QuantumPhysics.dvi

Next, we are interested in taking the size of the boxLvery large compared to all other scales in
the problem. The levels innxandnythen become very closely spaced and the sums overnxand
nymay be well-approximated by integrals, with the help of the following conversion process,

dnx=

Ldkx

2 π

dny=

Lky

2 π

(19.93)

so that

E(a)−E 0 =

L^2

4 π^2

∫+∞

−∞

dkx

∫+∞

−∞

dky

[

1

2

ωθ(ωc−ω) (19.94)

+

∑∞

nz=1

(

ω(2)θ(ωc−ω(2)) +ω(3)θ(ωc−ω(3))− 2 ω(1)θ(ωc−ω(1))

)]

(19.95)

Now, forω(1)andω(3), the frequency levels are also close together and the sum overnzmay also
be approximated by an integral,

∑∞

nz=1

ω(2)θ(ωc−ω(2)) =

L−a

π

∫∞

0

dkz

√

k^2 x+k^2 y+k^2 zθ(ωc−

√

kx^2 +k^2 y+k^2 z)

∑∞

nz=1

2 ω(1)θ(ωc−ω(1)) =

L

2 π

2

∫∞

0

dkz

√

k^2 x+ky^2 +kz^2 θ(ωc−

√

k^2 x+k^2 y+k^2 z)

Introducingk^2 =k^2 x+ky^2 , and in the last integral the continuous variablen=akz/π, we have

E(a)−E 0 =

L^2

2 π

∫∞

0

dk k

[

1

2

kθ(ωc−k) +

∑∞

n=1

√

k^2 +

π^2 n^2

a^2

θ(ω^2 c−k^2 −

π^2 n^2

a^2

)

−

∫∞

0

dn

√

k^2 +

π^2 n^2

a^2

θ(ωc^2 −k^2 −

π^2 n^2

a^2

)

]

Introducing the function

f(n)≡

∫∞

0

dk

2 π

k

√

k^2 +

π^2 n^2

a^2

θ(ω^2 c−k^2 −

π^2 n^2

a^2

) (19.96)

we have

E(a)−E 0

L^2

=

1

2

f(0) +

∑∞

n=1

f(n)−

∫∞

0

dnf(n) (19.97)

Now, there is a famous formula that relates a sum of the valuesof a function at integers to the
integral of this function,

∫∞

0

dnf(n) =

1

2

f(0) +

∑∞

n=1

f(n) +

∑∞

p=1

B 2 p

(2p)!

f(2p−1)(0) (19.98)