# Change of variables in partial derivative

I am stuck on a simple exercise in quantum mechanics because I can't figure out how to modify a partial derivative under a change in variables. If I have a Hamiltonian in two variables $x_1$ and $x_2$, and I introduce two new variables $u = x_1 - x_2$ and $v = x_1+x_2$, how to I change the partial derivatives $\frac{\partial^2}{\partial x_1^2}$ and $\frac{\partial^2}{\partial x_2^2}$ to be expressed in terms of $u$ and $v$?

I have the following Hamiltonian:

$$H = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x_1^2} - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x_2^2} + \frac{1}{2}m\omega^2 x_1^2 + \frac{1}{2}m\omega^2 x_1^2 + \frac{1}{2}\epsilon(x_1-x_2)^2$$ I tried a change of variables $u = x_1-x_2$ and $v =x_1+x_2$. The potential part of the Hamiltonian becomes $$\frac{1}{4}m\omega^2 (u^2+v^2) + \frac{1}{2}\epsilon u^2$$ My question is what happens to the kinetic part, $- \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x_1^2} - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x_2^2}$? How do these derivative change under this transformation?

We have $$\frac{\partial f}{\partial x_1}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x_1}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x_1}$$ and so $$\frac{\partial }{\partial x_1}=\frac{\partial u}{\partial x_1}\frac{\partial }{\partial u}+\frac{\partial v}{\partial x_1}\frac{\partial }{\partial v}=\frac{\partial }{\partial u}+\frac{\partial }{\partial v}.$$ You can proceed similarly for ${\partial f}/{\partial x_2}$.

For the second derivatives you apply this procedure twice: $$\begin{split} \frac{\partial^2 f}{\partial x_1^2} &=\frac{\partial}{\partial x_1}\left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial x_1}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x_1}\right)=\frac{\partial}{\partial x_1}\left(\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}\right)\\ &=\frac{\partial^2 f}{\partial u^2}\frac{\partial u}{\partial x_1} +\frac{\partial^2 f}{\partial u\partial v}\frac{\partial v}{\partial x_1} +\frac{\partial^2 f}{\partial v\partial u}\frac{\partial u}{\partial x_1} +\frac{\partial^2 f}{\partial v^2}\frac{\partial v}{\partial x_1}\\ &=\frac{\partial^2 f}{\partial u^2} +2\frac{\partial^2 f}{\partial u\partial v} +\frac{\partial^2 f}{\partial v^2}, \end{split}$$ assuming that $f$ is $C^2$. So $$\frac{\partial^2}{\partial x_1^2} =\frac{\partial^2}{\partial u^2} +2\frac{\partial^2}{\partial u\partial v} +\frac{\partial^2}{\partial v^2}.$$ Again you can proceed similarly for ${\partial^2 f}/{\partial x_2^2}$.

I'll add a note about how I keep myself from getting confused in these situations (where abuse of notation is rife), after enforcing the change of coordinates $$(x_1,x_2)\mapsto (u,v)$$ we rewrite ($C^2$) functions $f(x_1,x_2)$ as $$f(x_1,x_2)=g(u(x_1,x_2),v(x_1,x_2))$$ Then let's figure out what the familiar partials are in terms of the new partials, $$f_{x_1^2}=g_{u^2}+2g_{uv}+g_{v^2}\\ f_{x_1^2}=g_{u^2}-2g_{uv}+g_{v^2}$$ by the same computation as in the other answer. Since this holds for any (sufficiently nice) function $f$ we transform, we have $$\partial_{x_1}^2=\partial_{u^2}+2\partial_{uv}+\partial_{v^2}\\ \partial_{x_2}^2=\partial_{u^2}-2\partial_{uv}+\partial_{v^2}$$

• The simple fact of making explicit that the function of the transformed variables $g$ is not $f$ itself helped me understand a family of reasonings that have literally been haunting me. In particular now I see that, indeed, a book I'm reading is sloppy when it says that $\Box f(x,t) = 0$ implies, for $x_{\pm} = x \pm c t$, that $\partial_{x_+}\partial_{x_-}f(x_+, x_-) = 0$. It should say there is a function $g$ such that by definition $g(x_+, x_-) := f(x,t)$ and that, therefore, $\Box f(x,t) = \partial_{x_+}\partial_{x_-}g(x_+, x_-) = 0$ isn't it? (I certainly see how to build such $g$) May 13, 2021 at 15:25
• That is exactly how I think of things, and unfortunately what you described is typically how things are written. Being careful about the coordinate change has kept me from getting confused quite a few times. I'm glad it was helpful to you as well. May 13, 2021 at 19:48
• @Hal I should add that a good exercise to practice this thinking is to rewrite the laplacian in polar or spherical coordinates May 13, 2021 at 19:50

I know this is an old question, but for future passer-by I'd like to point out that it is actually a very useful exercise that helps to understand (reduced) density matrices in 1st quantised formalism. See (Peschel 1999): https://arxiv.org/abs/cond-mat/9906224

Consider a purely bosonic model, a chain of $$L$$ harmonic oscillator with frequency $$\omega_0$$, coupled together by springs. It has a gap in the phonon spectrum and is a non-critical integrable system. The Hamiltonian reads $$$$H = \sum_{i=1}^L \left( -\frac{1}{2} \frac{\partial ^2}{\partial x_i^2} + \frac{1}{2} \omega_0^2 x_i^2 \right) + \sum_{i=1}^{L-1} \frac{1}{2}\kappa(x_{i+1} - x_i)^2$$$$ Peschel parameterized it by $$\omega_0 = 1 - \kappa$$, so that if $$\kappa = 0$$ the Hamiltonian is digonal under boson occupation number, and there is no dispersion (only one mode $$\omega_0$$) and the system is gapped. If $$\kappa \rightarrow 1$$ (thus $$\omega_0 \rightarrow 0$$), there will only be acoustic phonon excitations and the system become gapless.

As the simplest example let us scrutinize the 2-particle problem. Its Hamiltonian reads $$$$H = \frac{1}{2} \frac{\partial^2 }{\partial x_1^2} + \frac{1}{2} \frac{\partial ^2}{\partial x_2^2} + \frac{1}{2}\omega_0^2 x_1^2 + \frac{1}{2} \omega_0^2 x_2^2 + \frac{\kappa}{2}(x_1 - x_2)^2$$$$ We don't want off-diagonal terms like $$x_1 - x_2$$, so we do the following transformation: $$$$v = (x_1 + x_2)/\sqrt{2},\;\;u = (x_1 - x_2)/\sqrt{2}\;\;\iff\;\; x_1 = (v + u)/\sqrt{2},\;\; x_2 = (v - u)/\sqrt{2}$$$$ I like the factor of $$\sqrt{2}$$ because of its reciprocal symmetry (also the transformation belongs to $$O(2)$$ so that $$\sum_{i} x_i^2$$ remain the same form). Then the potential energy becomes $$$$\frac{1}{2} \omega_0 x_1^2 + \frac{1}{2}\omega_0 x_2^2 + \frac{\kappa}{2} (x_1 - x_2)^2 = \frac{1}{2}\omega_0 v^2 + \frac{1}{2} \omega_0 u^2 + \frac{\kappa}{4} u^2$$$$ Now one question worthy of asking is how this transformation affect momentum terms? $$$$\frac{\partial }{\partial x_1}=\frac{\partial u}{\partial x_1}\frac{\partial }{\partial u}+\frac{\partial v}{\partial x_1}\frac{\partial }{\partial v}=\frac{1}{\sqrt{2}}\left(\frac{\partial }{\partial u}+\frac{\partial }{\partial v}\right)$$$$ $$$$\frac{\partial }{\partial x_2}=\frac{\partial u}{\partial x_2}\frac{\partial }{\partial u}+\frac{\partial v}{\partial x_2}\frac{\partial }{\partial v}= \frac{1}{\sqrt{2}}\left(-\frac{\partial }{\partial u}+\frac{\partial }{\partial v}\right)$$$$ so the second derivative gives $$$$\begin{split} 2\frac{\partial ^2}{\partial x_1^2} &= \sqrt{2}\frac{\partial }{\partial x_1} \left( \frac{\partial }{\partial u} + \frac{\partial }{\partial v}\right) = \left( \frac{\partial }{\partial u} + \frac{\partial }{\partial v}\right)\left( \frac{\partial }{\partial u} + \frac{\partial }{\partial v}\right) \\ &= \frac{\partial ^2}{\partial u^2} + 2 \frac{\partial ^2}{\partial u \partial v} + \frac{\partial ^2}{\partial v^2} \end{split}$$$$ $$$$\begin{split} 2\frac{\partial ^2}{\partial x_2^2} &= \sqrt{2}\frac{\partial }{\partial x_2} \left( -\frac{\partial }{\partial u} + \frac{\partial }{\partial v}\right) = \left( -\frac{\partial }{\partial u} + \frac{\partial }{\partial v}\right)\left( -\frac{\partial }{\partial u} + \frac{\partial }{\partial v}\right) \\ &= \frac{\partial ^2}{\partial u^2} - 2 \frac{\partial ^2}{\partial u \partial v} + \frac{\partial ^2}{\partial v^2} \end{split}$$$$ So the kinetic term after transformation is $$$$\frac{1}{2}\frac{\partial ^2}{\partial x_1^2} + \frac{1}{2} \frac{\partial ^2}{\partial x_2^2} = \frac{1}{2}\frac{\partial ^2}{\partial u^2} + \frac{1}{2} \frac{\partial ^2}{\partial v^2}$$$$ which is of the same form! We can actually understand this intuitively by perceiving the Hamiltonian as a single oscillator in a 2D plane with $$x_1$$ and $$x_2$$ standing for the two axes. The symmetric matrix is orthogonally diagonalizable, which is essentially a rotation of axes, thus shouldn't change the form of momentum. Then the original Hamiltonian is rotated into $$$$H = \frac{1}{2} \frac{\partial ^2}{\partial u^2} + \frac{1}{2}\left( \omega_0^2 + \frac{\kappa}{2} \right)u^2 + \frac{1}{2} \frac{\partial ^2}{\partial v^2} + \frac{1}{2}\omega_0^2 v^2 \equiv H_u + H_v$$$$ which describes two de-coupled harmonic oscillators. Since $$[H, H_u] = [H, H_v] = [H_u, H_v] = 0$$, wavefunctions of two harmonic modes can be measured simultanously, and their corresponding wavefunctions become separable. The ground state of a 1D harmonic oscillator with angular frequency $$\omega$$ is $$$$\Psi(x) = \left( \frac{\omega}{\pi} \right)^{1/4}\exp(-\frac{\omega}{2} x^2) \exp(-i\frac{\omega}{2}t)$$$$ therefore, if define $$\Omega^2 \equiv (1/2)(\omega_0^2 + \kappa/2)$$, the joint wavefunction of normal modes is $$$$\Psi(u,v) = C \exp(-\frac{\Omega}{2}u^2-\frac{\omega_0}{2}v^2)$$$$ where $$C$$ is a normalization constant. Then following Peschel1999 you can derive the reduced density matrix for a single oscillator.