[Click here for a PDF of this post with nicer formatting]
Here’s a simple problem, a lot like the problem set 6 variational calculation.
Q: [1] 5.21
Estimate the lowest eigenvalue \lambda of the differential equation
\begin{equation}\label{eqn:absolutePotentialVariation:20} \frac{d^2}{dx^2}\psi + \lr{ \lambda – \Abs{x} } \psi = 0. \end{equation}
Using \alpha variation with the trial function
\begin{equation}\label{eqn:absolutePotentialVariation:40} \psi = \left\{ \begin{array}{l l} c(\alpha – \Abs{x}) & \quad \mbox{\(\Abs{x} < \alpha \) } \\ 0 & \quad \mbox{\(\Abs{x} > \alpha \) } \end{array} \right. \end{equation}
A:
First rewrite the differential equation in a Hamiltonian like fashion
\begin{equation}\label{eqn:absolutePotentialVariation:60} H \psi = -\frac{d^2}{dx^2}\psi + \Abs{x} \psi = \lambda \psi. \end{equation}
We need the derivatives of the trial distribution. The first derivative is
\begin{equation}\label{eqn:absolutePotentialVariation:80} \begin{aligned} \frac{d}{dx} \psi &= -c \frac{d}{dx} \Abs{x} \\ &= -c \frac{d}{dx} \lr{ x \theta(x) – x \theta(-x) } \\ &= -c \lr{ \theta(x) – \theta(-x) + x \delta(x) + x \delta(-x) } \\ &= -c \lr{ \theta(x) – \theta(-x) + 2 x \delta(x) }. \end{aligned} \end{equation}
The second derivative is
\begin{equation}\label{eqn:absolutePotentialVariation:100}
\begin{aligned}
\frac{d^2}{dx^2} \psi
&=
-c \frac{d}{dx} \lr{
\theta(x) – \theta(-x)
+
2 x \delta(x)
} \\
&=
-c \lr{
\delta(x) + \delta(-x)
+
2 \delta(x)
+
2 x \delta'(x)
} \\
&=
-c \lr{
4 \delta(x)
+
2 x \frac{-\delta(x) }{x}
} \\
&=
-2 c \delta(x).
\end{aligned}
\end{equation}
This gives
\begin{equation}\label{eqn:absolutePotentialVariation:120} H \psi = -2 c \delta(x) + \Abs{x} c \lr{ \alpha – \Abs{x} }. \end{equation}
We are now set to compute some of the inner products. The normalization is the simplest
\begin{equation}\label{eqn:absolutePotentialVariation:140} \begin{aligned} \braket{\psi}{\psi} &= c^2 \int_{-\alpha}^\alpha ( \alpha – \Abs{x} )^2 dx \\ &= 2 c^2 \int_{0}^\alpha ( x – \alpha )^2 dx \\ &= 2 c^2 \int_{-\alpha}^0 u^2 du \\ &= 2 c^2 \lr{ -\frac{(-\alpha)^3}{3} } \\ &= \frac{2}{3} c^2 \alpha^3. \end{aligned} \end{equation}
For the energy
\begin{equation}\label{eqn:absolutePotentialVariation:160}
\begin{aligned}
\braket{\psi}{H \psi}
&=
c^2 \int dx \lr{ \alpha – \Abs{x} } \lr{ -2 \delta(x) + \Abs{x} \lr{ \alpha – \Abs{x} } } \\
&=
c^2 \lr{ – 2 \alpha + \int_{-\alpha}^\alpha dx \lr{ \alpha – \Abs{x} }^2 \Abs{x} } \\
&=
c^2 \lr{ – 2 \alpha + 2 \int_{-\alpha}^0 du u^2 \lr{ u + \alpha } } \\
&=
c^2 \lr{ – 2 \alpha + 2 \evalrange{\lr{ \frac{u^4}{4} + \alpha \frac{u^3}{3} }}{-\alpha}{0} } \\
&=
c^2 \lr{ – 2 \alpha – 2 \lr{ \frac{\alpha^4}{4} – \frac{\alpha^4}{3} } } \\
&=
c^2 \lr{ – 2 \alpha + \inv{6} \alpha^4 }.
\end{aligned}
\end{equation}
The energy estimate is
\begin{equation}\label{eqn:absolutePotentialVariation:180} \begin{aligned} \overline{{E}} &= \frac{\braket{\psi}{H \psi}}{\braket{\psi}{\psi}} \\ &= \frac{ – 2 \alpha + \inv{6} \alpha^4 }{ \frac{2}{3} \alpha^3} \\ &= – \frac{3}{\alpha^2} + \inv{4} \alpha. \end{aligned} \end{equation}
This has its minimum at
\begin{equation}\label{eqn:absolutePotentialVariation:200}
0 = -\frac{6}{\alpha^3} + \inv{4},
\end{equation}
or
\begin{equation}\label{eqn:absolutePotentialVariation:220}
\alpha = 2 \times 3^{1/3}.
\end{equation}
Back subst into the energy gives
\begin{equation}\label{eqn:absolutePotentialVariation:240} \begin{aligned} \overline{{E}} &= – \frac{3}{4 \times 3^{2/3}} + \inv{2} 3^{1/3} \\ &= \frac{3^{4/3}}{4} \\ &\approx 1.08. \end{aligned} \end{equation}
The problem says the exact answer is 1.019, so the variation gets within 6 %.
References
[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.