Question

Recall that the time evolution of a wavefunction y(x, t) is determined by the Schrödinger equation, which in position space reads iħ 4(x, t) = -24(x, t) + V(x, t)(x, t). ih vrt - h ? a) Consider any two normalized solutions to the Schrödinger equation, 41(2, t) and 02(3,t). Prove that their inner product is independent of time, doo 1 Vi (2, t)u2(x, t) dc = 0. dt J-00 Hint: prove the useful intermediate result, a 202 201 - - 2202 2241 an2-" 02 1 1 а d. да b) Say you have a wavefunction at time t=0, (C, t = 0), which you have normalized to total probability one. Given your results, what happens to this normalization at later times? Why is this important?

Answer 1

a) The time derivative of inner product of the two normalized solutions of Schroedinger equation is

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Now using the Schroedinger equation and its complex conjugate

$i\hbar\frac{\partial }{\partial t}\psi(x,t) = -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2}\psi(x,t)+V(x,t)\psi(x,t) \quad (2)$

$-i\hbar\frac{\partial }{\partial t}\psi^{*}(x,t) = -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2}\psi^{*}(x,t)+V(x,t)\psi^{*}(x,t) \quad (3)$

we get

$\frac{d}{dt}\int_{-\infty}^{\infty}\psi_{1}^{*}(x,t)\psi_{2}(x,t)dx = \int_{-\infty}^{\infty}\psi_{2}\left(\frac{\hbar^{2}}{2mi}\frac{\partial^2 \psi_{1}^{*}}{\partial x^2}+\frac{iV}{\hbar}\psi_{1}^{*} \right ) + \int_{-\infty}^{\infty}\psi^{*}_{1}\left(\frac{-\hbar^{2}}{2mi}\frac{\partial^2 \psi_{2}}{\partial x^2}-\frac{iV}{\hbar}\psi_{2} \right )\Rightarrow$

$\frac{d}{dt}\int_{-\infty}^{\infty}\psi_{1}^{*}(x,t)\psi_{2}(x,t)dx = \frac{\hbar^{2}}{2mi}\int_{-\infty}^{\infty}\left(\psi_{2}\frac{\partial^2 \psi_{1}^{*}}{\partial x^2}- \psi_{1}^{*}\frac{\partial^2 \psi_{2}}{\partial x^2}\right ) dx = \frac{\hbar^{2}}{2mi}\int_{-\infty}^{\infty}\frac{\partial }{\partial x}\left(\psi_{2}\frac{\partial \psi_{1}^{*}}{\partial x}- \psi_{1}^{*}\frac{\partial \psi_{2}}{\partial x}\right ) dx \quad (4)$

Now the integrand is a full derivative of x, therefore it can be easily integrated to get

$\frac{d}{dt}\int_{-\infty}^{\infty}\psi_{1}^{*}(x,t)\psi_{2}(x,t)dx = \frac{\hbar^{2}}{2mi}\left(\psi_{2}\frac{\partial \psi_{1}^{*}}{\partial x}- \psi_{1}^{*}\frac{\partial \psi_{2}}{\partial x}\right )_{x=\infty}-\frac{\hbar^{2}}{2mi}\left(\psi_{2}\frac{\partial \psi_{1}^{*}}{\partial x}- \psi_{1}^{*}\frac{\partial \psi_{2}}{\partial x}\right )_{x=-\infty} \quad (5)$

For a normalized wave function we must have

$\lim_{x\to\pm\infty}\psi(x,t) = 0$

or else the integral

$\int_{-\infty}^{\infty}|\psi(x,t)|^{2}dx$

will not converge.

Hence we get

$\frac{d}{dt}\int_{-\infty}^{\infty}\psi_{1}^{*}(x,t)\psi_{2}(x,t)dx = 0 \quad (6)$

b) The normalization of the initial wave function is

$\int_{-\infty}^{\infty}\psi^{*}(x,0)\psi(x,0)dx = \int_{-\infty}^{\infty}|\psi(x,0)|^{2}dx = 1 \quad (7)$

From part (a) we can deduce that

$\frac{d}{dt}\int_{-\infty}^{\infty}\psi^{*}(x,t)\psi(x,t)dx = 0\Rightarrow \int_{-\infty}^{\infty}\psi^{*}(x,t)\psi(x,t)dx = 1 \quad (8)$

this normalization is conserved. This is important because in quantum mechanics $|\psi(x,t)|^{2}$ is interpreted as probability density and a probability density function should always be normalized.

c) Consider again the Schroedinger equation and its complex conjugate

$i\hbar\frac{\partial }{\partial t}\psi(x,t) = -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2}\psi(x,t)+V(x,t)\psi(x,t) \quad (9)$

$-i\hbar\frac{\partial }{\partial t}\psi^{*}(x,t) = -\frac{\hbar^{2}}{2m}\frac{\partial^2 }{\partial x^2}\psi^{*}(x,t)+V(x,t)\psi^{*}(x,t) \quad (10)$

Multiply (9) by $\psi^{*}$ and (10) by $\psi$ then subtract (10) from (9)

$i\hbar\frac{\partial }{\partial t}\left(\psi^{*}\psi \right ) = \frac{\hbar^{2}}{2m}\left(\psi\frac{\partial^2 \psi^{*}}{\partial x^2}-\psi^{*}\frac{\partial^2 \psi}{\partial x^2} \right ) \quad (11)$

Defining

$\mathcal{P}(x,t) = |\psi(x,t)|^{2} \quad (12)$

and

$J(x,t) = \frac{i\hbar}{2m}\left(\psi\frac{\partial \psi^{*}}{\partial x}-\psi^{*}\frac{\partial \psi}{\partial x} \right ) \quad (13)$

Equation (11) can be written as

$\frac{\partial \mathcal{P}(x,t)}{\partial t}+\frac{\partial J(x,t)}{\partial x} = 0 \quad (14)$

Integrating equation (14) from a to b we get

$\int_{a}^{b}\frac{\partial \mathcal{P}(x,t)}{\partial t}dx = \frac{\mathrm{d} }{\mathrm{d} t}\int_{a}^{b}\mathcal{P}(x,t)dx = \frac{\mathrm{d} \mathcal{P}_{ab}}{\mathrm{d} t} = -\int_{a}^{b}\frac{\partial J(x,t)}{\partial x}dx = J(a,t)-J(b,t) \quad (15)$