Quantum Mechanics for Mathematicians

(lily) #1
The Bargmann transform is then given by

(Bψ)(z) =

∫+∞

−∞

〈δz|q〉〈q|ψ〉dq

=


π

) (^14)
e−
z^2
2


∫+∞

−∞

e−

ω
2 q
2
e


2 ωzqψ(q)dq (23.14)

(hereψ(q) is the position space wavefunction) while the inverse Bargmann trans-
form is given by


(B−^1 φ)(q) =


C

〈q|δu〉〈δu|φ〉e−|u|

2
d^2 u

=


π

)^14

e−

ω 2 q^2


C

e−

u 22
e

√ 2 ωuq
φ(u)e−|u|

2
d^2 u

(hereφ(z) is the Bargmann-Fock wavefunction).
As a check of equation 23.14, consider the case of the lowest energy state in
the Schr ̈odinger representation, where| 0 〉has coordinate space representation


ψ(q) = (

ω
π

)

(^14)
e−
ωq^2
2
and
(Bψ)(z) =

π
) 14 (ω
π
) (^14)
e−
z 22


∫+∞

−∞

e−
ω 2 q^2
e


2 ωzqe−ωq (^22) dq


=


π

)^12 ∫+∞

−∞

e−

z 22
e−ωq

2
e


2 ωzqdq

=


π

)^12 ∫+∞

−∞

e
−ω

(
q−√z 2 ω

) 2
dq

=1

which is the expression for the state| 0 〉in the Bargmann-Fock representation.
For an alternate way to compute the harmonic oscillator propagator, the
kernel corresponding to applying the Bargmann transform, then the time evo-
lution operator, then the inverse Bargmann transform can be calculated. This
will give


〈qT|e−iHT|q 0 〉=


C

〈qT|δu〉〈δu|e−iTH|q 0 〉e−|u|

2
d^2 u

=e−i

ωT 2


C

〈qT|δu〉〈δue−iωT|q 0 〉e−|u|

2
d^2 u

from which 23.12 can be derived by a (difficult) manipulation of Gaussian inte-
grals.

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