Question:
Computing the Gaussian Variational Estimate for the ground state of this potential.
$V(x) = - V_0 a \delta(x)$
Solution:
First, let's compute the expected energy for the Gaussian 'solution'.
$ \begin{eqnarray*} \psi_d(x) &=& \frac{1}{\pi^{1/4}\sqrt{d}} \exp\left[-\frac{x^2}{2d^2}\right] \\ E_d &=& \langle \psi_d(x) | \hat{H}|\psi_d(x)\rangle \\ &=& \int\limits_{-\infty}^{\infty}{\psi_d^*(x)(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_d(x) + V(x) \psi_d(x))dx} \\ &=& \int\limits_{-\infty}^{\infty}{\psi_d(x)(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_d(x) + V(x) \psi_d(x))dx} \\ &=& \int\limits_{-\infty}^{\infty}{\psi_d(x)(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_d(x))dx} + \int\limits_{-\infty}^{\infty}{\psi_d(x)(V(x) \psi_d(x))dx} \\ &=& \frac{-\hbar^2}{2m}\int\limits_{-\infty}^{\infty}{\psi_d(x)\frac{\partial^2}{\partial x^2}\psi_d(x)dx} + \int\limits_{-\infty}^{\infty}{(V(x) \psi_d^2(x))dx} \\ \end{eqnarray*} $
To compute that, let's start with the second term.
$ \begin{eqnarray*} \int\limits_{-\infty}^{\infty}{(V(x) \psi_d^2(x))dx} &=& \int\limits_{-\infty}^{\infty}{(-V_0a\delta(x) \psi_d^2(x))dx} \\ &=& -V_0a \psi_d^2(0) \\ &=& -V_0a (\frac{1}{\pi^{1/4}\sqrt{d}})^2 \\ &=& \frac{-V_0a}{\pi^{1/2}d} \end{eqnarray*} $
Next, we deal with the second derivative
$ \begin{eqnarray*} \frac{\partial^2}{\partial x^2}\psi_d(x) &=& \frac{\partial^2}{\partial x^2} \frac{1}{\pi^{1/4}\sqrt{d}} \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{1}{\pi^{1/4}\sqrt{d}} \frac{\partial^2}{\partial x^2} \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{1}{\pi^{1/4}\sqrt{d}} \frac{\partial}{\partial x} \frac{-x}{d^2} \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{-1}{\pi^{1/4}d^{5/2}} \frac{\partial}{\partial x} x \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{-1}{\pi^{1/4}d^{5/2}} (\exp\left[-\frac{x^2}{2d^2}\right] + \frac{-x^2}{d^2} \exp\left[-\frac{x^2}{2d^2}\right]) \\ &=& \frac{-1}{\pi^{1/4}d^{5/2}} \exp\left[-\frac{x^2}{2d^2}\right](1 - \frac{x^2}{d^2}) \\ \end{eqnarray*} $
Now we substitute this back into the first term and see how it works out:
$ \begin{eqnarray*} & & \frac{-\hbar^2}{2m}\int\limits_{-\infty}^{\infty}{\psi_d(x)\frac{\partial^2}{\partial x^2}\psi_d(x)dx} \\ &=& \frac{-\hbar^2}{2m}\int\limits_{-\infty}^{\infty}{(\frac{1}{\pi^{1/4}\sqrt{d}} \exp\left[-\frac{x^2}{2d^2}\right])\frac{-1}{\pi^{1/4}d^{5/2}} \exp\left[-\frac{x^2}{2d^2}\right](1 - \frac{x^2}{d^2})dx} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{(\exp\left[-\frac{x^2}{2d^2}\right]) \exp\left[-\frac{x^2}{2d^2}\right](1 - \frac{x^2}{d^2})dx} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{\exp\left[-\frac{x^2}{d^2}\right](1 - \frac{x^2}{d^2})dx} \\ \end{eqnarray*} $
Now it looks simpler, let's further simplify it further by letting $ y = \frac{x}{d} $, so $ d dy = dx $
$ \begin{eqnarray*} & & \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{\exp\left[-\frac{x^2}{d^2}\right](1 - \frac{x^2}{d^2})dx} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{\exp\left[-y^2\right](1 - y^2)ddy} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{2}} \int\limits_{-\infty}^{\infty}{\exp\left[-y^2\right](1 - y^2)dy} \\ \end{eqnarray*} $
The integral value is $ \frac{\sqrt{\pi}}{2} $. it can be seen as computing the total probability minus the variance of a standard Gaussian distribution with variance $ \sqrt{\frac{1}{2}} $. So the final value of the first term is
$ \begin{eqnarray*} & & \frac{\hbar^2}{2\pi^{1/2}md^{2}} \int\limits_{-\infty}^{\infty}{\exp\left[-y^2\right](1 - y^2)dy} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{2}} \frac{\sqrt{\pi}}{2} \\ &=& \frac{\hbar^2}{4md^{2}} \end{eqnarray*} $
Now we combine them back together, the expected energy for the Gaussian 'solution' would be $ \frac{\hbar^2}{4md^{2}} - \frac{V_0a}{\pi^{1/2}d} $. To minimize the quantity, we compute the derivative.
$ \begin{eqnarray*} 0 &=& \frac{\partial}{\partial d} \frac{\hbar^2}{4md^{2}} - \frac{V_0a}{\pi^{1/2}d} \\ &=& \frac{-2\hbar^2}{4md^{3}} + \frac{V_0a}{\pi^{1/2}d^2} \\ &=& \frac{-2\hbar^2\pi^{1/2}}{4\pi^{1/2}md^{3}} + \frac{4mV_0ad}{4\pi^{1/2}md^3} \\ 4mV_0ad -2\hbar^2\pi^{1/2} &=& 0 \\ d &=& \frac{2\hbar^2\pi^{1/2}}{4mV_0a} \\ &=& \frac{\hbar^2\pi^{1/2}}{2mV_0a} \\ \end{eqnarray*} $
Last but not least, we just substitute this back to the energy to get the minimum possible energy for the 'solution'
$ \begin{eqnarray*} & & \frac{\hbar^2}{4md^{2}} - \frac{V_0a}{\pi^{1/2}d} \\ &=& \frac{\hbar^2}{4m(\frac{\hbar^2\pi^{1/2}}{2mV_0a})^2} - \frac{V_0a}{\pi^{1/2}(\frac{\hbar^2\pi^{1/2}}{2mV_0a})} \\ &=& \frac{\hbar^2(2mV_0a)^2}{4m(\hbar^2\pi^{1/2})^2} - \frac{V_0a(2mV_0a)}{\pi^{1/2}(\hbar^2\pi^{1/2})} \\ &=& \frac{4m^2\hbar^2V_0^2a^2}{4m\hbar^4\pi} - \frac{2mV_0^2a^2}{\hbar^2\pi} \\ &=& \frac{mV_0^2a^2}{\hbar^2\pi} - \frac{2mV_0^2a^2}{\hbar^2\pi} \\ &=& - \frac{mV_0^2a^2}{\hbar^2\pi} \end{eqnarray*} $
Can you believe it all simplify to this? I am shocked now when I get this result! The final answer is to compare this with $ E_0 $ and get a numerical answer, which is $ \frac{2}{\pi} $.
Computing the Gaussian Variational Estimate for the ground state of this potential.
$V(x) = - V_0 a \delta(x)$
Solution:
First, let's compute the expected energy for the Gaussian 'solution'.
$ \begin{eqnarray*} \psi_d(x) &=& \frac{1}{\pi^{1/4}\sqrt{d}} \exp\left[-\frac{x^2}{2d^2}\right] \\ E_d &=& \langle \psi_d(x) | \hat{H}|\psi_d(x)\rangle \\ &=& \int\limits_{-\infty}^{\infty}{\psi_d^*(x)(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_d(x) + V(x) \psi_d(x))dx} \\ &=& \int\limits_{-\infty}^{\infty}{\psi_d(x)(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_d(x) + V(x) \psi_d(x))dx} \\ &=& \int\limits_{-\infty}^{\infty}{\psi_d(x)(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi_d(x))dx} + \int\limits_{-\infty}^{\infty}{\psi_d(x)(V(x) \psi_d(x))dx} \\ &=& \frac{-\hbar^2}{2m}\int\limits_{-\infty}^{\infty}{\psi_d(x)\frac{\partial^2}{\partial x^2}\psi_d(x)dx} + \int\limits_{-\infty}^{\infty}{(V(x) \psi_d^2(x))dx} \\ \end{eqnarray*} $
To compute that, let's start with the second term.
$ \begin{eqnarray*} \int\limits_{-\infty}^{\infty}{(V(x) \psi_d^2(x))dx} &=& \int\limits_{-\infty}^{\infty}{(-V_0a\delta(x) \psi_d^2(x))dx} \\ &=& -V_0a \psi_d^2(0) \\ &=& -V_0a (\frac{1}{\pi^{1/4}\sqrt{d}})^2 \\ &=& \frac{-V_0a}{\pi^{1/2}d} \end{eqnarray*} $
Next, we deal with the second derivative
$ \begin{eqnarray*} \frac{\partial^2}{\partial x^2}\psi_d(x) &=& \frac{\partial^2}{\partial x^2} \frac{1}{\pi^{1/4}\sqrt{d}} \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{1}{\pi^{1/4}\sqrt{d}} \frac{\partial^2}{\partial x^2} \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{1}{\pi^{1/4}\sqrt{d}} \frac{\partial}{\partial x} \frac{-x}{d^2} \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{-1}{\pi^{1/4}d^{5/2}} \frac{\partial}{\partial x} x \exp\left[-\frac{x^2}{2d^2}\right] \\ &=& \frac{-1}{\pi^{1/4}d^{5/2}} (\exp\left[-\frac{x^2}{2d^2}\right] + \frac{-x^2}{d^2} \exp\left[-\frac{x^2}{2d^2}\right]) \\ &=& \frac{-1}{\pi^{1/4}d^{5/2}} \exp\left[-\frac{x^2}{2d^2}\right](1 - \frac{x^2}{d^2}) \\ \end{eqnarray*} $
Now we substitute this back into the first term and see how it works out:
$ \begin{eqnarray*} & & \frac{-\hbar^2}{2m}\int\limits_{-\infty}^{\infty}{\psi_d(x)\frac{\partial^2}{\partial x^2}\psi_d(x)dx} \\ &=& \frac{-\hbar^2}{2m}\int\limits_{-\infty}^{\infty}{(\frac{1}{\pi^{1/4}\sqrt{d}} \exp\left[-\frac{x^2}{2d^2}\right])\frac{-1}{\pi^{1/4}d^{5/2}} \exp\left[-\frac{x^2}{2d^2}\right](1 - \frac{x^2}{d^2})dx} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{(\exp\left[-\frac{x^2}{2d^2}\right]) \exp\left[-\frac{x^2}{2d^2}\right](1 - \frac{x^2}{d^2})dx} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{\exp\left[-\frac{x^2}{d^2}\right](1 - \frac{x^2}{d^2})dx} \\ \end{eqnarray*} $
Now it looks simpler, let's further simplify it further by letting $ y = \frac{x}{d} $, so $ d dy = dx $
$ \begin{eqnarray*} & & \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{\exp\left[-\frac{x^2}{d^2}\right](1 - \frac{x^2}{d^2})dx} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{3}} \int\limits_{-\infty}^{\infty}{\exp\left[-y^2\right](1 - y^2)ddy} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{2}} \int\limits_{-\infty}^{\infty}{\exp\left[-y^2\right](1 - y^2)dy} \\ \end{eqnarray*} $
The integral value is $ \frac{\sqrt{\pi}}{2} $. it can be seen as computing the total probability minus the variance of a standard Gaussian distribution with variance $ \sqrt{\frac{1}{2}} $. So the final value of the first term is
$ \begin{eqnarray*} & & \frac{\hbar^2}{2\pi^{1/2}md^{2}} \int\limits_{-\infty}^{\infty}{\exp\left[-y^2\right](1 - y^2)dy} \\ &=& \frac{\hbar^2}{2\pi^{1/2}md^{2}} \frac{\sqrt{\pi}}{2} \\ &=& \frac{\hbar^2}{4md^{2}} \end{eqnarray*} $
Now we combine them back together, the expected energy for the Gaussian 'solution' would be $ \frac{\hbar^2}{4md^{2}} - \frac{V_0a}{\pi^{1/2}d} $. To minimize the quantity, we compute the derivative.
$ \begin{eqnarray*} 0 &=& \frac{\partial}{\partial d} \frac{\hbar^2}{4md^{2}} - \frac{V_0a}{\pi^{1/2}d} \\ &=& \frac{-2\hbar^2}{4md^{3}} + \frac{V_0a}{\pi^{1/2}d^2} \\ &=& \frac{-2\hbar^2\pi^{1/2}}{4\pi^{1/2}md^{3}} + \frac{4mV_0ad}{4\pi^{1/2}md^3} \\ 4mV_0ad -2\hbar^2\pi^{1/2} &=& 0 \\ d &=& \frac{2\hbar^2\pi^{1/2}}{4mV_0a} \\ &=& \frac{\hbar^2\pi^{1/2}}{2mV_0a} \\ \end{eqnarray*} $
Last but not least, we just substitute this back to the energy to get the minimum possible energy for the 'solution'
$ \begin{eqnarray*} & & \frac{\hbar^2}{4md^{2}} - \frac{V_0a}{\pi^{1/2}d} \\ &=& \frac{\hbar^2}{4m(\frac{\hbar^2\pi^{1/2}}{2mV_0a})^2} - \frac{V_0a}{\pi^{1/2}(\frac{\hbar^2\pi^{1/2}}{2mV_0a})} \\ &=& \frac{\hbar^2(2mV_0a)^2}{4m(\hbar^2\pi^{1/2})^2} - \frac{V_0a(2mV_0a)}{\pi^{1/2}(\hbar^2\pi^{1/2})} \\ &=& \frac{4m^2\hbar^2V_0^2a^2}{4m\hbar^4\pi} - \frac{2mV_0^2a^2}{\hbar^2\pi} \\ &=& \frac{mV_0^2a^2}{\hbar^2\pi} - \frac{2mV_0^2a^2}{\hbar^2\pi} \\ &=& - \frac{mV_0^2a^2}{\hbar^2\pi} \end{eqnarray*} $
Can you believe it all simplify to this? I am shocked now when I get this result! The final answer is to compare this with $ E_0 $ and get a numerical answer, which is $ \frac{2}{\pi} $.
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