# Tag Info

8

Let $|n'\rangle$ be a basis of the Hilbert space, then $$\textrm{tr}\Big[|\alpha\rangle\langle\alpha|A\Big]=\sum_{n'}\langle n'|\alpha\rangle\langle\alpha|A|n'\rangle=\sum_{n'}\langle\alpha|A|n'\rangle\langle n'|\alpha\rangle = \langle\alpha|A\left(\sum_{n'}|n'\rangle \langle n'|\right)|\alpha\rangle=\langle\alpha|A|\alpha\rangle$$

2

Another way to see this is to observe that any state $|\psi⟩\in\mathcal H$ can be extended to an orthonormal basis of the Hilbert space, and in that basis the trace $\operatorname{Tr}\left(|\psi⟩⟨\psi|\hat A\right)$ is exactly $⟨\psi|\hat A|\psi⟩$. More explicitly, for any $|\psi⟩\in\mathcal H$ there exists a sequence ...

2

It depends on what exactly you are asking. Suppose we take 64g of copper i.e. one mole of copper. Each copper atom contributes one conduction electron, so our chunk of copper contains $6.023 \times 10^{23}$ (Avagadro's number) conduction electrons with a total charge of 96488 coulombs. John's answer involves removing those electrons by a chemical reaction. ...

2

Perhaps the easiest way to see that there can't be a potential difference between $A$ & $B$ is a symmetry argument. You're tempted to say that $A$ is at a higher potential than $B$ so that current will flow from $A$ to $B$. But continuing along the loop, I find that current must also flow from $B$ to $A$, which would lead me to conclude that $B$ is at a ...

2

We don't control the allowed energies $E_i$ independently of the potential: the energies must be the eigenvalues of the Hamiltonian. The "inputs" are the shape and height of the barrier between the two wells. You can kinda sorta think of the energy difference between the symmetric state (with energy $E_1$ in your diagram) and the antisymmetric state (with ...

2

The EMF created by a changing magnetic field is not considered to arise from a potential. This can easily be seen because when there is an emf, a charge can move around in a complete circle and dissipate energy the whole way around, but a potential cannot drive a charge around in a circle, because potentials are conservative. The two pieces of the electric ...

2

Substitute two values for $\mu$ and $\nu$. If $\mu=\nu$, then using $(\gamma^\mu)^2 =\pm1$ and $tr(\gamma ^5) =0$ you have finished. If $\mu \neq\nu$, then use $tr(\gamma ^\mu\gamma ^\nu) =0$ with the two remaining $\gamma$ .

2

A double well with a high or wide barrier will have a smaller $\Delta E=E_2−E_1$ than one with a low or narrow barrier. (Less coupling.) I think we can understand this intuitively as follows but first it has to be said that rob is right: the energies $E_i$ are NOT inputs but the eigenvalues of the Schrödinger equation. Width, height and potential of the ...

2

Start noticing that ${(\gamma^{\alpha})}^2 =1\cdot g^{\alpha \alpha}$ and that $$\textrm{tr} (\gamma^{\mu}\gamma^{\nu}\gamma^5)= \textrm{tr}\left(\frac{1}{g^{\alpha \alpha}}{(\gamma^{\alpha})}^2\gamma^{\mu}\gamma^{\nu}\gamma^5\right)=\frac{1}{g^{\alpha \alpha}}\textrm{tr} (\gamma^{\alpha}\gamma^{\alpha}\gamma^{\mu}\gamma^{\nu}\gamma^5).$$ Now choose ...

2

Torque $\vec{\tau}(\vec{r})$ is an vector multiplication of radius-vector $\vec{r}$ on applied force vector $\vec{F}$, i.e. $\vec{\tau}=[\vec{r}\times\vec{F}]$ Here the radius-vector (or position vector) $\vec{r}$ is the vector from the point where the torque is defined to the point where the force is applied (see image). On the picture you shown ...

2

So let's start from the relations you gave and transform one of them from ket to bra. $$\left|i\right> = \mathcal{ CPT}\left | \bar{i}\right>$$ $$\left<f\right| = \left< \bar{f}\right| (\mathcal{ CPT})^{\dagger}$$ Using the CPT invariance condition, $\left(\mathcal{ CPT} \right)T \left(\mathcal{ CPT}\right)^{-1}= T^{\dagger}$, It is easy ...

2

Which is right? So let's first establish what is correct. Suppose we have a circle of radius $r$ and mass $m$ centered on a circle of radius $\bar R$ (not quite what you have defined; in your case $\bar R = R + r$). We'll say that the position of the $r$-circle on the $\bar R$-circle is angle $\phi$ and, from the angle $\phi$, we will measure an angle ...

2

In my (ancient) copy of Hecht and Zajac (1980), the answer is found in figure 10.18. It shows that for slit spacing $a$ and slit width $b$, peaks in the diffraction pattern are spaced $\lambda/d$ while the first zero due to the finite width is at $\lambda/b$. In the figure, $a = 3b$ and the third peak is suppressed:

2

A free body diagram on the $2m$ mass would have $2mg$ down and $T$ up. This would give a Newton's 2nd Law equation, assuming up to be the positive vertical direction, of $$T-2mg=2ma_{2v}$$. The $m$ mass free-body diagram would yield two downward forces, $T$ and $mg$ with a Newton's 2nd Law equation of $$-T-mg=ma_{1v},$$ assuming the tension magnitude in the ...

2

Just use snell's law, that is, $\mu \,\sin\theta$=constant, where $\mu$ denotes the refractive index and $\theta$ is the angle between the ray and the normal between a generic point and the point of incidence.The rest is math, you need to express $\sin\theta$ in terms of the slope at that point and solve the resulting differential equation.

1

calculate centre of charge by the process of integration and then find potential and fields.

1

You are certainly missing something out. $$U_p |x_0 \rangle = \exp(-i b \hat{x}) |x_0 \rangle =\exp(-i b x_0) |x_0 \rangle$$ Your equations are clearly wrong which you can see from $$\langle x_0|U_p^\dagger p U_p |x_0\rangle =\langle x_0|p |x_0\rangle+b$$ using your second equation. Using your third equation one would get $$\langle x|U_p^\dagger p U_p ... 1 It's even easier to see what's going on by simplifying further. Consider a rod of length L and uniform linear density \lambda. If you try to calculate the acceleration caused by the rod on a point at one end of the rod, you should get the integral:$$|a| = \left|\int_{r=0}^L\frac{G\lambda \, dr}{r^2}\right|$$Obviously, this diverges, and the ... 1 You can transform to the rotating frame as follows: $$\psi_{\mathrm{rot}} (t) = \hat{U}(t)\psi(t),$$ where the time-dependent unitary transformation U(t) is defined by U(t) \equiv \exp\left(i\omega t S_{z}/\hbar\right) = \begin{pmatrix} e^{i\omega t}&0&0\\ 0&1&0\\0&0&e^{-i\omega ... 1 You've computed the acceleration of the block once it is moving. However, if the block starts stopped there is a minimum force needed to get it started that is larger than the force needed to maintain it's motion. That minimum force to start the motion is computed using the maximum force of static friction and the assumption of equilibrium. The working ... 1 In the rules of quantum mechanics, every state |\psi\rangle is a "vector" which has a "dual", which is usually a complex conjugate \langle \psi| and every measurement in some state is described by an average \langle A\rangle and an operator \hat A which is its own conjugate transpose: together these say that in state |\psi\rangle the average ... 1 I'll risk moderatorial opprobium with a partial answer because you have come so close. You correctly use the SUVAT equation v^2 = u^2 + 2as to find that the velocity of the ball just before it strikes the ground is v_i = -7 m/s (using the sign convention that upwards is positive). So far so good. Now you know the ball rises back up to a height of 1.8m, ... 1 Your expression for the acceleration due to the kinetic friction is incorrect. Remember,$$ f_k = \mu_kn, $$where n is the normal force. To find the normal force, you have to use what you know about the centripetal acceleration. Draw a free-body diagram, and label the weight of the car and the normal force, and then you know that ... 1 An accelerating object has a changing velocity. Obviously so since the object starts with zero velocity and the velocity increases with time according to the SUVAT equation:$$ v = u + at $$So your equation 1.1 is no use here. It calculates the average velocity. This could actually be used to calculate the acceleration, but the working is a bit involved ... 1 Both are right. The first approach gives the compression where the net force on the object is zero. The second approach gives the compression when the velocity of the object is zero. When the block falls on the spring, it oscillates between x=\frac{2mg}{k} and x = 0. Since the spring is ideal and the air resistance is negligible, this oscillation does ... 1 Your equation 1.1 can be used with constant velocity. Here you have to use the 2^{\text{nd}} equation. ie a = 2d/(t^2). So, the answer is 118.4 \, \text{cm}/s^2. 1 The work-energy theorem is certainly the easiest way to do the problem, but you can also solve it by calculating the force. In any situation where you need to calculate the response of an object to a force you use Newton's second law. This tells us (after a minor rearrangement):$$ \frac{d^2x}{dt^2} = \frac{F}{m} \tag{1} $$In this case the force on the ... 1 "As accurate as possible" is a fuzzy concept. Given that you ask this question, I expect that a few simplifying assumptions are justified. For an object in a circular orbit of constant radius R, orbiting a perfectly spherical earth of constant density, the kinetic and potential energy can be calculated. Their relationship is beautifully simple, as derived ... 1 Your answer is perfectly fine. As you can see one can choose an abritrary phase \exp(i\phi) for c_n in the equation$$E_n + \frac{\hbar \omega}{2} = |c_n|^2 and it will still hold. This relates to the fact that you can always choose an arbitrary phase for the eigenfunctions $\psi_n$. All physical observables (e.g. $A_{nn} =\langle ... 1 Recall that states or wave-functions are only defined up to an overall phase, i.e.$\psi(x)$and$e^{i \alpha(x)} \psi(x)$are both wave-functions that describe the same state. The wave-function generically is a complex function of the form$\psi = f(x) e^{i h(x)}$where$f(x)$and$h(x)\$ are real functions. It is then often convenient to make a choice of ...

Only top voted, non community-wiki answers of a minimum length are eligible