Styg
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You might be assuming the matrix element $T_{ii}$ to be real. If so, then $$\lvert S_{ii} \rvert^2 = 1 + \lvert T_{ii} \rvert^2 > 1$$ Without such an assumption, \begin{align*} \lvert S_{ii} ... View answer 4 votes Writing out Craig's answer explicitly (I personally find explicit examples helpful): \rho = \frac{1}{2} \lvert 0 \rangle_{_A} \lvert 0 \rangle_{_B} \langle 0 \rvert_{_A} \langle 0 \rvert_{_B} + \...

The derivation of the time-independent Schrödinger equation doesn't assume both sides equal a constant. It begins with the assumption that the wavefunction can be written as a product of two functions:...

They're (in principle) different quantities, that coincide for velocity-independent potentials in a Cartesian coordinate system. The Lagrangian and the generalized momenta in such a system are $$\... View answer 1 votes No interaction with an external environment is assumed when deriving the allowed frequencies for a string. The basic assumptions are that tension is the same at every point on the string, and that ... View answer 1 votes The Bohr model does not limit the number of possible orbitals that hydrogen has. At any given time, the single electron is in any one of the possible orbits, sure. But there is no restriction on which ... View answer 1 votes What you're looking for is a density operator. Writing a quantum state as a Dirac state \lvert \psi \rangle  implies a well-defined state, that is, that there is a set of Hermitian operators with ... View answer 1 votes We do take into account the field due to both plates in the cylindrical geometry. The field due to an infinite uniformly charged cylinder is zero in the interior of the cylinder, as can be shown by ... View answer Accepted answer 1 votes You're right in saying that the work performed on an object with constant velocity is 0. The work-energy theorem says that the net work done on an object equals change in its kinetic energy.$$ W_{net}...

The work energy theorem says that the net work done on a point mass equals the change in kinetic energy. $$\int \vec F_{net} \cdot d\vec r = \frac{1}{2} m \Delta (v^2)$$ In the hypothetical ...

It might be better to not think of $\frac{mv^2}{r}$ as the expression for centripetal force, and instead to think of $\frac{v^2}{r}$ as the acceleration of a particle moving in a circle of radius $r$ ...

If you are beginning with electrostatics, you must have been introduced to Coulomb's Law for force between two point charges $q_1$ and $q_2$: $$F \propto \frac{q_1 q_2}{r^2}$$ where $r$ is the ...

Post the discussion in comments, here is my answer: It is not correct to say that $g = -\frac{\partial V}{\partial r}$, where $g$ is the magnitude of the gravitational field. The full statement for a ...

The emf as defined between two electrodes is the difference in electric potential $V$ between the two electrodes. If $A$ and $B$ denote the locations of the two electrodes, then emf can be defined as ...

As Chris mentioned in his comment, it depends on how the mass is lost. Specifically, it depends on the velocity of the mass that is lost. I'm assuming that the entire system, consisting initially of ...

You would have to include the tension as a generalized force if you did away with the constraint that $r$ is constant when writing your Lagrangian (since it acts inward towards the support, that is, ...

Short answer: measurements would not change if the earth is taken to be at $100V$ instead of $0V$. The physically measurable quantity in electrostatics is the electric field, derived from the ...

To add on to Billy's answer, there is no deeper meaning because $R$ is not dimensionless. Let's take $R = 1\Omega$ for the rest of the answer. Let upright letters denote numerical values of the ...