# Calculation of Lagrangian from Hamiltonian $\frac{1}{2}(-i\partial_\phi -A)^2$

The Hamiltonian for particle on a ring is claimed to be (Eq. 9.1 of Altland2010a Condensed Matter Field Theory, pp. 498): $$H = \frac{1}{2}(-i\partial_\phi -A)^2 = \frac{1}{2}(p-A)^2$$

The book claims that $$L = \frac{1}{2}\dot{\phi}^2 - iA \dot{\phi}$$ I am quite confused, especially about the appearance of $\dot{\phi}$. Can any explain a bit?

What I tried:

Since the inverse of a Legendre transformation is Legendre transformation itself, \begin{align} \text{Denote }x &\equiv \frac{\partial H}{\partial p} = p-A,\text{ so,} \\ p &= x + A,\quad H = \frac{1}{2}x^2 ,\text{ so,}\\ L = x p - H &= x(x+A) - \frac{1}{2}x^2 = \frac{1}{2}x^2 + x A \end{align} So my calculation found that the Lagrangian of above Hamiltonian is: $$L = \frac{1}{2}x^2 + x A$$ where $$x = \frac{\partial H}{\partial p}$$

• ...why do you call the Legendre-transformed variable associated to $p$ $x$ and not $\dot{\phi}$? – ACuriousMind Feb 22 '17 at 8:33
• @ACuriousMind Right,I am not familiar with field theories, so I just choose $x$ out of convenience. But even when I tried to compare my answer with that in the book, I cannot get the Legendre-transformed them consistent term any choice. – taper Feb 22 '17 at 8:52
• $\partial_\phi$ in your Hamiltonian is surely a type, isnt it? – lalala Feb 22 '17 at 10:13
• @lalala you mean typo? I can't see. BTW, $\phi$ denotes the angle, not the field. – taper Feb 22 '17 at 10:17
• @taper ok. but then it looks like your question has nothing to do with field theory. Also is H then the classical Hamiltonian or quantized? – lalala Feb 22 '17 at 10:23

The Legendre transformation is almost correctly done in the question. $x$ should be replaced with $\dot{\phi}$. The book asks for the Lagrangian in the imaginary time path integral, derived from the Hamiltonian. This can be different. (Although the book alludes that this can be done by Legendre Transform, I believe this is more subtle).