I have an hydrogenic atom, knowing that its ground-state wavefunction has the standard form $$ \psi = A e^{-\beta r} $$ with $A = \frac{\beta^3}{\pi}$, I have to find the best value for $\beta$ (using the variational method). After having included the Darwin correction $$ H_D = D\delta(r) $$ with $D = \frac{\alpha^2\pi Z}{2}$, into the Hamiltonian $$ H_0 = -\frac12\nabla^2-\frac{Z}{r} $$

I calculated $$\langle\psi(\beta)|-\frac{\nabla^2}{2}|\psi(\beta)\rangle = \langle \psi(\beta)|T|\psi(\beta)\rangle = \frac{\beta^2}{2}$$ then I thought to use the virial theorem to calculate the part $\langle V\rangle $: $$ \langle \psi(\beta)|-\frac{Z}{r}|\psi(\beta)\rangle = \langle \psi(\beta)|V|\psi(\beta)\rangle = -2\langle \psi(\beta)|T|\psi(\beta)\rangle = -\beta^2 $$ with the Darwin part left to be easily calculated.

Looking at the solutions that my professor wrote, I've just found a different result:

$$\langle V\rangle = \langle \psi(\beta)|-\frac{Z}{r}|\psi(\beta)\rangle = \frac{Z}{\beta}\langle \psi(\beta)|-\frac{\beta}{r}|\psi(\beta)\rangle = -2\frac{Z}{\beta}\frac{\beta^2}{2} = -\beta Z$$

Why's that? Is this somehow related to the fact that I'm going to use the variational method later? Please, help me understand.

  • $\begingroup$ Thanks to Kyle Kanos for the edit: I'm new on StackExchange! $\endgroup$
    – a Shy Guy
    Dec 24, 2013 at 18:35
  • $\begingroup$ Good question :-) It may take a while for someone to come along and answer it, especially since it's so close to Christmas, so don't panic if you don't get an answer right away. $\endgroup$
    – David Z
    Dec 24, 2013 at 21:18

1 Answer 1


In summary, you can either find a $|\psi\rangle$ that solves the schrodinger equation (and hence the virial theorem) or does not solve the schrodinger equation and then minimize $E(\beta)=\langle \psi(\beta)|H|\psi(\beta)\rangle$.

If you've solved the schrodinger equation then there is nothing to minimize. If you have guessed the right functional form of $|\psi(\beta)\rangle$ then that function will solve the schrodinger equation only at the minimum value $E(\beta)$.

What you have is (ignoring the constant $A$ for now) the right functional form of $|\psi(\beta)\rangle$ which does not satisfy the conditions the virial theorem for all values of $\beta$ because it does not solve the schrodinger equation for all values of $\beta$. We can illustrate this last point.

Calculating $H|\psi\rangle$ for arbitrary $\beta$ (and ignoring the darwin term),

$$H|\psi\rangle= \left(\frac{\beta}{r}-\frac{\beta^2}{2}-\frac{Z}{r}\right)|\psi\rangle$$

(up to a multiplicative constant.) Since $E$ must be independent of $r$ we find that $\beta$ must equal $Z$. This is what you've found by demanding that the virial theorem holds. This leaves you with the familiar result that


Notice that if you try to apply the virial theorem and then the variational method, the total energy $E(\beta) \sim -\beta^2$, which as you can see is unbounded from below. You can't minimize this! Your professor's approach is correct -- show yourself that you can minimize $\langle T \rangle + \langle V \rangle$ to conclude that $\beta = Z$.


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