Relativistic Oscillator
المؤلف:
Sidney B. Cahn, Gerald D. Mahan And Boris E. Nadgorny
المصدر:
A GUIDE TO PHYSICS PROBLEMS
الجزء والصفحة:
part 2 , p 68
21-8-2016
1579
Relativistic Oscillator
Consider a spineless particle in a one-dimensional harmonic oscillator potential:
a) Calculate leading relativistic corrections to the ground state to first order in perturbation theory.
b) Consider an anharmonic classical oscillator with
For what values of α will the leading corrections be the same as in (a)?
SOLUTION
a) The classical Hamiltonian is given by H0 = p2/2m + V(x), whereas the relativistic Hamiltonian may be expanded as follows:
(1)
The perturbation to the classical Hamiltonian is therefore
First solution: For the nonrelativistic quantum harmonic oscillator, we have
(2)
where x, p are operators. Defining new operators Q, P,
and noting the commutation relations
we may rewrite (2) as
(3)
Introducing the standard creation and annihilation operators:
(4)
we find that
Using these results, we may express the first-order energy shift ∆0 as
(5)
where
The expansion of ⟨0|P4|0⟩ is simplified by the fact that a|0⟩ = 0, so
(6)
Finally, we obtain
Second solution: Instead of using operator algebra, we can find a wave function a(p) in the momentum representation, where
(6)
The Hamiltonian then is
(7)
The Schrodinger equation for a(p) becomes
(8)
This equation has exactly the same form as the standard oscillator Schrodinger equation:
(9)
We then obtain for the momentum probability distribution for the ground state:
(10)
Therefore
(11)
where A ≡ -1/8m3c2. Using the old “differentiate with respect to an innocent parameter method” of simplifying an integral, we may rewrite ∆0 as
(12)
where we substituted (10) into (11) and let ζ ≡ 1/mωh. Finally,
(13)
as found in the first solution.
b) The first-order energy shift from ax3 would be zero (no diagonal elements in the Q3 matrix). The leading correction would be the second-order shift as defined by the formula
where ∑' means sum over m ≠ n = 0. From (3) and (4), we have
So,
As for any second-order correction to the ground state, it is negative. To make this expression equal to the one in part (a), we require that
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