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



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