Download Path integral in quantum mechanics

Path integral in quantum mechanics
let’s look at one piece first:
based on S-6
Consider nonrelativistic quantum mechanics of one particle in one
dimension with the hamiltonian:
Campbell-Baker-Hausdorf formula
P and Q obey:
Probability amplitude for the particle to start at q’ at time t’ and end up at
position q’’ at time t’’ is
where
and
complete set of momentum states
are eigenstates of the position operator Q.
In the Heisenberg picture:
and we can define instantaneous eigenstates:
Probability amplitude is then:
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=
=
to evaluate the transition amplitude:
to evaluate the transition amplitude:
let’s divide the time interval T = t’’ - t’ into N+1 equal pieces
let’s divide the time interval T = t’’ - t’ into N+1 equal pieces
insert N complete sets of position eigenstates
insert N complete sets of position eigenstates
important for general form of hamiltonians
with terms containing both P and Q
in our case, it doesn’t make any difference
we find:
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What is it good for?
Consider e.g.:
taking the limit
we get:
the result can be simply written using path integral as:
should be understood as integration over all paths in phase space
that start at
and end at
(with arbitrary initial and final momenta)
Similarly:
In simple cases when hamiltonian is at most quadratic in momenta, the integral over p
is gaussian and can be easily calculated:
lagrangian
time ordering is crucial!
Time-ordered products appeared in LSZ formula for scattering amplitudes!
prefactor can be absorbed into the definition of measure
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Functional derivatives:
taking the limit
we get:
Dirac delta function
“continuous generalization” of
they are defined to satisfy all the usual rules of derivatives (product rule, ...)
Consider modifying hamiltonian to:
should be understood as integration over all paths in phase space
that start at
and end at
Then we have:
(with arbitrary initial and final momenta)
In simple cases when hamiltonian is at most quadratic in momenta:
And we find, e.g.:
where
is calculated by finding the stationary point of the p-integral by solving:
for p and plugging the solution back to
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thus we have:
more examples:
Similarly, for
the replacement
picks up the ground state as the final state in the limit
.
we can integrate over q’ and q’’ which leads to a constant factor that can
be absorbed into the normalization of the path integral.
Thus, with the replacement
the boundary conditions and we have:
we don’t have to care about
after we bring down as many qs and ps as we want we can set
and return to the original hamiltonian:
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Adding perturbations:
Finally, we want both initial and final states to be ground states and take
the limits
and
:
we can simply write (suppressing the
):
is the ground-state wave function
looks complicated, we will use the following trick instead:
eigenstate of H
Finally, if perturbing hamiltonian depends only on q, and we want to
corresponding eigenvalue
calculate only time-ordered products of Qs, and if H is no more than
quadratic in P and if the term quadratic in P does not involve Q, then the
equation above can be written as:
is wave function of n-th state
let’s replace
with
and take the limit
:
every state except the ground state is multiplied by a vanishing factor!
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Path integral for harmonic oscillator
based on S-7
Consider a harmonic oscillator:
ground state to ground state transition amplitude is:
it is convenient to change integration variables:
external force
then we get:
a shift by a constant
equivalent to
and the transition amplitude is:
thus going to lagrangian formulation (integrating over p) we get:
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using Furier-transformed variables:
and setting
But
since, if there is no external force, a system in its
ground state remain in its ground state.
for simplicity, we get
thus we have:
or, in terms of time-dependent variables:
and thus:
E ! = −E
using inverse Fourier transformation
where:
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Comment:
For even number of Qs we pair up Qs in all possible ways:
is a Green’s function for the equation of motion of the harmonic oscillator:
you can evaluate it explicitly, treating the integral as a contour integral in
the complex E-plane and using the residue theorem. Make sure you are
careful about closing the contour in the correct half-plane for t > t’ and t <
t’ and that you pick up the correct pole.
in general:
you should find:
We can now easily generalized these results to a free field theory...
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Path integral for free field theory
Let’s calculate “correlation functions” of Q operators:
based on S-8
Hamiltonian density of a free field theory:
for harmonic oscillator we find:
similar to the hamiltonian of the harmonic oscillator
dictionary between QM and QFT:
classical field
operator field
classical source
we repeat everything we did for the path integral in QM but now for fields;
we divide space and time into small segments; take a field in each segment
to be constant; the differences between fields in neighboring segments
become derivatives; use the trick: multiplying
by
is
equivalent to replacing
with
which we often don’t write
explicitly; ... eventually we can integrate over “momenta” and
For odd number of Qs there is always one f(t) left-over and the result is 0!
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obtain path integral (functional integral) for our free field theory:
path in the space of field configurations
change of integration variables:
Comments:
a shift by a constant
lagrangian is manifestly Lorentz invariant and all the symmetries of a
lagrangian are preserved by path integral
lagrangian seems to be more fundamental specification of a quantum
field theory
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to evaluate
we can closely follow the procedure we did for the
harmonic oscillator:
Fourier transformation:
But
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Thus we have:
For even number of
we pair up
in all possible ways:
we used inverse Fourier transformation to go back to position-functions
where
in general:
is the Feynman propagator, a Green’s function for the Klien-Gordon equation:
integral over zero’s component can be calculated explicitly by completing the contour
and using the residue theorem, the three momentum integral can be calculated in
terms of Bessel functions
Wick’s theorem
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Now we can calculate correlation functions:
we find:
For odd number of
there is always one J left-over and the result is 0!
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