# von Neumann on the Hamiltonian

I'm reading von Neumann's book on QM and I'm slightly confused by a simple point. He writes,

"The energy is a given function of the coordinates and their time derivatives: $$E = L(q_1,..,q_k;\dot{q}_1,...,\dot{q}_k) = H(q_1,...,q_k;p_1,...,p_k)$$. (This $$H$$ is the Hamiltonian function.)

This quote comes from chapter 1. Does $$L(....)$$ denote the Lagrangian? Why does the second equality hold?

• It seems to me that the statement is written in this way to emphasize that energy can be written as a function of the generalized coordinates and velocities, and by finding the generalized momenta we can also express energy as a function of generalized coordinates and momenta. The second equality should only underscore the fact that energy itself, is a given function of the coordinates and their derivatives (though it might be reached via different formalizations) as stated in the first part of the expression. – Gulce Kardes Sep 5 at 12:33
• I have no deeper reasoning for this, we already know L and H are different functions that stand for different things at the simplest level. – Gulce Kardes Sep 5 at 12:38
• if $H$ does not explicitly depend on time, as you have written it, then it is a constant of time hence $E=H$ – hyportnex Sep 5 at 13:44
• @hyportnex Aren't the $q_i$ and $p_i$ typically functions of time, so $H$ is in fact, implicitly, a function of time? – user193319 Sep 5 at 14:22
• yes they are, but the question if it is explicitly a function of time (same for the lagrangian $\mathcal L$) en.wikipedia.org/wiki/Lagrangian_mechanics – hyportnex Sep 5 at 14:41

No, L does not stand for the Lagrangian $$\cal L$$: it stands for the "energy function" , ("as a rule, a quadratic function of the $$\dot{q}$$s"), an evident constant of the motion, $$L(q_i, \dot{q}_i) = \frac{\partial \cal{L}}{\partial\dot{q}_i}\dot{q}_i - \cal{L},$$ so, basically, $$\sim {\cal L} + 2V(q)$$ for the quadratic function he is considering, a deeply unfortunate convention, ipso facto... (the energy function in the Lagrangian formalism), is all.

The momenta p are then the same gradients of $$\cal L$$ or L ! Storm in a teacup. Are you overthinking it?

The equality $$L(q,\dot{q})=H(q,p)$$ holds as usual in the Legendre transform transition from Lagrangian to Hamiltonian mechanics.

PS. von Neumann was famously too quick to imagine he'd have to explain things fully, preferring to rely on his proverbial well-meaning intelligent reader. Wait until you get to III.5.F, which has victimized generations of readers, and required a famous theorem to be declared confusing, after all.