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2

It's a convention. You may also occasionally find it without the $\frac{1}{2}$ factor. Observe that $\tilde{H}^{ab} = -\tilde{H}^{ba}$ anyways, so the minus in the definition translates into a sign/index ordering convention in all equations involving it. The convention is basically related to the question whether you want the Hodge dual $\star H$ to have ...


0

The author is working by analogy with the 3D box case. The 3D box worked out easily because $\epsilon\propto k^2$, so that the number of states up to each energy could use the volume of an ellipsoid, which is well known. He/she makes the volume of an ellipsoid work in a "rough way" for the harmonic oscillator. He/she puts a state where each 3D box state is ...


0

The right hand rule and cross product of matrices is a convenient and useful tool for describing natural phenomena in three dimensional space. Like all conceptual tools, it was devised by humans to acquire knowledge, and to further our ability to predict and to manipulate aspects of our world. We use it because it fits with nature. Examples ("x" denotes ...


-1

Why should the resultant cross-product necessarily point in the orthogonal direction given by the right hand rule? Because, we made it to. Other math tools give other results. We made them to do that too. We could make any mathematical method to bring any result we want. But some methods, tools and definitions turn out to fit with the real world. ...


3

Vectors (which kets are) don't have adjoints, they have duals. Whether the dual of $\lvert n_1,\dots,n_n\rangle$ is denoted by $\langle n_1,\dots,n_n\rvert$ or $\langle n_n,\dots,n_1 \rvert$ is entirely conventional.


1

If q is the momentum for positron, then the propagator for it is still $i\frac{\not{q}+m}{\not{q}^2 +m^2}$.


2

For reference, the fermion propagator is $$ \left\langle 0 \right| T\psi(x)\overline\psi(y) \left|0\right \rangle= S(x-y) = \int \frac{d^4k}{(2\pi)^4} \frac{i}{\not k-m}e^{-ik\cdot(x-y)}$$ Depending on the time ordering, this describes a particle moving from $y$ to $x$, or an antiparticle moving from $x$ to $y$. Now, consider a one-loop diagram in which ...


4

"The number of degrees of freedom can be defined as the MINIMUM number of independent coordinates that can specify the position of the system completely" (wikipedia) In your case the number is ONE, because you only need to know the position of the particle along the curve. It doesn't matter if the curve is not a line, or even contained on a plane, because ...


0

To find the solution to your problem: Area 1: U1,E>U1 The Schrodinger equation and it's solution is: $$ψ'' + {2m \over \hbar ^2}(E-U1)ψ=0 ---> ψ_A=e^{ik_1 x} + c_1e^{-i k_1 x} $$ where $k_1 ^2 = {2m \over \hbar ^2}(E - U1) $. Area 2: U2, E As in the above: $$ψ'' - {2m \over \hbar ^2}(U2-E)ψ=0--> Ψ_Β = c_2 e^{k_2 x} + c_3 e^{-k_2 x} $$ where $k_2 ^2 ...


2

The solution you give is unphysical, because it cannot be normalized (as you noticed). This happens a lot in physics (e.g. try solving the wave equation in cylindrical coordinates: you'll get Bessel functions that rise to infinity at the origin, which you'll also discard if the origin is part of the solution domain). Unphysical solutions are discarded, and ...


3

You have a sign error. The potential is defined by see: http://en.wikipedia.org/wiki/Electric_potential


0

The quantity that you can measure in the lab is the force experienced by a test charge (in reality sensor of some kind). This force is obtained via lorentz force that depends on the electric field. So the physically measurable quantity is the Electric field and not the potential. Remember that $ \vec{E} =-\nabla \phi $ so the electric field decreases with r, ...


2

A negatively charged particle has an electric field, $$\mathbf{E} =-\vert\mathbf{E}\vert \, \hat{\mathbf{r}} =\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}$$ Gauss's law gives, \begin{align*} \int \mathbf{E}\cdot d\mathbf{A} &= \int \Big(\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}\Big) \cdot (r^2 \sin \theta d\theta d\phi \, \hat{\mathbf{r}}) = ...


1

It is said to be perpendicular to the surface in the outward direction.


1

I have highlighted some key word lacking in your revision. Also, work has a very specific definition. The difference in gravitational potential difference between $\vec{r}_1$ and $\vec{r}_2$ is the negative of the work done on a unit mass by the external gravitational field as the unit mass moves from $\vec{r}_1$ to $\vec{r}_2$. As an example, consider a ...


1

Words are imprecise. Your wording is much less imprecise than the one from the textbook. As the work is positive or negative (depending on the direction taken), there is no need to define a "positive direction", but the potential difference depends on the order of $\vec r_1$ and $\vec r_2$ (that is: if the potential difference moving from 1 to 2 is positive, ...


1

All of them in fact mean the same thing: given $\mathbf{A}$ and $\mathbf{B}$, you are finding (/deciding) the direction of $\mathbf{A}\times\mathbf{B}$. For the situations in which Fleming's rules are of interest, we are interested in just one relation: $$\mathbf{F} = q\mathbf{v}\times\mathbf{B}$$ or its equivalent: $$\mathbf{F} = ...


3

Have a look at the Wikipedia article on the left hand rule. It says: The direction of the electric current is that of conventional current: from positive to negative.


0

Potential has a negative sign as work has to be done against the external force to bring the mass from infinity to the point (definition of potential )


4

Because of a convention wherein zero gravitational potential is said to be at infinity. See Wikipedia: $V(x) = \frac{W}{m} = \frac{1}{m} \int\limits_{\infty}^{x} F \ dx = \frac{1}{m} \int\limits_{\infty}^{x} \frac{G m M}{x^2} dx = -\frac{G M}{x}$ "By convention, it is always negative where it is defined, and as x tends to infinity, it approaches zero." ...


1

Hmm. I think there should be a minus sign and it doesn't matter where you set the zero of potential. Gravitational field strength is the negative gradient of the potential. For a spherically symmetric field $$ g(r) = -\frac{dV}{dr}.$$ If $$V = -\frac{GM}{r} + V_0,$$ where $V_0$ is an arbitrary constant, then $$g = -\frac{GM}{r^2}.$$ i.e. your book is ...


4

They are obviously talking about "conventional" current, not the flow of electrons. When you want to know the direction of magnetic force on a current you need to use "conventional" current direction. There is an interesting corollary to this relating to the Hall effect - if current in a semiconductor is carried by "holes" the polarity of the Hall effect ...


0

Work is defined by the line integral $W=\int_\gamma F\cdot dr$, where $W$ is work, $\gamma$ is some particular path, and $F = F(x,y,z)$ is a vector field. Now let's consider a fact we know from calculus I: $$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$$ Likewise, when we take the path in reverse (I suppose we could call it $-\gamma$), we'd get $$\int_\gamma ...


2

Electrons flow through the wire from the battery's negative terminal to the battery's positive terminal. If said wire is actually a capacitor, the electrons still flow the same way - from the battery's negative terminal. But since it's a capacitor, the electrons are pushed into the capacitor's negative plate instead of making it all the way to the battery's ...


2

No. Step number two already mentions electrons from the battery flowing into the negative terminal of the capacitor, giving the negative terminal a negative charge. Step number three is talking about electrons flowing out of the positive terminal of the capacitor, giving the positive terminal a net positive charge.


0

The total work is simply the algebraic sum of the negative and positive work.


2

Just a guess... The purpose is to reproduce the nice features of $SU(2)$. With that convention, the generators of $SU(2)$ are, in terms of Pauli matrices $$T^i = \frac{1}{2}\sigma^i$$ So a transformation with parameters $\theta_i$ is given by $$U=\exp\left(-i\frac{1}{2}\theta_i\sigma^i\right)$$ Things get interesant when you realize that the elements of ...


2

The Lie algebra $\mathfrak{su}(N)$, viewed as a vector space of matrices, can be equipped with the following standard inner product: \begin{align} \langle X,Y\rangle = \mathrm{tr}(X^\dagger Y), \end{align} where $X^\dagger Y$ is the matrix product of $X^\dagger$ and $Y$, and $\mathrm{tr}$ is the trace. Since $X^\dagger = X$ for all $X\in\mathfrak{su}(N)$, ...


1

For a classical point charge, the field is divergent at $r=0$, and if you were to take the potential to be zero there, it would be infinite everywhere else. Meanwhile, you can approximate $r=\infty$ as the region with no interaction, so it's reasonably naturally to treat it in the way you would treat ground.


4

You evidently understand that any constant can be added to a potential without affecting the physics -- or equivalently, any place can be taken to have zero potential. You also suggest, rightly, that there are really only two "natural" places to define the zero of the potential: either $r=\infty$ or $r=0$. For example, there's no particular reason to ...


1

This is accurate, and it ultimately comes down to the fact that we can get arbitrarily close to an electric point charge in classical E&M. That means that the field right up next to the point charge could be arbitrarily large. So you get these huge, singular potentials close to point charges, which is really more-or-less fine. For instance, that huge ...


3

There are at least two answers possible to give, but both, in the end, amount to the same thing: There is no "right" way to fix the energy scale of a process, but that doesn't matter, except that your perturbation theory will probably break if you choose the scale badly. The old answer: The renormalization scale is arbitrarily defined to fix some parameters ...


1

You go to the center of mass frame to find that $\sum_i \vec{p}=\vec{0}$, and the total momentum four vector is thus $$P_{\text{tot}}^{\mu}=\left(\frac{1}{c}\sum_{i}E_i^{\text{COM}}, \vec{0}\right)$$ then we define the energy scale covariantly as $\mu=\sqrt{-s}$ where $s$ is the mandelstam variable $s\equiv P_{\text{tot}}^{\mu}P_{\text{tot}\mu}$ in units of ...


1

I don't agree with Ben Crowell. I think that the reason that the moment is directed from negative to positive is because of the definition of moment: $$ \mathbf{p} = \sum\nolimits q*\mathbf{d} $$ q: charge, d:distance from the origin of coordinates to the carge If you have a negative and positive charge, this relation gives you the direction from negative ...


2

You might see it from the notion of Fourier transformation. For example, you could express an arbitrary quantum state of your Hilbert space in momentum representation by applying a Fourier transformation on your position representation. More explicitly, for $\left|\psi\right>\in\mathcal{H}$, you might define $$\left< \mathbf{q} |\psi\right> = ...



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