[\phi,J]}{\delta \phi(x)} e^{iS[\phi,J]}=0, \end{equation} but we already know that  \begin{equation} \frac{\delta S[\phi, J]}{\delta \phi(x)}=F(\phi(x))-J(x), \end{equation} where $F(\phi(x))$ is the equation of motion.   So if we go back to the first equation and use the identification  \begin{equation} \phi(x)\rightarrow -i\frac{\delta}{\delta J(x)}, \end{equation}  we conclude that  $$ \int \mathcal{D}\phi\left(F(\phi(x))-J(x)\right)e^{iS[\phi,J]}=\left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)\int \mathcal{D}\phi e^{iS[\phi,J]} $$ $$ \left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)Z[J]=\left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)e^{iW[J]}=0. $$  But is here where I get lost, how did he pass from the above equation for  $$ e^{-iW[J]}\left(F\left(-i\frac{\delta }{\delta J(x)}\right)-J(x)\right)e^{iW[J]}=F\left(\frac{\delta W[J]}{\delta J(x)}-i\frac{\delta}{\delta J(x)}\right)-J(x)=0. $$  A more important question is: what does this last equation mean at all, mathematically speaking, because the functional derivative now is acting on nothing.
101251
This is a question I've been asked several times by students and I tend to have a hard time phrasing it in terms they can understand. This is a natural question to ask and it is not usually well covered in textbooks, so I would like to know of various perspectives and explanations that I can use when teaching.  The question comes up naturally in what is usually students' second course in quantum physics / quantum mechanics. At that stage one is fairly comfortable with the concept of wavefunctions and with the Schrödinger equation, and has had some limited exposure to operators. One common case, for example, is to explain that some operators commute and that this means the corresponding observables are 'compatible' and that there exists a mutual eigenbasis; the commutation relation is usually expressed as $[A,B]=0$ but no more is said about that object.  This naturally leaves students wondering     what is, exactly, the physical significance of the object $[A,B]$ itself?   and this is not an easy question. I would like answers to address this directly, ideally at a variety of levels of abstraction and required background. Note also that I'm much more interested in the object $[A,B]$ itself than what the consequences and interpretations are when it is zero, as those are far easier and explored in much more depth in most resources.    One reason this is a hard question (and that commutators are such confusing objects for students) is that they serve a variety of purposes, with only thin connecting threads between them (at least as seen from the bottom-up perspective).   Commutation relations are usually expressed in the form $[A,B]=0$ even though, a priori, there appears to be little motivation for the introduction of such terminology. A lot of stock is placed behind the canonical commutation relation $[x,p]=i \hbar$, though it is not always clear what it means.  (In my view, the fundamental principle that this encodes is essentially de Broglie's relation $\lambda=h/p$; this is made rigorous by the  but that's quite a bit to expect a student to grasp at a first go.) From this there is a natural extension to the Heisenberg Uncertainty Principle, which in its general form includes a commutator (and an anticommutator, to make things worse). Canonically-conjugate pairs of observables are often introduced, and this is often aided by observations on commutators. (On the other hand, the energy-time and angle-angular momentum conjugacy relations cannot be expressed in terms of commutators, making things even fuzzier.) Commutators are used very frequently, for example, when studying the angular momentum algebra of quantum mechanics. It is clear they play a big role in encoding symmetries in quantum mechanics but it is hardly made clear how and why, and particularly why the combination $AB-BA$ should be important for symmetry considerations.  This becomes even more important in more rigorous treatments of quantum mechanics, where the specifics of the Hilbert space become less important and the algebra of observable operators takes centre stage. The commutator is the central operation of that algebra, but again it's not very clear why that combination should be special. An analogy is occasionally made to the Poisson brackets of hamiltonian mechanics, but this hardly helps - Poisson brackets are equally mysterious. This also ties the commutator in with time evolution, both on the classical side and via the Heisenberg equation of motion.   I can't think of any more at the moment but they are a huge number of opposing directions which can make everything very confusing, and there is rarely a uniting thread. So: what are commutators, exactly, and why are they so important?
41151
The Euler-Lagrange equation for particles is given by   $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q},\tag{1}$$  and for fields  it is   $$ \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} = \frac{\partial \mathcal{L}}{\partial \phi}.\tag{2} $$  Comparing the two equations, the first one has a total time derivative $\frac{d}{dt}$ but the other one appears to have partial derivatives $\partial_{\mu}$. These derivatives come from integration by parts in the derivation of the EL equation. I was wondering why the field version has partial derivatives and the particle version has total derivatives?   I have also seen for the specific example (in Quantum Field Theory for the gifted amateur) of 1 dimensional waves on a string, the corresponding Euler-Lagrange equation is  $$ \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \left(\frac{d\phi}{dt}\right)} + \frac{d}{dx} \frac{\partial \mathcal{L}}{\partial \left(\frac{d\phi}{dx}\right)} = \frac{\partial \mathcal{L}}{\partial \phi}, \tag{3}$$  which uses total derivatives, so I am a bit confused.
104966
For the definition of the momentum operator $$\hat{P } = -i \hbar \nabla$$ in quantum mechanics, as I understand you can derive this by either considering a more general definition of momentum, i.e. 'canonical momentum' which is an operator and then apply this operator to wave functions. This is shown here in this . We can alternatlively start with translations and use the face that the momentum operator is a generator of translations as is done here in this .   What I am interested in is what is the more fundamental derivation for the position operator: $$\hat{X} = x.$$ To this point I have considered that the motivation for defining position operator is from the definition of the expectation value $$\langle x \rangle = \int dx x |\psi(x)|^2 = \langle \psi|x | \psi \rangle$$ where $| \psi \rangle$ is normalised. Is this the full extent of the motivation or the main point to consider? Is there another motivation for the position operator or is it taken from this motivation and confirmed by experiments to be acceptable?
43963
In the Schrödinger equation, at any given time $t$ we should jointly add another sub equation, like $$||\psi_t(x)|| = 1$$ where $\psi_t(x) = \Psi(x,t)$, and then try to solve the two equations simultaneously. Why not? I know it does not yield, but I am always baffled, who is doing the normalization of the wave function? Observers, the system, the measuring process, God?
81203
I've been studying the postulates of QM and seeing how to derive important ideas from them. One thing that I haven't been able to derive from them, however, is the identity of the momentum operator.  For simplicity, I'm only thinking about no relativistic effects, no spin, no time-dependent potentials, and one spatial dimension. Also I'm assuming the position operator is simply multiplication by $x$, as in, I'm in position space. So the Hamiltonian operator is  $ H = -\frac{\hbar^2}{2m}\nabla^2+V$.  I know that the momentum operator is $p = -i\hbar \frac{\partial}{\partial x}$.  But how do I get there from the postulates? I know that it makes sense, as it results in the Ehrenfest Theorem, the De Broglie wavelength hypothesis, the Heisenberg Uncertainty Principle (for $x$ and $p$), the momentum operator being the generator of the translation operator, and possibly many other desirable theorems, and correlations with classical momentum.  But none of these are postulates (at least, not in the various formalisms I encountered), so you can't derive $p = -i\hbar \frac{\partial}{\partial x}$ from them. Rather, they are consequences of it. You need to know the operator beforehand to see that they are correct. Yes, this is just semantics, but that is the core issue for me:  Regardless of how much sense it makes, is the identity $p = -i\hbar \frac{\partial}{\partial x}$ (under the assumptions I made) a Postulate, meaning that you can't derive it from other postulates, or can it in fact be obtained from them? And in the latter case, could you show me how?  Note: I know that there are many different and equivalent sets of postulates for QM. But in none that I saw did they name it as a postulate nor properly derived it.
57771
We know that in a wave equation $y = A \sin (\omega t + kx + \phi)$, $y$ = Displacement of particle on y-axis (assuming transverse wave) $A$ = Amplitude $\omega$ = Angular velocity $k$ = wave number and $\phi$ = phase difference  I understand that the everything within the bracket in the equation is an angle. But I dont understand what is the difference between those terms. When do we use $\omega t$ and $kx$ and $\phi$?  As much as I understand $\phi$ is taken into account when we need to find the angle difference between two waves. But I am really confused between the other two. Is there even any difference between both of them?  NOTE: I read  and  Phys.SE questions but they don't take into account the $kx$ term in their phase.
118042
I'm currently referring to the wave equation derivation given in "Introduction to Electrodynamics" by David J. Griffiths. It follows something like this:  The electromagnetic wave equations are given by the equations: \begin{equation}  v^2_{ph}\nabla^2\textbf{E} = \frac{\partial^2 \textbf{E}}{\partial t^2} \tag{1}\label{eq1} \end{equation} \begin{equation} v^2_{ph}\nabla^2\textbf{B} = \frac{\partial^2 \textbf{B}}{\partial t^2} \tag{2}\label{eq2} \end{equation} Where, $v_{ph} = \frac{1}{\sqrt{\mu\epsilon}} $, is the velocity at which the wave travels in the medium    We have the Heaviside' form of Maxwell's equations in differential form, $$\nabla . \textbf{E} = \frac{\rho}{\epsilon_0}$$ $$\nabla . \textbf{B} = 0$$ $$\nabla \times \textbf{E} = -\frac{\partial\textbf{B}}{\partial t}$$ $$\nabla \times \textbf{B} = \mu_0 \left(\textbf{J} + \epsilon_0\frac{\partial\textbf{E}}{\partial t}\right)$$ In a vacuum and charge-free space ($\rho = 0, \textbf{J} = 0$), these equations are:, \begin{equation} \nabla . \textbf{E} = 0 \tag{3}\label{eq3} \end{equation} \begin{equation} \nabla . \textbf{B} = 0 \tag{4}\label{eq4} \end{equation} \begin{equation} \nabla \times \textbf{E} = -\frac{\partial\textbf{B}}{\partial t} \tag{5}\label{eq5} \end{equation} \begin{equation} \nabla \times \textbf{B} = \mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t} \tag{6}\label{eq6} \end{equation} We also have the vector identity for any vector function of space, \begin{equation}  \nabla \times \textbf{V} = \nabla.(\nabla\textbf{V}) - \nabla^2\textbf{V} \tag{7}\label{eq7} \end{equation} We thus try to find the curl of $ (5) $ \begin{equation} \nabla \times (\nabla \times \textbf{E}) = \nabla \times -\frac{\partial\textbf{B}}{\partial t} = -\frac{\partial}{\partial t}(\nabla \times \textbf{B}) \tag{8}\label{eq8} \end{equation} Plugging $(6)$ into $(8)$, we get \begin{equation} \nabla \times (\nabla \times \textbf{E}) = -\frac{\partial}{\partial t}\left(\mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t}\right) = -\mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2} \tag{9}\label{eq9} \end{equation} Using the identity (7), \begin{equation} \nabla.(\nabla\textbf{E}) - \nabla^2\textbf{E} = -\mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2} \tag{10}\label{eq10} \end{equation} Plugging (3) into (10), we get \begin{equation} 0 - \nabla^2\textbf{E} = -\mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2} \tag{11}\label{eq11} \end{equation} With a slight bit of rearrangement we can see that, $$- \nabla^2\textbf{E} = -\mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2} $$ $$\nabla^2\textbf{E} = \mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2}$$ \begin{equation}  \frac{1}{\mu_0\epsilon_0}\nabla^2\textbf{E} = \frac{\partial^2\textbf{E}}{\partial t^2} \tag{12}\label{eq12} \end{equation} (12) and (1) correspond to the same equation under the conditions imposed. Following a similar line of reasoning for equation (6), we will get the equation; \begin{equation}   \frac{1}{\mu_0\epsilon_0}\nabla^2\textbf{B} = \frac{\partial^2\textbf{B}}{\partial t^2} \tag{13}\label{eq13} \end{equation} which corresponds to equation $(2)$. Thus, we have obtained the electromagnetic wave equation from Heaviside' form of Maxwell's equations in differential form.   Is there a way to derive the same wave equations without assuming to be in a vacuum and charge-free space i.e $ \ \textbf{J} = 0, \rho = 0$?
51101
In non-relativistic QM, on one hand we have the following relations:  $$\langle x | P | \psi \rangle ~=~ -i \hbar \frac{\partial}{\partial x} \psi(x),$$  $$\langle p | X | \psi \rangle ~=~ i \hbar \frac{\partial}{\partial p} \psi(p).$$  On the other hand, despite the similarities, the relations cannot be directly applied to energy and time:  $$\langle t | H | \psi \rangle ~=~ i \hbar \frac{\partial}{\partial t} \psi(t),$$  $$\langle E | T | \psi \rangle ~=~ -i \hbar \frac{\partial}{\partial E} \psi(E).$$  Just wondering, how can one mathematically prove that the "classical time" (which means no QFT or relativity involved), unlike its close relative "position", is not a Hermitian operator?  I ask your pardon if you feel the question clumsy or scattered. But to be honest, if I can clearly point out where the core issue of the problem is, I may have already answered it by myself :/
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.