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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?

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    $\begingroup$ 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.) $\endgroup$
    – NDewolf
    Aug 9, 2021 at 17:08
  • $\begingroup$ @NDewolf The Hamiltonian are different so in the Schrodinger picture their equation of motion are different $\endgroup$ Aug 9, 2021 at 17:29
  • $\begingroup$ @NDewolf $\sin x$ can take any value if the argument is not restricted to be real. $\endgroup$
    – my2cts
    Aug 9, 2021 at 19:06
  • $\begingroup$ @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) $\endgroup$
    – NDewolf
    Aug 9, 2021 at 19:11
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    $\begingroup$ @NDewolf I am too young to be blamed for the complex nature of $\phi$. :-). $\endgroup$
    – my2cts
    Aug 9, 2021 at 19:24

2 Answers 2

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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 $\tilde {\mathscr H}(\tilde \pi,\eta,\nabla \eta)$.

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  • $\begingroup$ Is this a general result that in quantum field if we change variables the phisics will be the same? $\endgroup$ Aug 9, 2021 at 18:32
  • $\begingroup$ J. Murray if it so could you give some reference $\endgroup$ Aug 9, 2021 at 18:39
  • $\begingroup$ @amiltonmoreira, there is the notion of "canonical transformations", i.e. the transformations that leaves the Hamiltonian equations of motion invariant (in classical mechanics, but of course an extension to quantum exists). $\endgroup$
    – NDewolf
    Aug 9, 2021 at 19:00
  • $\begingroup$ @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. $\endgroup$
    – J. Murray
    Aug 9, 2021 at 19:17
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Suppose a Lagrangian gives the correct physical model, while after a particular change of variables it no longer does. Thus the two forms of the Lagrangian contradict each other. This must mean that the variable change was incorrectly executed.

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.

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