time-independent Schrodinger equation.?

Question

ψ1 = A sin kx

ψ2 = B exp(-kx)

Show by direct substitution into the given version of the Schrödinger equation that ψ1 and ψ2 can indeed be possible solutions of the time-independent Schrödinger equation.

Answer 1

The time-independent SE is:

E u(x) = - (h-bar)^2/(2m) (d/dx)^2 u(x) + V(x) u(x)

where E is the constant energy.

a) Note that u1(x) = A sin(kx) satisfies the equation
u1''(x) = - k^2 u1(x). If you plug that into the SE, you get:
E u1(x) = -(h-bar)^2 (-k^2)/(2m) u1(x) + V(x) u1(x), and if you divide out the u1(x), you get:
E = (h-bar)^2(k^2)/(2m) + V(x)
so if V(x) is constant = Vo, this is a solution.
E = (h-bar)^2(k^2)/(2m) + Vo is > 0 if Vo is not too negative.

b) Note that u2(x) = B exp(-kx) satisfies the equation
u2''(x) = k^2 u2(x). So in a similar way, you get:
E = -(h-bar)^2(k^2)/(2m) + V(x)
so if V(x) is a constant Vo, this can also work.
E = Vo - (h-bar)^2(k^2)/(2m) > 0 if Vo is high enough.

Answer 2

Differential when you consider which you're in basic terms staring on the spatial not time dependence. Why 2d order?? The equation is relatively an power equation. Kinetic power is given by using p^2/2m the place p is the momentum. P in turn is given by using an operator i (hbar) d by using dx. So once you sq. it you get d by using dx ^2.