Question

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.

Answer 1

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 = (<< 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!