# Why change of variables in the Hamiltonian give us the same physics while changes of variables in the Lagrangian does not?

Suppose we have the Lagrangian density

$$L=\partial_\mu\phi\partial^\mu\phi-m^2\phi^2\tag 1$$ With $$\phi$$ a scalar field and $$\pi=\frac{\partial L}{\partial \dot{\phi}}=2\dot{\phi}$$ we can show that the Hamiltonian density $$H$$ is given by

$$H=\pi+\nabla \phi \cdot \nabla \phi+m^{2} \phi^2 \tag 2$$

Now suppose we make a change of variables $$\phi=\sin \eta \tag 3$$ we obtain $$L(\phi)=L'(\eta)=\partial_\mu\eta\partial^\mu\eta\cos^2\eta-m^2\sin^2\eta$$ From $$L'(\eta)$$ we can construct the another Hamiltonian density $$H'=\pi'+ \nabla\eta\cdot\nabla\eta\cos^2\eta+m^2\sin^2\eta \tag 4$$

where $$\pi'=2\dot{\eta}\sin^2\eta$$

Now if we make the substitution $$(3)$$ in $$(2)$$ we obtain $$H(\phi)=H''(\eta)=\pi+ \nabla\eta\cdot\nabla\eta\cos^2\eta+m^2\sin^2\eta \tag 5$$

Comparing $$(4)$$ and $$(5)$$ we see that $$H'\neq H''$$

Now the evolution of states in the Schrodinger picture $$|\psi(t)\rangle=U\left(t\right)\left|\psi\left(0\right)\right\rangle$$

Where $$U(t)=\mathrm{T} \exp \left(-\frac{i}{\hbar} \int_{0}^{t} H\left(t^{\prime}\right) d t^{\prime}\right)$$ So probability amplitudes using $$H'$$ and $$H=H''$$ will give different results.

My question is why change of variables in the Hamiltonian give us the same physics while changes of variables in the Lagrangian does not?

• And why exactly do you say the Lagrangian equations of motion are different? (Also note that your transformation is extremely restrictive, a priori there is no reason for $\phi$ to be limited to the unit interval.) Aug 9, 2021 at 17:08
• @NDewolf The Hamiltonian are different so in the Schrodinger picture their equation of motion are different Aug 9, 2021 at 17:29
• @NDewolf $\sin x$ can take any value if the argument is not restricted to be real. Aug 9, 2021 at 19:06
• @my2cts That's of course true, but then you are unnecessarily complicating things. My point was just that it was an awkward thing to do (unless you have really good reasons) Aug 9, 2021 at 19:11
• @NDewolf I am too young to be blamed for the complex nature of $\phi$. :-). Aug 9, 2021 at 19:24

Your Legendre transformation isn't right. Given some Lagrangian density $$\mathscr L(\phi,\dot \phi, \nabla \phi)$$, the canonical momentum is $$\pi := \frac{\partial \mathscr L}{\partial \dot \phi}$$ and the Hamiltonian density is $$\mathscr H = \pi \dot \phi -\mathscr L$$. In your first example, you should have $$\mathscr H(\pi,\phi, \nabla \phi) = \frac{\pi^2}{4} + (\nabla \phi)^2 + m^2 \phi^2, \qquad \pi = 2\dot \phi$$ After the transformation $$\phi = \sin(\eta)$$ in the Lagrangian and performing the Legendre transformation, the Hamiltonian density becomes $$\tilde{\mathscr H}(\tilde \pi,\eta,\nabla \eta) = \frac{\tilde \pi^2}{4\cos^2(\eta)}+ (\nabla \eta)^2\cos^2(\eta) + m^2\sin^2(\eta), \qquad \tilde \pi = 2\cos^2(\eta) \dot \eta$$ Inserting $$\phi = \sin(\eta)$$ into the expressions for $$\pi$$ and $$\mathscr H(\pi,\phi,\nabla\phi)$$ reproduces the expressions for $$\tilde \pi$$ and $$\tilde {\mathscr H}(\tilde \pi,\eta,\nabla \eta)$$.
• @amiltonmoreira $\uparrow\$Yes, as per NDewolf's comment. Note that none of this is quantum yet; Legendre transformations and the Lagrangian/Hamiltonian equations of motion are part of classical field theory. Aug 9, 2021 at 19:17
A a side note: usually this Lagrangian carries a factor of 1/2. This will avoid cluttering expressions unnecessarily. Also in your Hamiltonians you should write $$\pi^2$$, $$\pi'^2$$ etc.