Assignment 1
... the phase trajectories are then given by cx2 + 2dxy − by 2 = constant. 4. The relativistic linear harmonic oscillator is given by the equation of motion d m0 c2 (p ) + kx = 0, (k > 0, v = ẋ) dt 1 − v 2 /c2 ...
... the phase trajectories are then given by cx2 + 2dxy − by 2 = constant. 4. The relativistic linear harmonic oscillator is given by the equation of motion d m0 c2 (p ) + kx = 0, (k > 0, v = ẋ) dt 1 − v 2 /c2 ...
![[2014 solutions]](http://s1.studyres.com/store/data/008881817_1-9b59b3a488861f750cad1da068f01a69-300x300.png)
[2014 solutions]
... (c) In first order perturbation theory, the energy change is ⟨ψ0 |∆V |ψ0 ⟩. Taking |ψ0 ⟩ = | ↓1 , ↓2 ⟩ and ∆V = −Jσ1x σ2x , we see that the energy change is ⟨ψ0 |∆V |ψ0 ⟩ = 0, since ⟨↓ |σ x | ↓⟩ = 0. ...
... (c) In first order perturbation theory, the energy change is ⟨ψ0 |∆V |ψ0 ⟩. Taking |ψ0 ⟩ = | ↓1 , ↓2 ⟩ and ∆V = −Jσ1x σ2x , we see that the energy change is ⟨ψ0 |∆V |ψ0 ⟩ = 0, since ⟨↓ |σ x | ↓⟩ = 0. ...
Systems of Equations
... To “solve the system” is to find the ordered pair (x, y) that make all equations true. This is where the graphs of the equations intersect. For a system of linear equations, there is: one solution if the lines intersect; no solution if the lines never intersect (i.e. the lines are parallel); and inf ...
... To “solve the system” is to find the ordered pair (x, y) that make all equations true. This is where the graphs of the equations intersect. For a system of linear equations, there is: one solution if the lines intersect; no solution if the lines never intersect (i.e. the lines are parallel); and inf ...
