Discrepancy in Lagrangian to Hamiltonian transformation?

[JALAJ CHATURVEDI]

I know,
$$ L=T-V \;\;\; \; \;\;\; [1]\;\;\; \; \;\;\; ( Lagrangian) $$
$$ H=T+V \;\;\; \; \;\;\;[2] \;\;\; \; \;\;\; (Hamiltonian)$$
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!

 

[Marc Rindermann]

The Langrangian is an equation of the generalised coordinates $q$ and velocities $\dot q$. To get from the Lagrangian to the Hamiltonian you perform a Legendre Transform and the Hamiltonian becomes a function of $q$ and $p = \frac{\partial L}{\partial \dot q}$. So you need to consider that $H$ and $L$ are functions of different arguments.