- #1

JALAJ CHATURVEDI

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$$ L=T-V \;\;\; \; \;\;\; [1]\;\;\; \; \;\;\; ( Lagrangian) $$

$$ H=T+V \;\;\; \; \;\;\;[2] \;\;\; \; \;\;\; (Hamiltonian)$$

and logically, this leads to the equation,

$$ H - L= 2V \;\;\; \; \;\;\; [3a]$$

$$ H + L= 2T \;\;\; \; \;\;\; [3b]$$

but,when it comes to :

$$\widehat{T}=\frac{-{\left(2\pi h\right)}^{2 } }{2 m} \frac{\partial^2 }{\partial x^2 }\ \;\;\; \; \;\;\; [4]$$

and

$$\widehat{V}=V\left(x\right)\ \;\;\; \; \;\;\; [5]$$

equation [3] doesn't seem to hold. And we need an altogether different equation which is :

$$ H=\dot{q}\frac{\partial L}{\partial \dot{q} }-L\ \;\;\; \; \;\;\; [6]$$

**1. Why is this happening?2. Even if equation [6] does hold, the results obtained by equation [3] & [6] must coincide, which is certainly not the case?**For example, given a Hamiltonian

$$\widehat{H}=\frac{-{\left(2\pi h\right)}^{2 } }{2 m} \frac{\partial^2 }{\partial x^2 }\ + V\left(x\right)\ \;\;\; \; \;\;\; [7]$$

why not can I use equation [3a] and get the corresponding Lagrangian instead of going through equation [6] ?

An example in favour of my question :

$$\ H=\frac{p_{\theta}^{2}}{2 ml^{2 }}+mgl\left(1-\cos\theta\right)\ $$and was asked to find $$ \frac{\text{d}L}{\text{d}t}\ $$ and the answers didn't match!