# 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.