# Tag Info

9

The principle of superposition comes from the fact that equations you solve most of the time are made of Linear operators (just like the derivative). So as long as you are using these operators you can write something like $$\mathcal L\cdot \psi = 0$$ where $\mathcal L$ is a linear operator and, let say, $\psi$ is a function that depends on coordinates ...

7

It is true up to very high filed strengths. For too high strengths the field itself is not stable, it can create real pairs. It is like a limit on a field strength in a capacitor. The capacitor dielectric can break. EDIT: Classical Maxwell equations are linear indeed so the principle of superposition is implemented into them. But break of a dielectric can ...

6

When one uses complex variables in this way one never multiplies two variables because the whole system is linear: if $z$ is the oscillating variable and you choose to represent it by a complex number, then things like $z\times z$ don't arise in a linear equation, so you don't get the kind of contradiction you astutely and clearly pointed out above. If the ...

5

The second solution is there to allow for arbitrary start and stop times. Using standard trig identities you can convert an arbitrary linear combination of $\sin$ and $\cos$ into a time-displaced sinusoidal function: $$A\sin(\omega t)+B\cos(\omega t)=R\cos(\omega(t-t_0)),$$ where $R=\sqrt{A^2+B^2}$ and $\tan(\omega t_0)=A/B$.

5

Schrodinger's equation is homogeneous -- so if $\phi_1,\phi_2,\cdots,\phi_n$ are solutions, $c_1\phi_1 + c_2\phi_2 + \cdots +c_n\phi_n$ is a solution. More importantly, if $\phi$ is a solution, $A\phi$ is a solution as well. If $A$ is the normalization constant, we see that both non-normalized and normalized versions are valid solutions of Schrodinger's ...

5

Well, surely you can compute it using matrix operations. But it won't be very natural. Let me instead provide you with a very similar solution (based on a similar matrix) that you'll hopefully find useful. It's not new at all (Kirchhoff, 1847) but I think it's not very well known. I first learned about it in this Wu's review paper of Potts model, p. 252. Let ...

5

The problem you describe is (mathematically) similar to blind deconvolution. Given a signal which is the result of blurring an image (a linear operation) and adding noise, blind deconvolution tries to estimate the blur and the image. As described here, the blind deconvolution process consists roughly of: Guess the blurring function (transfer function) ...

4

Assuming you mean "linear" in the mathematical sense of "the sum of two solutions to the relevant equation is also a solution," there's no particular reason why macroscopic objects are inherently non-linear. In fact, there is a large body of work in the quantum foundations community on ways to have macroscopic objects behave in a linear manner but look ...

4

I guess this was prompted by one of my comments on this question. The point I was making is that quantization, even of a system which is defined by a Lagrangian encapsulating a nonlinear set of equations of motion (such as in interacting QED) proceeds by defining a Hilbert space of states and operators on this space which evolve unitarily $$|\Psi(t)\rangle ... 4 The no-cloning theorem states that it is not possible to have a quantum state |\psi> evolve into two separable (non-entangled) copies described by the tensor product state |\psi>|\psi>. The proof boils down to the simple observation that when expressing |\psi> in some basis {|0>, |1>, |2>, ...}:$$|\psi> = \alpha_0 |0> + ...

4

I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian $L=L(q,\dot{q})$ of $n$ generalized coordinates $$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$ One of the first questions one would like to ask is, if a specific ...

4

I think your qualification of "most" systems needs some clarification because really almost all of the classical universe is described by second-order, nonlinear partial differential equations. Fluids/liquids/gases and solids are described by the same set of second-order, nonlinear PDE's. Linear equations, both linear PDE's and linear ODE's, show up often ...

3

"By the linearity of quantum mechanics" is actually a reference to the linearity of the operators used it quantum mechanics. It means that, for a linear operator $A$ (by the very definition of linearity), $$A\bigl(\alpha\lvert\Psi\rangle +\beta\lvert\Phi\rangle\bigr)=\alpha A\lvert\Psi\rangle + \beta A\lvert\Phi\rangle,$$ where $\alpha$ and $\beta$ are ...

3

Gearboxes belong to a class of linear system that conserves a product of observable quantities by dint of the principle of conservation of energy. For a gearbox, the product $\tau \omega$, where $\tau$ is the torque exerted on or by a driveshaft and $\omega$ the shaft's angular speed. For a lossless gearbox at steady state, we have: $\tau_{in} \omega_{in} = ... 3 Hints: We are evidently only supposed to solve for$z$-dependence (as opposed to$x$- and$y$-dependence). Note that the two variables$E_z$and$H_z$can be eliminated. In the reduced coupled ODE system of four first-order ODEs and four variables$(E_x,E_y,H_x,H_y)$, note that the variables couple two and two together. Which pairs? Within one such pair, ... 3 In a linear wave equation, there is nothing to pull a pulse or envelope of running waves apart. But there is nothing to hold it together, either. A minor disturbance such as a small obstacle or some dispersion, will change the waveshape, or break it up, such as losing some of its energy to outward spherical waves from the obstacle. Two or more pulses in ... 3 Within the realm of Maxwell's equations, the principle of superposition is exactly true because Maxwell's equations are linear in both the sources and the fields. So if you have two solutions to Maxwell's equations for two different sets of sources then the sum of those two solutions will be a solution to the case where you add together the two sets of ... 2 Linear in the quantum mechanics has nothing to do with its complexity. A two-state spin can be described by a simple 2-by-2 matrix; however, 30 interacting spin, in general, must be described by a 1 billion by 1 billion matrix. It grows exponentially as the number of spins increases, for$10^{23}$spin, you may need a matrix of size$2^{10^{23}}$. It is not ... 2 Math: If you have an operator$D$with $$D(\Psi+\Phi)=D(\Psi)+D(\Phi),$$ then if$D(\Psi)=0$and$D(\Phi)=0$, you can also conclude that$D(\Psi+\Phi)=0$. This is the case for the Schrödinger equation, as it reads $$D(\Psi):=(i\hbar\tfrac{\partial}{\partial t}-H)\Psi=0,$$ where$H$is linar. For example you certainly have linearity for the derivatives: ... 2 What's$V_{ab}$? Well,$V_{ab}$is the "symmetric, positive definite potential energy matrix". Ok lol I'm trolling here, but as the name suggests,$V_{ab}$describes the strength of the (linearized) interaction between particles$a$and$b$. To be precise, it is the second derivative of the potential energy function of the system with respect to$u_a$and ... 2 Notice that there are three equations for the quantities$(E_y,H_x,H_z)$: $$\frac{\partial E_y}{\partial z} = -i\omega\mu_0 H_x$$ $$i\beta E_y = i\omega\mu_0 H_z$$ $$\frac{\partial H_x}{\partial z} -i\beta H_z = -i\omega\varepsilon_0\varepsilon E_y$$ These are totally independent of the equations for the three other components,$(H_y,E_x,E_z)$. If ... 2 Bare with me, I don't remember every little step, but I hope this derivation helps you. First remember how a wave travels through a waveguide (dielectric). $$E(x,y,z) = E^{0}(x,y)e^{-\gamma z}$$ $$H(x,y,z) = H^{0}(x,y)e^{-\gamma z}$$ Then consider Ampere's and Faraday's Laws for a source-free region. $$\triangledown \times H = j\omega\epsilon E$$ $$... 2 For a linear system, the superposition principle holds since, be definition, a linear system has the following property: (1) if y_1 is the output for input x_1 and (2) if y_2 is the output for input x_2 then (3) the output is ay_1 + by_2 for input ax_1 + bx_2 In other words, the output for a superposition of inputs is the superposition of the ... 1 The reason why complex variable method is feasible is actually equivalent representation of Cartesian coordinate and polar coordinate. Usually when describing revolution, trigonometric functions are involved in Cartesian coordinate. So as to avoid them, we introduce polar coordinate so that only azimuth angle and amplitude are used. Your given example ... 1 Quantization is done for the free theory, without interactions, and this free theory is linear. For a scalar boson, for instance, each composant like \phi(\vec k,t) is an harmonic oscillator (so the equation for \phi(\vec k,t) is a linear equation), and you know how to quantize an harmonic oscillator. When you calculate the transition amplitudes, you ... 1 What is the physical significance of a second sinusoidal function in the solution? The physical significance is that two initial conditions are required for a unique solution. Ignoring the constant, the general solution for this system is: x(t) = x(0) \cos \omega t + \dfrac{v(0)}{\omega} \sin \omega t Where x(0), v(0) are the initial position ... 1 A rule of thumb would be to get rid of unwanted variables. For example, since we're only interested in \frac{m_1}{m_2} and \theta, we can get rid of F_T.$$F_T \sin \theta = m_1 gF_T \cos \theta = m_1 a$$Rearrange to get$$\frac{m_1 g}{\sin \theta}=\frac{ m_1 a}{\cos \theta} \hspace{20mm}\text{ ...Eq 5}$$I don't know whether getting ... 1 It is not completely clear what you mean by the approach 2. What one can do is to calculate the current via$$j^{\mu}(\phi,A)=\frac{\delta S_{eff}[\phi,A]}{\delta A_{\mu}}.$$Here the effective action$S_{eff}$is a functional of both the source$A$and the phase$\phi\$ of the condensate in the superconductor. Imagine now that you solve the linearized ...

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