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- https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Essential_Graduate_Physics_-_Quantum_Mechanics_(Likharev)/04%3A_Bra-ket_Formalism/4.08%3A_Exercise_ProblemsIn a certain basis, the Hamiltonian of a two-level system is described by the matrix \[\mathrm{H}=\left(E100E2\right), \quad \text { with } E_{1} \neq E_{...In a certain basis, the Hamiltonian of a two-level system is described by the matrix H=(E100E2), with E1≠E2, while the operator of some observable A of this system, by the matrix A=(1111). For the system’s state with the energy definitely equal to E1, find the possible results of measurements of the observable A…
- https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Essential_Graduate_Physics_-_Classical_Electrodynamics_(Likharev)/06%3A_ElectromagnetismThis chapter discusses two major effects that arise when electric and magnetic fields are changing in time: the “electromagnetic induction” of an additional electric field by changing magnetic field, ...This chapter discusses two major effects that arise when electric and magnetic fields are changing in time: the “electromagnetic induction” of an additional electric field by changing magnetic field, and the reciprocal effect of the “displacement currents”- actually, the induction of an additional magnetic field by changing electric field.
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Essential_Graduate_Physics_-_Classical_Mechanics_(Likharev)/01%3A_Review_of_Fundamentals/1.02%3A_Kinematics-_Basic_NotionsIf the frames move versus each other by translation only (no mutual rotation!), similar relations are valid for the velocity and the acceleration as well: \[\begin{aligned} &\left.\mathbf{v}\right|_{\...If the frames move versus each other by translation only (no mutual rotation!), similar relations are valid for the velocity and the acceleration as well: v|in 0′=v|in 0+v0|in 0′a|in 0′=a|in 0+a0|in 0′ Note that in the case…
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Essential_Graduate_Physics_-_Classical_Mechanics_(Likharev)/07%3A_Deformations_and_Elasticity/7.05%3A_Rod_BendingNow proceeding in the same fashion to Eq. (76), we get φy=−ρgA2EIy(z−l)33+ const =−ρgA6EIy[(z−l)3+l3], where th...Now proceeding in the same fashion to Eq. (76), we get φy=−ρgA2EIy(z−l)33+ const =−ρgA6EIy[(z−l)3+l3], where the integration constant is selected to satisfy the clamping condition at the left end of the rod: φy =0 at z=0. (Note that this is different from the support condition illustrated on the lower panel of Figure 9 b, which allows the angle at z=0 to be different f…
- https://phys.libretexts.org/Bookshelves/Thermodynamics_and_Statistical_Mechanics/Essential_Graduate_Physics_-_Statistical_Mechanics_(Likharev)/05%3A_Fluctuations/5.07%3A_The_Fokker-Planck_equationwhere jq (which was called jw in the last section) is the probability current density in the coordinate space, and ∇q (which was denoted as ∇ in that sec...where jq (which was called jw in the last section) is the probability current density in the coordinate space, and ∇q (which was denoted as ∇ in that section) is the usual vector operator in the space, while jp is the current density in the momentum space, and ∇p is the similar vector operator in that space:
- https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Essential_Graduate_Physics_-_Classical_Electrodynamics_(Likharev)/02%3A_Charges_and_Conductors/2.09%3A_Variable_Separation__Polar_Coordinates15a (E0>0), the surface charge is positive on the right-hand side of the cylinder and negative on its left-hand side, thus creating a field directed from the right to the left, whi...15a (E0>0), the surface charge is positive on the right-hand side of the cylinder and negative on its left-hand side, thus creating a field directed from the right to the left, which exactly compensates the external field inside the conductor, where the net field is zero. (Please take one more look at the schematic Fig.
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Essential_Graduate_Physics_-_Classical_Mechanics_(Likharev)/08%3A_Fluid_Mechanics/8.04%3A_Dynamics-_Ideal_FluidsThe result shows that the pressure reaches its maximum at the ends of the longitudinal diameter y=0, while at the ends of the transverse diameter x=0, where the velocity is largest, it is lowe...The result shows that the pressure reaches its maximum at the ends of the longitudinal diameter y=0, while at the ends of the transverse diameter x=0, where the velocity is largest, it is lower by 2ρv02. (Here ρ is the fluid density again - sorry for the notation jitters!) Note that the distributions of both the velocity and the pressure are symmetric with respect to the transverse axis x =0, so that the fluid flow does not create any net drag force in i…
- https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Essential_Graduate_Physics_-_Classical_Mechanics_(Likharev)/02%3A_Lagrangian_Analytical_Mechanics/2.05%3A_Exercise_Problems(i) At not very high frequencies (whose quantum ℏω is lower than the binding energy 2Δ of the Cooper pairs), the Josephson effect in a sufficiently small junction may be descri...(i) At not very high frequencies (whose quantum ℏω is lower than the binding energy 2Δ of the Cooper pairs), the Josephson effect in a sufficiently small junction may be described by the following coupling energy: U(φ)=−EJcosφ+ const where the constant EJ describes the coupling strength, while the variable φ (called the Josephson phase difference) is connected to the voltage V across the junction by the famous …
- https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Essential_Graduate_Physics_-_Classical_Electrodynamics_(Likharev)/02%3A_Charges_and_Conductors/2.03%3A_Exercise_ProblemsCalculate the mutual capacitance between the terminals of the semi-infinite lumped-capacitor circuit shown in the figure on the right, and the law of decay of the applied voltage along the system. Use...Calculate the mutual capacitance between the terminals of the semi-infinite lumped-capacitor circuit shown in the figure on the right, and the law of decay of the applied voltage along the system. Use the method of images to find the Green’s function of the system shown in the figure on the right, where the bulge on the conducting plane has the shape of a semi-sphere of radius R.
- https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Essential_Graduate_Physics_-_Quantum_Mechanics_(Likharev)/02%3A_1D_Wave_Mechanics/2.05%3A_Motion_in_Soft_PotentialsAs a result, we get the following (still approximate!) result:\[\begin{gathered} \frac{d \Phi_{1}}{d x}=\frac{i}{2} \frac{d^{2} \Phi_{0}}{d x^{2}} / \frac{d \Phi_{0}}{d x}=\frac{i}{2} \frac{d}{d x}\le...As a result, we get the following (still approximate!) result:\[\begin{gathered} \frac{d \Phi_{1}}{d x}=\frac{i}{2} \frac{d^{2} \Phi_{0}}{d x^{2}} / \frac{d \Phi_{0}}{d x}=\frac{i}{2} \frac{d}{d x}\left(\ln \frac{d \Phi_{0}}{d x}\right)=\frac{i}{2} \frac{d}{d x}[\ln k(x)]=i \frac{d}{d x}\left[\ln k^{1 / 2}(x)\right], \\ \left.i \Phi\right|_{\mathrm{WKB}} \equiv i \Phi_{0}+i \Phi_{1}=\pm i \int^{x} k\left(x^{\prime}\right) d x^{\prime}+\ln \frac{1}{k^{1 / 2}(x)}, \\ \psi_{\mathrm{WKB}}(x)=\frac{…
- https://phys.libretexts.org/Bookshelves/Thermodynamics_and_Statistical_Mechanics/Essential_Graduate_Physics_-_Statistical_Mechanics_(Likharev)/02%3A_Principles_of_Physical_Statistics/2.04%3A_Canonical_ensemble_and_the_Gibbs_distributionThe relations (???) and (???) play the key role in the connection of statistics to thermodynamics, because they enable the calculation, from Z alone, of the thermodynamic potentials of ...The relations (???) and (???) play the key role in the connection of statistics to thermodynamics, because they enable the calculation, from Z alone, of the thermodynamic potentials of the system in equilibrium, and hence of all other variables of interest, using the general thermodynamic relations – see especially the circular diagram shown in Figure 1.4.2, and its discussion in Sec.
