Ex 1.2-1 Stationary States

Quantum Methods with Mathematica


Assume a wavefunction of the form psi[x, t] == f[t] psi[x] and perform a separation of variables on the wave equation.

Show that f[t] = E^(-I w t) where h w is the separation constant. Try the built-in function DSolve.

Equate h w to the Energy by evaluating the [expected] value of hamiltonian[V] in the state psi[x, t].



hbar := \[HBar]

H[V_] @ psi_  := -hbar^2/(2m) D[psi,{x,2}] + V psi

psi[x_,t_] := f[t] psi[x]

I hbar D [psi[x,t],t] == H[V] @ psi[x, t]

I hbar D [psi[x,t],t] / psi[x,t] == H[V] @ psi[x,t] / psi[x,t]

\displaystyle{i \hbar  \psi (x) f'(t)=V f(t) \psi (x)-\frac{\hbar ^2 f(t) \psi ''(x)}{2 m}}

\displaystyle{\frac{i \hbar  f'(t)}{f(t)}=\frac{V f(t) \psi (x)-\frac{\hbar ^2 f(t) \psi ''(x)}{2 m}}{f(t) \psi (x)}}

E1 := I hbar D [psi[x,t],t] / psi[x,t] == H[V] @ psi[x,t] / psi[x,t]


\displaystyle{\frac{1}{2} \hbar  \left(\frac{\hbar  \psi ''(x)}{m \psi (x)}+\frac{2 i f'(t)}{f(t)}\right)=V}

E2 := - 1/2 hbar hbar (D[D[psi[x],x],x]/(m psi[x])) == hbar omega

DSolve[E2, psi[x], x]

E3 := 1/2 hbar 2 i D[f[t],t] / f[t] == hbar omega

DSolve[E3, f[t], t]

\displaystyle{\left\{\left\{\psi (x)\to c_1 \cos \left(\frac{\sqrt{2} \sqrt{m} \sqrt{\omega } x}{\sqrt{\hbar }}\right)+c_2 \sin \left(\frac{\sqrt{2} \sqrt{m} \sqrt{\omega } x}{\sqrt{\hbar }}\right)\right\}\right\}}

\displaystyle{\left\{\left\{f(t)\to c_1 e^{\frac{\omega  t}{i}}\right\}\right\}}


psi[x_] := c E^(I k x)


f[t_] := E^(-I omega t)


psi[x_,t_] := f[t] psi[x]


\displaystyle{  \left\{k,c e^{i k x},e^{-i \omega  t},c e^{i k x-i \omega  t}\right\}  }

E4 := Conjugate[psi[x,t]] H[0] @ psi[x,t]


E5 := Simplify[E4]


k := Sqrt[2 m omega / hbar]

Refine[E5, {Element[{c, omega, m, t, hbar, k, x}, Reals]}]

\displaystyle{  \frac{c k^2 \hbar ^2 c^* \exp \left(-i \left(-(\omega  t-k x)^*-k x+\omega  t\right)\right)}{2 m}  }

\displaystyle{  = c^2 \omega  \hbar  }

E6 := Conjugate[psi[x,t]] psi[x,t]


\displaystyle{  c c^* \exp \left(-i \left(-(\omega  t-k x)^*-k x+\omega  t\right)\right)  }

\displaystyle{  = c^2  }


\displaystyle{\begin{aligned}            \langle E \rangle     &= \frac{\int_{-\infty}^{\infty} \psi^* H_{V=0} \psi dx}{\int_{-\infty}^{\infty} \psi^* \psi dx} \\ \\     &= \frac{c^2 \omega  \hbar \int_{-\infty}^{\infty} dx}{c^2 \int_{-\infty}^{\infty} dx} \\ \\    &= \omega  \hbar \\    \end{aligned}}


— Me@2022-11-26 07:17:29 PM



2022.11.28 Monday (c) All rights reserved by ACHK

Why does the universe exist? 7.2

“There is nothing in that region of space”


“there is nothing outside the universe”

have different meanings.


there is nothing except quantum fluctuations in that region of space

= the best detector detects nothing but quantum fluctuations


there is nothing outside the universe

= whatever detected, label the whole collection as “the universe”


“There is nothing outside the universe” does not (!!!) mean that “we go outside the universe to keep searching, but find nothing”.

— Me@2012-10-15 08:33:01 AM

— Me@2022-11-27 09:09:53 PM



2022.11.28 Monday (c) All rights reserved by ACHK