# Using psi_0 = ((2c)/(pi))^("1/4")e^(-cx^2) as the normalized ground-state wave function, find c, and the trial energy E_phi such that E_phi = (<< phi_0 | hatH | phi_0 >>)/(<< phi_0 | phi_0 >>) >= E_0, where E_0 is the exact ground-state energy?

## The system is the simple harmonic oscillator, and this is called the variational method. I am super happy that I figured out how to do this, so I'm sharing this.

Oct 16, 2016

DISCLAIMER: Really long answer!

The variational method states that from a guess wave function $\phi$, we can approximate the ground-state energy for a system if we set ${\phi}_{0} = {\psi}_{0}$ and acquire the constant $c$ by minimizing the energy expression obtained from the expectation value equation:

${E}_{\phi} = \frac{\left\langle{\phi}_{0} | \hat{H} | {\phi}_{0}\right\rangle}{\left\langle{\phi}_{0} | {\phi}_{0}\right\rangle} \ge {E}_{0}$

where $\left\langley \left(x\right) | \hat{A} | y \left(x\right)\right\rangle = {\int}_{\text{allspace" y(x)^"*}} \hat{A} y \left(x\right) \mathrm{dx}$ is Dirac notation for an integral, hatH = (-ℏ^2)/(2mu)d/(dx^2) + 1/2kx^2 is the Hamiltonian operator for the harmonic oscillator, and ${\psi}_{0} = {\left(\frac{2 c}{\pi}\right)}^{\text{1/4}} {e}^{- c {x}^{2}}$ is the normalized ground-state wave function for the harmonic oscillator.

So the general steps are:

1. Evaluate the numerator integral.
2. Evaluate the denominator integral.
3. To minimize ${E}_{\phi}$, take $\frac{{\mathrm{dE}}_{\phi}}{\mathrm{dc}}$ and set it equal to $0$.
4. Find $c$ and plug it back into ${E}_{\phi}$ to see if it is greater than or equal to ${E}_{0}$.

FINDING THE TRIAL ENERGY IN TERMS OF C

The denominator goes to $1$, because $\int {\psi}_{0} {\left(x\right)}^{\text{*}} {\psi}_{0} \left(x\right) \mathrm{dx} = 1$ when ${\psi}_{0}$ is normalized. So we just evaluate the numerator and get:

int_(-oo)^(oo) ((2c)/(pi))^(1/4) e^(-cx^2) [(-ℏ^2)/(2mu)d/(dx^2) + 1/2kx^2] ((2c)/(pi))^(1/4) e^(-cx^2)dx

Move the constants out front and move the rightmost ${e}^{- c {x}^{2}}$ into the square brackets.

= ((2c)/(pi))^(1/4)((2c)/(pi))^(1/4) int_(-oo)^(oo) e^(-cx^2) [(-ℏ^2)/(2mu)(d^2)/(dx^2)(e^(-cx^2)) + 1/2kx^2e^(-cx^2)] dx

Now we take the second derivative of ${e}^{- c {x}^{2}}$ to get $\frac{d}{\mathrm{dx}} \left[- 2 c x {e}^{- c {x}^{2}}\right]$:

= ((2c)/(pi))^(1/2) int_(-oo)^(oo) e^(-cx^2) [(-ℏ^2)/(2mu)(d)/(dx)(-2cxe^(-cx^2)) + 1/2kx^2e^(-cx^2)] dx

From the product rule, we get:

= ((2c)/(pi))^(1/2) int_(-oo)^(oo) e^(-cx^2) [(-ℏ^2)/(2mu)(4c^2x^2e^(-cx^2) - 2ce^(-cx^2)) + 1/2kx^2e^(-cx^2)] dx

= ((2c)/(pi))^(1/2) int_(-oo)^(oo) e^(-cx^2) [(-ℏ^2)/(2mu)(4c^2x^2 - 2c)e^(-cx^2) + 1/2kx^2e^(-cx^2)] dx

Factor out the ${e}^{- c {x}^{2}}$, and combine them so that ${e}^{- c {x}^{2}} {e}^{- c {x}^{2}} = {e}^{- 2 c {x}^{2}}$:

= ((2c)/(pi))^(1/2) int_(-oo)^(oo) e^(-2cx^2) [(-ℏ^2)/(2mu)(4c^2x^2 - 2c) + 1/2kx^2] dx

Now it's a matter of distributing terms and getting things down to the tabled integrals

${\int}_{0}^{\infty} {x}^{2 n} {e}^{- \alpha {x}^{2}} \mathrm{dx} = \frac{1 \cdot 3 \cdot 5 \cdots \left(2 n - 1\right)}{{2}^{n + 1} {\alpha}^{n}} {\left(\frac{\pi}{\alpha}\right)}^{\text{1/2}}$, and

${\int}_{0}^{\infty} {e}^{- \alpha {x}^{2}} \mathrm{dx} = \frac{1}{2} {\left(\frac{\pi}{\alpha}\right)}^{\text{1/2}}$.

So we simplify to get:

=> ((2c)/(pi))^(1/2) int_(-oo)^(oo) e^(-2cx^2) [-(4ℏ^2c^2x^2)/(2mu) + (2cℏ^2)/(2mu) + 1/2kx^2] dx

= ((2c)/(pi))^(1/2) int_(-oo)^(oo) (cℏ^2)/mue^(-2cx^2) - (2ℏ^2c^2)/(mu) x^2e^(-2cx^2) + 1/2kx^2e^(-2cx^2) dx

Now, we plug in the tabled integrals, noting that ${\int}_{- \infty}^{\infty} \mathrm{dx} = 2 {\int}_{0}^{\infty} \mathrm{dx}$ for an even function like ${e}^{- \alpha {x}^{2}}$ or ${x}^{2} {e}^{- \alpha {x}^{2}}$, and that $\alpha = 2 c$ and $n = 1$, to get:

=> ((2c)/(pi))^(1/2) 2{ (cℏ^2)/mu[1/2(pi/(2c))^(1/2)] - (2ℏ^2c^2)/(mu) [1/(2^2(2c)^1) (pi/(2c))^(1/2)] + 1/2k[1/(2^2(2c)^1) (pi/(2c))^(1/2)]}

Now, if you notice, the constant out front can be cancelled out if we manage to factor out ${\left(\frac{\pi}{2 c}\right)}^{\frac{1}{2}}$. So, we proceed to simplify terms, to get:

= cancel(((2c)/(pi))^(1/2)) [(cℏ^2)/(mu)cancel((pi/(2c))^(1/2)) - (ℏ^2c)/(2mu)cancel((pi/(2c))^(1/2)) + k/(8c) cancel((pi/(2c))^(1/2))]

= (cℏ^2)/(mu) - (ℏ^2c)/(2mu) + k/(8c)

So finally, our trial energy is

=> color(green)(E_phi = (cℏ^2)/(2mu) + k/(8c))

Sorry, that was a really long process. This is a lot shorter.

MINIMIZING THE TRIAL ENERGY AND FINDING C

(dE_phi)/(dc) = 0 = d/(dc)[ℏ^2/(2mu)c + k/8 1/c]

= (ℏ^2)/(2mu) - k/(8c^2)

So, the value of $c$ that minimizes ${E}_{\phi}$ is acquired like so:

(2mu)/(ℏ^2) = (8c^2)/k

c^2 = (2kmu)/(8ℏ^2)

color(blue)(c = pm1/2 (sqrt(kmu))/ℏ)

See, this part wasn't so bad. We take the positive ${c}^{2}$ root to ensure that ${E}_{\phi} \ge {E}_{0}$.

CALCULATING THE TRIAL ENERGY

Finally, when we find what ${E}_{\phi}$ actually is after we minimize it, we plug $c$ back in to check that ${E}_{\phi} \ge {E}_{0}$:

E_phi = (ℏ^2)/(2mu)(1/2 (sqrt(kmu))/ℏ) + k/8 ((2ℏ)/(sqrt(kmu)))

= ℏ/4 sqrt(k/mu) + ℏ/4 sqrt(k/mu)

= 1/2 ℏ sqrt(k/mu)

If you recall from physics, $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency in terms of the mass $m$. For the harmonic oscillator, it is a two-body problem reduced down to a one-body problem, with a reduced mass $\mu = \frac{{m}_{1} {m}_{2}}{{m}_{1} + {m}_{2}}$.

So, we still have $\omega = \sqrt{\frac{k}{\mu}}$, where $\mu$ stands in for $m$. Therefore:

color(blue)(E_phi = 1/2ℏomega = 1/2hnu) >= E_0

Since E_(upsilon) = ℏomega(upsilon + 1/2) for a one-dimensional harmonic oscillator, we have that:

E_0 = [E_upsilon]|_(upsilon = 0) = 1/2ℏomega = 1/2hnu,

and we have exactly calculated the ground-state energy. That is, $\textcolor{b l u e}{{E}_{\phi} = {E}_{0}}$. Success!