THE HELLMANN - FEYNMAN THEOREM

Consider a system with, time - independent Hamiltonian operator Ĥ that's involving parameters. The example is the molecular electronic Hamiltonian, which depending parametrically on the nuclear coordinates. However the Hamiltonian operator of any system contains parameters. For example, in the one-dimensional harmonic oscillator and Hamiltonian operator −(h²/2m) (d²/dx²) +kx², the force constant k is the parameters, as is the mass m. Although ħ is a constant, now we can consider it as a parameter also. The stationary - state energies Eₙ are functions of the same parameter as Ĥ. For example, for the harmonic oscillator,

Eₙ = (ʋ+1/2) hʋ = (ʋ+ 1/2 )ħ (k/m)¹/²......(1)

The stationary state, wave function also depending on the parameters in Ĥ. We now investigate how Eₙ varies with each of the parameters. More specifically, if λ is one of these parameters, we ask for ∂Eₙ/∂λ, where the partial derivatives is taken with all other parameters, held constant.

We begin with the Schrodinger equation,

Ĥψₙ= Eₙψₙ....................... (2)

Where the ψₙ's are the normalized stationery state eigenfunctions. Because of normalization, we have

Eₙ= ∫ ψₙ* Ĥψₙ dτ................(3)

∂Eₙ  =   ∂   ∫ ψₙ* Ĥ ψₙ dτ..... (4)

∂λ        ∂λ

The integral equation (3) is a definite integral over all space, and it's value depending para metrically on λ since Ĥ and ψₙ depend on λ. Provides the integrand is well behaved. We can find the integral's derivatives with the respect to a parameters by differentiating the integrand with respect to the parameters and then integrating. Thus

∂Eₙ  = ∫  ∂      ( ψₙ* Ĥ ψₙ) dτ = ∫   ∂ψ*ₙ Ĥψₙ 

∂λ          ∂λ                                     ∂λ

dτ + ∫ ψ*ₙ   ∂     ( Ĥψₙ) dτ......... (5)

                   ∂λ

We have, 

∂∕∂λ (Ĥψₙ) = ∂∕∂λ ( T̂ψₙ) + ∂∕∂λ ( V̂ψₙ)...... (6)

The potential operator is just multiplication by V, so

∂∕∂λ ( V̂ψₙ) = ∂V/∂λψₙ + V ∂ψₙ/∂λ........ (6)

The parameter ∂ will occurs in the kinetic energy operator as part of the factor multiplying one or more of the derivatives with respect to the coordinates. For example, taking λ as the mass of the particles, we have for a one - particle problem. 

T̂ = − ħ²/2λ ( ∂²/∂x² + ∂²/∂y² + ∂²/∂ʐ²) 

∂∕∂λ(T̂ψ) = − h²/2 ∂∕∂λ   [¹/λ ( ∂²ψ/∂x² + ∂²ψ/∂y² + ∂²ψ/dʐ²) ]

= ħ²/2λ² ( ∂²ψ/∂x² + ∂²ψ/∂y² + ∂²ψ/∂ʐ²) − ħ²/2λ ( ∂²/∂x² + ∂²/∂y² + ∂²/∂ʐ²) ( ∂ψ/∂λ) 

Since we can change the order of the partial differentiations without affecting the result. 

We can write this last equation as, 

∂∕∂λ ( T̂ψₙ) = ( ∂T̂/∂λ ) ψₙ + T̂ ( ∂ψₙ/∂λ)......... (7)

Where ∂T̂ /∂λ is found by differentiating T̂ with respect to λ just as if it were a function of operator. Although we got equation (7) by considering a specific T̂ and λ, the same argument shows it to be generally valid. The combining of equation (6) and (7) we can write, 

∂∕∂λ ( Ĥψₙ) = ( ∂Ĥ/∂λ) ψₙ + Ĥ (∂ψₙ/∂λ)......... (8)

Equation (4) becomes, 

 ∂Eₙ = ∫ ∂ψ*ₙ   Ĥψₙ dτ + ∫ ψ*ₙ  ∂Ĥ   ψₙdτ +

∂λ          ∂λ                                 ∂λ 

∫ ψ*ₙ Ĥ   ∂ψₙ  dτ............... (9)

               ∂λ

For the first integral equation (9) we have, 

∫ ∂ψ*ₙ  Ĥψₙdτ  = Eₙ ∫ ∂ψ*ₙ ψₙ dτ...... (10)

  ∂λ                              ∂λ

The Hermitian property of Ĥ and equation (1) give for the last integral in equation (9)

∫ ψ*ₙ Ĥ  ∂ψₙ  dτ = ∫ ∂ψₙ (Ĥψₙ)* d

              ∂λ              ∂λ 

Therefore, 

 ∂En = ∫ ψ*ₙ ∂Ĥ  ψₙ dτ + Eₙ ∫ ∂ψ*ₙ ψₙ dτ +

 ∂λ                ∂λ                      ∂λ 

Eₙ ∫ ψ*ₙ  ∂ψₙ  dτ................(11)

               ∂λ

The wave function is normalized, so

∫ ψ*ₙψₙdτ =1,   ∂    ∫ ψ*ₙ ψₙ dτ = 0

                         ∂λ

∫   ∂ψ*ₙ  ψₙ dτ + ∫ ψ*ₙ  ∂ψₙ  dτ =0........(12)

     ∂λ                             ∂λ

Using equation (12) in (11), we obtain 

∂Eₙ  = ∫ ψ*ₙ ∂Ĥ  ψₙ  dτ............... (13)

∂λ                ∂λ

Equation (13) is the generalized, Hellmann - Feynman theorem.