# Tag Info

11

In many cases our systems are described by linear differential equations, and these have the property that any linear combination of solutions to the differential equation is also a solution to the differential equation. This is useful because usually any arbitrary solution can be Fourier transformed to express it as a sum of plane waves. So if we can find ...

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 ...

7

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 ...

6

"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 ...

6

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 ...

6

As user1104 commented, you use Euler's identity: $$e^{ix} = cos(x) + i \space sin(x)$$ so: $$sin(kx-wt) = \frac{ e^{i(kx-wt)} - e^{-i(kx-wt)}}{2i}$$ But we wouldn't normally procede by replacing sin by this expression. Both the sin form and the exponential form are mathematically valid solutions to the wave equation, so the only question is their ...

6

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. $$\triangledown \times H = j\omega\epsilon E$$ $$\... 6 I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian L=L(q,\dot{q}) of n generalized coordinates$$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$One of the first questions one would like to ask is, if a specific ... 5 The problem you describe is (mathematically) similar to blind deconvolution. Given a signal which is the result of blurring an image (a linear operation) and adding noise, blind deconvolution tries to estimate the blur and the image. As described here, the blind deconvolution process consists roughly of: Guess the blurring function (transfer function) ... 5 Assuming you mean "linear" in the mathematical sense of "the sum of two solutions to the relevant equation is also a solution," there's no particular reason why macroscopic objects are inherently non-linear. In fact, there is a large body of work in the quantum foundations community on ways to have macroscopic objects behave in a linear manner but look non-... 5 Well, surely you can compute it using matrix operations. But it won't be very natural. Let me instead provide you with a very similar solution (based on a similar matrix) that you'll hopefully find useful. It's not new at all (Kirchhoff, 1847) but I think it's not very well known. I first learned about it in this Wu's review paper of Potts model, p. 252. Let ... 5 The second solution is there to allow for arbitrary start and stop times. Using standard trig identities you can convert an arbitrary linear combination of \sin and \cos into a time-displaced sinusoidal function:$$A\sin(\omega t)+B\cos(\omega t)=R\cos(\omega(t-t_0)),$$where R=\sqrt{A^2+B^2} and \tan(\omega t_0)=A/B. 5 Gearboxes belong to a class of linear system that conserves a product of observable quantities by dint of the principle of conservation of energy. For a gearbox, the product \tau \omega, where \tau is the torque exerted on or by a driveshaft and \omega the shaft's angular speed. For a lossless gearbox at steady state, we have: \tau_{in} \omega_{in} = ... 5 The no-cloning theorem states that it is not possible to have a quantum state |\psi\rangle evolve into two separable (non-entangled) copies described by the tensor product state |\psi\rangle|\psi\rangle. The proof boils down to the simple observation that when expressing |\psi\rangle in some basis {|0\rangle, |1\rangle, |2\rangle, ...}:$$|\psi\...

4

You asked about the second equation. See below: $e^{ix}{}= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \frac{(ix)^8}{8!} + \cdots \\[8pt] {}= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \frac{x^8}... 4 In a linear wave equation, there is nothing to pull a pulse or envelope of running waves apart. But there is nothing to hold it together, either. A minor disturbance such as a small obstacle or some dispersion, will change the waveshape, or break it up, such as losing some of its energy to outward spherical waves from the obstacle. Two or more pulses in ... 4 What's$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$and$...

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 ...

4

John has answered this partially, however the fundamental mathematical idea is missing: If we think of a function being member of some vectorspace, then basisvectors exist. This concept will surely be familiar to you from Quantum mechanics. But also from there we remember, that problems were alot easier to handle, if we know the Eigenbasis of the Operators ...

4

No, you can't do that. First of all because it doesn't make much sense to talk about the wave function corresponding to a spatial dimension. A wave function $\psi(x_1,\dots,x_n,t)$ gives you the probability amplitude of the system being at the position $(x_1,\dots,x_n)$ at the time $t$. It must by definition be dependent on all the coordinates necessary to ...

4

A linear system is one where, if we have two inputs $x_1$ and $x_2$, producing outputs $y_1$ and $y_2$, then the output for an input of $\alpha{}x_1 + \beta{}x_2$ is $\alpha{}y_1 + \beta{}y_2$. This is precisely the property we rely on when we apply superposition to solve a problem with inputs composed of sums of easier-to-analyze inputs. This means, ...

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 ...

3

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: ...

3

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 $$|\Psi(t)\rangle ... 3 Hints: We are evidently only supposed to solve for z-dependence (as opposed to x- and y-dependence). Note that the two variables E_z and H_z can be eliminated. In the reduced coupled ODE system of four first-order ODEs and four variables (E_x,E_y,H_x,H_y), note that the variables couple two and two together. Which pairs? Within one such pair, ... 3 It's a linear system due to the fact that if$$y_1{(n1,n2)} = C_{(n1,n2)} x_1{(n1,n2)}$$and$$y_2{(n1,n2)} = C_{(n1,n2)} x_2{(n1,n2)}$$then if$$x_3 = \alpha x_1 + \beta x_2$$it is the case that$$y_3 = C_{(n1,n2)}[\alpha x_1{(n1,n2)} + \beta x_2{(n1,n2)}] = C_{(n1,n2)}\alpha x_1{(n1,n2)} + C_{(n1,n2)}\beta x_2{(n1,n2)} = \alpha y_1 + \beta y_2$$3 This is a very generic system, because:$$\begin{align} \frac{\dot y}{y - \lambda} &= \alpha \\ ln(y-\lambda) &= \alpha t + \ln C\\ y &= \lambda + C e^{\alpha t} \end{align} Which is related to an arbitrary exponential function by a shift of the y coordinate. So, any exponential growth can be described by this equation. of course, a more ...

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