# 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?

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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.

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.

##### 1 Answer

**DISCLAIMER:** *Really long answer!*

The **variational method** states that from a guess wave function *approximate the ground-state energy* for a system if we set

#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:

- Evaluate the numerator integral.
- Evaluate the denominator integral.
- To minimize
#E_phi# , take#(dE_phi)/(dc)# and set it equal to#0# . - 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

#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

#= ((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

#= ((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

#= ((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

#=> ((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

#= 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

#(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

**CALCULATING THE TRIAL ENERGY**

Finally, when we find what

#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,

So, we still have

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

Since

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

and **we have exactly calculated the ground-state energy**. That is,