Question

1. The time-dependent Schrödinger equation The time-dependent Schrödinger equation is -R2 824(1,t) + V (1,t) (1,t) = in 2m 0:2 . (a) For V1, t) = 0, show that the wave function (1,t) = A sin (kr - wt) does not satisfy the time- dependent Schrödinger equation. (b) For VI,t) = 0, Show that I, t) = A cos(kr - wt) + i sin (kr - wt) does satisfy this equation. This is a simple demonstration that the wavefunction in quantum mechanics is complex.

Answer 1

Time dependent Schrodinger's equation:

04(x, t) +V(1, t)(x, t) = ih- at 2mar2

a) To show
vlr, t) = Asin(kr – wt)
does not satisfy equation 1.

We've been given that V(x,t)=0

thus LHS of equation 1 becomes:

h? a? Asin(kx - wt) 2m 2 +0 222

· LHS = -2 ha Akcos(kx - wt) де

...LHS = nk? Asin(kx – wt)

An2 2) LHS = 2m -Ut, t)

RHS of equation 1 becomes:

a Asin(kx - wt) RHS = ih

3) RHS = -Aih w cos(kx - wt)

From equations 2 and 3,

LHS + RHS

Thus, does not satisfy the Time Dependent Schrodinger's equation.

vlr, t) = Asin(kr – wt)

b) To show
(t, t) = Acos(kr – wt) + i Asin(kr - wt)
satisfies equation 1.

We've been given that V(x,t)=0

LHS of equation 1 becomes:

12 02[Acos(kx – wt) +iAsin(kx – wt)) .. 2m ar2

$\therefore LHS=-\frac{\hbar^2}{2m}\frac{\partial [-kAsin(kx-wt)+ikAcos(kx-wt)]}{\partial x}$

...LHS=- "[-k2 Acos(kr – wt) - ik? Asin(kx – wt)

$\therefore LHS=-\frac{\hbar^2(-k^2)}{2m} [Acos(kx-wt)+iAsin(kx-wt)]$

$4)\quad \therefore LHS=\frac{\hbar^2k^2}{2m} \psi(x,t)$

RHS of equation 1 becomes:

$RHS=i\hbar\frac{\partial [Acos(kx-wt)+iAsin(kx-wt)]}{\partial t}$

$\therefore RHS=i\hbar [wAsin(kx-wt)-iwAcos(kx-wt)]$

$\therefore RHS=w\hbar [Acos(kx-wt)+iAsin(kx-wt)]$

5) RHS = wh vzt)

Now, LHS will be equal to the RHS if

$6)\quad w\hbar=\frac{\hbar^2 k^2}{2m}\quad OR \quad w=\frac{\hbar k^2}{2m}$

Now we know;

$E=\hbar w$

$\therefore w=\frac{E}{\hbar}$

Now, De broglie wavelength is given by :

$\lambda=\frac{h}{mv}$

Also, $k=\frac{2\pi}{\lambda}$ and combining both we get:

$v=\frac{h}{2\pi} \frac{k}{m}=\frac{\hbar k}{m}$

$Now,\ \ Energy\ \ E=\frac{1}{2}mv^2$

$\therefore E=\frac{1}{2}m\frac{\hbar^2 k^2}{m^2}=\frac{\hbar^2 k^2}{2m}$

Thus, the expression for now becomes:

$\therefore w=\frac{1}{\hbar}\times \frac{\hbar^2 k^2}{2m}$

$7)\quad \therefore w= \frac{\hbar k^2}{2m}\ \ is \ true.$

From equations 4, 5, 6 and 7:

LHS=RHS.

Thus, does satisfy Time Dependent Schrodinger's equation.

(t, t) = Acos(kr – wt) + i Asin(kr - wt)