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?

DISCLAIMER: Really long answer!

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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:

$E_phi = (<< phi_0 | hatH | phi_0 >>)/(<< phi_0 | phi_0 >>) >= E_0$

where $<< y(x) | hatA | y(x) >> = int_"allspace" y(x)^"*" hatA y(x) dx$ 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 $psi_0 = ((2c)/pi)^"1/4"e^(-cx^2)$ 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 $(dE_phi)/(dc)$ 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 $int psi_0(x)^"*" psi_0(x)dx = 1$ 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 $e^(-cx^2)$ 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 $e^(-cx^2)$ to get $d/(dx)[-2cxe^(-cx^2)]$:

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

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 $e^(-cx^2)$, and combine them so that $e^(-cx^2)e^(-cx^2) = e^(-2cx^2)$:

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

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

$int_(0)^(oo) x^(2n)e^(-alphax^2)dx = (1*3*5cdots(2n-1))/(2^(n+1)alpha^n)(pi/alpha)^("1/2")$, and

$int_(0)^(oo) e^(-alphax^2)dx = 1/2(pi/alpha)^"1/2"$.

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 $int_(-oo)^(oo)dx = 2int_(0)^(oo)dx$ for an even function like $e^(-alphax^2)$ or $x^2e^(-alphax^2)$, and that $alpha = 2c$ 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 $(pi/(2c))^(1/2)$. 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 $E_phi >= E_0$.

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 $E_phi >= E_0$:

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

$= ℏ/4 sqrt(k/mu) + ℏ/4 sqrt(k/mu)$

$= 1/2 ℏ sqrt(k/mu)$

If you recall from physics, $omega = sqrt(k/m)$ 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 $mu = (m_1m_2)/(m_1 + m_2)$.

So, we still have $omega = sqrt(k/mu)$, 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, $color(blue)(E_phi = E_0)$. Success!