# 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

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) ...

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 $\tan(\omega t_0)=A/B$.

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 $$|\Psi(t)\rangle ... 4 I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian L=L(q,\dot{q}) of n generalized coordinates$$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$One of the first questions one would like to ask is, if a specific ... 4 I think your qualification of "most" systems needs some clarification because really almost all of the classical universe is described by second-order, nonlinear partial differential equations. Fluids/liquids/gases and solids are described by the same set of second-order, nonlinear PDE's. Linear equations, both linear PDE's and linear ODE's, show up often ... 4 The no-cloning theorem states that it is not possible to have a quantum state |\psi> evolve into two separable (non-entangled) copies described by the tensor product state |\psi>|\psi>. The proof boils down to the simple observation that when expressing |\psi> in some basis {|0>, |1>, |2>, ...}:$$|\psi> = \alpha_0 |0> + ...

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

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

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. $$\triangledown \times H = j\omega\epsilon E$$ $$... 3 "By the linearity of quantum mechanics" is actually a reference to the linearity of the operators used it quantum mechanics. It means that, for a linear operator A (by the very definition of linearity),$$A\bigl(\alpha\lvert\Psi\rangle +\beta\lvert\Phi\rangle\bigr)=\alpha A\lvert\Psi\rangle + \beta A\lvert\Phi\rangle,$$where \alpha and \beta are ... 3 Gearboxes belong to a class of linear system that conserves a product of observable quantities by dint of the principle of conservation of energy. For a gearbox, the product \tau \omega, where \tau is the torque exerted on or by a driveshaft and \omega the shaft's angular speed. For a lossless gearbox at steady state, we have: \tau_{in} \omega_{in} = ... 3 Within the realm of Maxwell's equations, the principle of superposition is exactly true because Maxwell's equations are linear in both the sources and the fields. So if you have two solutions to Maxwell's equations for two different sets of sources then the sum of those two solutions will be a solution to the case where you add together the two sets of ... 3 This question is very hard to answer at a fundamental level, because quantum mechanics seems to be exact so far, yet one cannot be sure in the scientific sense without confirmation that nontrivial quantum computation is possible. If this is so, then one would have to renounce any classical descriptions, at least within the bounds of scientific reason, and it ... 2 Linear in the quantum mechanics has nothing to do with its complexity. A two-state spin can be described by a simple 2-by-2 matrix; however, 30 interacting spin, in general, must be described by a 1 billion by 1 billion matrix. It grows exponentially as the number of spins increases, for 10^{23} spin, you may need a matrix of size 2^{10^{23}}. It is not ... 2 Math: If you have an operator D with$$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,$$where H is linar. For example you certainly have linearity for the derivatives: ... 2 Notice that there are three equations for the quantities (E_y,H_x,H_z):$$ \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

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

Superposition theorem is applicable only to linear and bilateral circuits. Diodes and trasistors are often not bilateral and linear.

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 ...

2

I think there is something wrong with your mapping. Looking at http://lpsa.swarthmore.edu/Analogs/ElectricalMechanicalAnalogs.html , I see the following table: This is inconsistent with the mapping you are showing. I can understand this table - I can't understand yours. I think an error crept in - which would reasonably explain your confusion. Looking ...

1

To add to Floris's reply, just to elaborate in a very basic manner in what way the flux linkage (analogous of displacement x in Force Current analogy) and magnetic flux differ. Consider the scenario where a magnetic field is present, and we have an open circuit that has been closed using a metalic rod. As the picture shows: We know that the induced EMF ...

1

Your confusion is well justified. TE and TM modes are not rigorous solutions to Maxwell's equations. This is why you are struggling to figure out why certain components can just be set to zero. They can't! This is just an approximation to make analysis easier. They are very good approximations, however. I've done rigorous modeling of waveguides and the ...

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 ...

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