In linear elasticity theory the stress tensor $\sigma$ is related to the strain tensor $\epsilon$ via the elastic tensor $C$. Specifically
$$ \sigma_{ij} = C_{ijkl} \epsilon_{kl} $$
Because $\sigma$ and $\epsilon$ are both symmetric second rank tensors, $\sigma_{ij} = \sigma_{ji}$ and $\epsilon_{kl} = \epsilon_{lk}$ so that $C$ has the so-called "minor symmetries":
$$ C_{ijkl} = C_{jikl} = C_{ijlk} $$
I can convince myself of that, but what I'm having trouble with is the "major symmetry": $$ C_{ijkl} = C_{klij} $$
This allegedly comes from symmetry of the Strain Energy Density $\psi(\epsilon)$,
$$ \psi = \frac{1}{2} C_{ijkl} \epsilon_{ij} \epsilon_{kl} $$
or some property of its second partial derivatives $$ C_{ijkl} = \frac{\partial^2 \psi}{\partial \epsilon_{ij}\partial \epsilon_{kl}} = \frac{\partial^2 \psi}{\partial \epsilon_{kl}\partial \epsilon_{ij}} = C_{klji} $$
To me it looks like its just switching or relabelling the indices rather than transposition. Any idea what I am missing?
** EXPANDED FOLLOWING Phoenix's POST **
Assuming a 2D solid in reduced ($IJ$) notation the strain density is
$$ \psi = \epsilon_I C_{IJ} \epsilon_J $$
$$ \psi = \begin{bmatrix} e & f \end{bmatrix} \begin{bmatrix} C_{11} & C_{12} \\ C_{21} & C_{22} \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} $$
$$ \psi = e[C_{11} a + C_{12} b] + f[C_{21} a + C_{22} b] $$
and the second derivatives are $$ \frac{\partial^2 \psi}{\partial e \partial b} = C_{12} $$ and $$ \frac{\partial^2 \psi}{\partial b \partial e} = C_{12} $$
Even in the special situation of $e=a$ and $f=b$ this still comes out as $$ \frac{\partial^2 \psi}{\partial a \partial b} = C_{12} + C_{21} $$ and $$ \frac{\partial^2 \psi}{\partial b \partial a} = C_{12} + C_{21} $$
i.e. nothing there requires $C$ to be symmetric.