# Liouville's Theorem derivation

As an example of the book "Introduction to Quantum Mechanics Schrodinger Equation and Path integral" by Harald J. W. Muller. We have to prove Liouville's theorem. Here I show the proof as the text presents it:

Show that $$\triangle q \triangle p$$ is independent of time $$t$$, which mean, this has the same value at >a time $$to$$, as at a time $$t_{0}^{\prime} \neq t_{0}$$.

Solution: We consider $$\frac{d}{d t} \ln (\triangle q \triangle p)=\frac{d(\triangle q)}{d t} \frac{1}{\triangle q}+\frac{d(\triangle p)}{d t} \frac{1}{\triangle p}\text{.}$$ Here $$d(\triangle q) / d t$$ is the rate at which the $$q$$-walls of the phase space element move away from the centre of the element, $$\dot{q}+\frac{\partial \dot{q}}{\partial q} \frac{\Delta q}{2} \quad \text { to the right } \quad \text { and } \quad \dot{q}-\frac{\partial \dot{q}}{\partial q} \frac{\Delta q}{2} \quad \text { to the left. }$$

Hence from the difference:

$$\frac{d(\Delta q)}{d t}=\frac{\partial \dot{q}}{\partial q} \triangle q, \quad \text { and similarly } \quad \frac{d(\Delta p)}{d t}=\frac{\partial \dot{p}}{\partial p} \Delta p$$

and with Hamilton's equations: $$\dot{q}_{i}=\frac{\partial H}{\partial p_{i}}, \quad \dot{p}_{i}=-\frac{\partial H}{\partial q_{i}}$$

we get:

$$\frac{d}{d t} \ln (\triangle q \triangle p)=\frac{\partial \dot{q}}{\partial q}+\frac{\partial \dot{p}}{\partial p}=\frac{\partial^{2} H}{\partial q \partial p}-\frac{\partial^{2} H}{\partial p \partial q}=0$$

My big problem is with the term $$\dot{q}+\frac{\partial \dot{q}}{\partial q} \frac{\Delta q}{2}$$ I can't be sure how to make sense of it. On the internet I looked at a alternative derivation in which they take the following approach:

For simplicity I will stick to a single particle in one dimension, so the phase space has only two dimensions, but the generalization to more dimensions is trivial. The region consists of all the systems with positions between $$x$$ and $$x+\delta x$$ and momentum between $$p$$ and $$p+\delta p$$, so its volume is $$V=\delta x \delta p$$. We can then write:

\begin{aligned} \frac{d V}{d t} &=\frac{d}{d t}(\delta x \delta p) \\ &=\delta x \frac{d(\delta p)}{d t}+\delta p \frac{d(\delta x)}{d t} \\ &=\delta x\left(\frac{d(p+\delta p)}{d t}-\frac{d p}{d t}\right)+\delta p\left(\frac{d(x+\delta x)}{d t}-\frac{d x}{d t}\right) \end{aligned}

Assume that $$\frac{dx}{dt}$$ and $$\frac{dp}{dt}$$ are linear in $$x$$ and $$p$$ respectively. This is an approximation for any finite sized region, but it becomes exact in the limit that $$\delta x$$ and $$\delta p$$ go to zero. That is,

$$\begin{array}{l} \frac{d(x+\delta x)}{d t}=\frac{d x}{d t}+\delta x \frac{\partial}{\partial x}\left(\frac{d x}{d t}\right) \\ \frac{d(p+\delta p)}{d t}=\frac{d p}{d t}+\delta p \frac{\partial}{\partial p}\left(\frac{d p}{d t}\right) \end{array}$$ [...]

In the second proof, the term is very familiar.I think they right wall is the one changing over time actually, whereas in the first proof both left and right are changing and they are separated by half the width. Now where does the term $$\delta x \frac{\partial}{\partial x}\left(\frac{d x}{d t}\right)$$ comes from? Are we treating the time derivative as a function that is Taylor expandable? i.e

$$f(x+\delta x)=f(x)+\delta x \frac{\partial f(x)}{\partial x}$$

but instead of $$f$$ we're treating the time derivative as a function? And is this the right way to think of it in the first proof? The left and right $$q$$-walls separating at a rate $$\frac{d(q-\frac{\triangle q}{2}) }{d t}= \dot{q}-\frac{\partial \dot{q}}{\partial q} \frac{\Delta q}{2} \quad and \quad \frac{d(q+\frac{\triangle q}{2}) }{d t}= \dot{q}+\frac{\partial \dot{q}}{\partial q} \frac{\Delta q}{2} \text{?}$$

The term $$\dot{q} + \frac{\partial \dot{q}}{\partial q}\frac{\Delta q}{2}$$ in your first proof is simply the Taylor expansion of the function $$q \mapsto \dot{q}$$, where the $$\frac{1}{2}$$ factor comes from distinguishing the right and left moves away from the centre of the element.
In the second proof your guess is correct. The term $$\delta x \frac{\partial}{\partial x}(\frac{dx}{dt})$$ comes indeed from the Taylor expansion of the time derivative function $$x \mapsto \dot{x}$$.
Below I will outline a third proof. We need to prove that any region in phase space is invariant under Hamiltonian evolution, i.e. $$\frac{d}{dt}V = 0$$. First, note that for a region $$V$$ in phase space, its volume is $$vol(V) = \int_{V}dq_1\dots dq_ndp_1\dots dp_n\,.$$ Now consider a map from phase space to itself given in coordinates by \begin{align*} {\bf q} \rightarrow {\bf Q} = {\bf Q}({\bf q},{\bf p},t) & & {\bf p}\rightarrow {\bf P}={\bf P}({\bf q},{\bf p},t)\,, \end{align*} then the region $$V$$ is mapped to $$\tilde{V}$$ with $$vol(\tilde{V}) = \int_{\tilde{V}} dQ_1\dots dQ_ndP_1\dots dP_n\,.$$ Actually, one has $$vol(\tilde{V}) = \int_{\tilde{V}} d\tilde{V} = \int_V \lvert det J \rvert dV\,,$$ where $$J$$ is the Jacobian matrix. If the transformation is canonical, one has $$(det J)^2 = 1$$, implying $$\lvert det J\rvert = 1$$. In this case we conclude that $$vol(\tilde{V}) = vol(V)$$.
Now, we can regard Hamiltonian flow as a one-parameter family of maps from phase space to itself, parametrized by time $$t$$. Since the Hamiltonian flow is a canonical transformation for all time $$t$$, starting with $$V = V(0)$$ and applying above the tranformation $$(q(0),p(0)) \rightarrow (q(t),p(t))$$ this region will change under Hamiltonian flow to $$V(t)$$. This means that $$\frac{d}{dt}V = 0$$.