# Orthonormality condition proof

1. Jan 2, 2008

### raintrek

I'm working through a proof of the orthormality condition for a complete set of states and am struggling with one element of it:

Consider the eigenstates of the Hamiltonian in the following way:

1: $$\int\Psi^{*}_{m}H\Psi_{n}dV = E_{n}\int\Psi^{*}_{m}H\Psi_{n}dV$$

and

2: $$\int\Psi^{*}_{n}H\Psi_{m}dV = E_{m}\int\Psi^{*}_{n}H\Psi_{m}dV$$

Taking the complex conjugate of the second (for H real) to obtain

3: $$\int\Psi_{n}H\Psi^{*}_{m}dV = E_{m}\int\Psi_{n}H\Psi^{*}_{m}dV$$

Now subtract them:

1-3: $$\int\Psi^{*}_{m}H\Psi_{n}dV - \int\Psi^{*}_{n}H\Psi_{m}dV = (E_{n} - E_{m}) \int\Psi^{*}_{m}H\Psi_{n}dV$$

I don't understand why the LHS of eq. 3 in this subtraction has had it's complex conjugate taken again, yet the right hand side has remained as it is in 3. Is it something to do with the Hermiticity of the Hamiltonian? Many thanks

2. Jan 2, 2008

### marcusl

$$H\Psi_{m} = E_{m}\Psi_{m}$$ and

$$H\Psi_{n} = E_{n}\Psi_{n}$$.

Multiply top by $$\Psi^{*}_{n}$$ and integrate, likewise with $$\Psi^{*}_{m}$$ for the second. Complex conjugate the resulting 2nd equation and use the Hermiticity of H

$$\int\Psi^{*}_{m}H\Psi_{n}dV =\int\Psi^{*}_{n}H\Psi_{m}dV$$

to rewrite the left hand side. Subtracting the two equations now gives

$$0=(E_{m}-E^{*}_{n})\int\Psi^{*}_{m}\Psi_{n}dV.$$

Now either

$$\int\Psi^{*}_{m}\Psi_{n}dV=0, m\neq n$$

because the eigenvalues of distinct states are different, establishing orthogonality, or

$$E_{m}=E^{*}_{m}$$

establishing that the eigenvalues are real.

Normality simply comes from normalizing the eigenfunctions so that

$$\int\Psi^{*}_{m}\Psi_{m}dV=1.$$