# I The Hamiltonian

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1. Mar 4, 2016

### Physgeek64

I'm a little confused about the hamiltonian.

Once you have the hamiltonian how can you find conserved quantities. I understand that if it has no explicit dependence on time then the hamiltonian itself is conserved, but how would you get specific conservation laws from this?

Many thanks

2. Mar 4, 2016

### vanhees71

Unfortunately you do not tell us about your level. Do you know Poisson brackets? If so, you question is quite easy to answer. Suppose you have an arbitrary phase-space function $f(t,q^k,p_j)$ ($k,j \in \{1,\ldots,f \}$) then the total time derivative is
$$\frac{\mathrm{d}}{ \mathrm{d} t} f=\dot{q}^k \frac{\partial f}{\partial q^k}+\dot{p}_j \frac{\partial f}{\partial p_j} + \partial_t f,$$
where the latter partial time derivative refers to the explicit time dependence of $f$ only. Now use the Hamilton equations of motion
$$\dot{q}^k=\frac{\partial H}{\partial p_k}, \quad \dot{p}_j=-\frac{\partial H}{\partial q^j}.$$
Plugging this in the time derivative you get
$$\frac{\mathrm{d}}{ \mathrm{d} t} f=\frac{\partial f}{\partial q^k} \frac{\partial H}{\partial p_k} - \frac{\partial f}{\partial p_j} \frac{\partial H}{\partial q^j} + \partial_t f=\{f,H \}+\partial_t f.$$
A quantity is thus obviously conserved by definition if this expression vanishes.

Applying this to the Hamiltonian itself you get
$$\frac{\mathrm{d}}{\mathrm{d} t} H=\{H,H\}+\partial_t H=\partial_t H,$$
i.e., $H$ is conserved (along the trajectory of the system) if and only if it is not explicitly time dependent.

Now an infinitesimal canonical transformation is generated by an arbitrary phase-space distribution function $G$,
$$\delta q^k=\frac{\partial G}{\partial p_k}\delta \alpha=\{q^k,G\} \delta \alpha, \quad \delta p_j=-\frac{\partial G}{\partial q^j} \delta \alpha =\{p_j,G \} \delta \alpha, \quad \delta H=\partial_t G \delta \alpha.$$
From this it is easy to show that
$$H'(t,q+\delta q,p+\delta p) = H(t,q,p),$$
i.e., that the infinitesimal canonical transformation is a symmetry of the Hamiltonian, if and only if
$$\{H,G \}+\partial_t G=0,$$
but that means that
$$\frac{\mathrm{d}}{\mathrm{d} t} G=0$$
along the trajectory of the system, i.e., the generator of a symmetry transformation is a conserved quantity, and also any conserved quantity is the generator of a symmetry transformation. That means that there's a one-to-one relation between the generators of symmetries and conserved quantities, which is one of Noether's famous theorems.

3. Mar 4, 2016

### Physgeek64

Sorry- no I don't know about Poisson brackets- I'm a complete novice. Haven't encountered hamiltonians before, nor do I know much about them.

Thank you for your response though! Unfortunately I can't see any of the maths you've included- For some reason my computer thinks its an error

4. Mar 4, 2016

### Edgardo

Right-click on the math error, go to Math Settings -> Math Renderer and try e.g. HTML-CSS (or some other renderer).