[CPMD-list] Hamiltonian vs. Lagrangian convention ?

[Serge Ivanov]

   Dear Lukasz,

Your confusion with signs comes from mixing the two "kinetic"
terms. The one in L=T-V results in springs that connect beads
(replicas) after performing some simple math. And then the whole
thing is considered as an effective potential used in simulation.
 The kinetic term with "wrong" sign comes from adding generalized 
momenta conjugated to the generalized coordinates in the standard 
MD fashion.

For more details, please consider, e.g., a nice lecture by Mark Tuckerman
in NIC series volume 10 entitled "Quantum simulations of complex many-body 
systems: From theory to algorithms" pp. 269-298.

I think you can dowload it from the web as well.

WBR, Serge

On Tue, 27 May 2008, [UTF-8] Å~Aukasz Walewski wrote:

> Date: Tue, 27 May 2008 11:54:40 +0200
> From: "[UTF-8] Å~Aukasz Walewski" <ljw at icm.edu.pl>
> To: cpmd-list <cpmd-list at cpmd.org>
> Subject: [CPMD-list] Hamiltonian vs. Lagrangian convention ?
> 
> Dear Colleagues,

I feel puzzled with the convention used through out the AIPI papers.
Although this post concerns rather the understanding of the principles
of path integrals implemented in the CPMD code, I hope Axel won't
redirect me to Santa Claus. First I will outline my assumptions (that
may be wrong) and then point out my doubts.

Lagrangian L(x,dx/dt) is a function defined in the "primitive"
variables, i.e. coordinates x and their time derivatives dx/dt.

Hamiltonian H(x,p) is a function of canonical variables, i.e.
coordinates x and momenta p.

One can relate the two functions by means of Legendre transform:
L(x,dx/dt) = dx/dt * p - H(x,p)
Here the knowledge of p(x,dx/dt) is required.

The high-school level introduction of Lagrangian is
L = T - V
and that of Hamiltonian is
H = T + V
each in its proper variables, and say for mechanical system only.

In statistical mechanics formulation the canonical partition function is
given by the integral:
Z = \int dx \int dp exp(-\beta H(x,p))

On the other hand in the path integral formulation the integration over
momenta p is replaced by the integration over all possible paths x(t)
Z = \int dx \int Dx(t) exp(-A[x(t)])
where action along the path x(t) is
A[x(t)] = \int dt L(x,dx/dt)

The first ambiguity comes from the paper by Berne and Thirumalai, Ann.
Rev. Phys. Chem. 37 (1986) 401-424, where action is defined as an
integral over imaginary time \tau
A[x(\tau)] = \int d\tau  H(x(\tau)) (see Eq. (2) in the above paper)
- why the Hamiltonian is integrated here instead of the Lagrangian ?
Even if we replace time with imaginary time \tau = it, i being the
imaginary unit, the minus sign should appear in front of kinetic energy
part and the integrand becomes -T - V.

The second confusion comes from Marx and Parrinello, J. Chem. Phys. 104
(1996) 4077-4082, where in Eq. (2.1) the path integral partition
function is defined for the nuclei and electrons. The integrand there
has the form of
exp( -\int d\tau ( T(dx/d\tau) + V(x) ) )
The quantity T + V is called an "Euclidean Lagrangian" and the whole
integral over imaginary time \tau is called "Euclidean action". I do not
understand the plus sign here. The same affects Eq. (310) in Marx and
Hutter, NIC Series, Vol. 3 (2000) 329-477, where the Euclidean
Lagrangian is defined in terms of Coulomb interactions. Deriving
Euler-Lagrange equations from this formula directly leads to wrong signs
in eqs. of motion.

In other words, the formal definition of Euclidean Lagrangian (which is
unclear for me at this stage) leads me to improper sign in the equations
of motion and contradicts my "naive" feeling of what Lagrangian is.

I would appreciate some remarks that would point me to the right
direction.

Kind regards,
Lukasz


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| Dr. Sergei Ivanov                  | Phone : ++49 (0)234-32 26751   |
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