Casimir Effect

[Dω]δ[∇ · A^ω], A^ω_m≡ Am+ ∂_mω (3.215) where Δ is the determinant which is independent of A_i, into the path integral, thereby making A purely transverse. Going through the same procedure as above, this will lead to a path integral of the form

Z[J ] =



DA⁰DA δ[∇ · A] e^{iS[A]+id4x JmAm} (3.216) when the field is coupled to a source. We have already seen in Sect. 3.1.5 that in the Coulomb gauge, the A⁰decouples from A in the action and can be integrated out separately. This will essentially add a phase containing the Coulomb energy of interaction which can be ignored. The rest of the integration is over the transverse degree of freedom which will lead to a term like

Z[J ]∝ exp 1

2



d⁴x d⁴y Jκ(x)G^κλ(x, y) Jλ(y) 

(3.217) where G^κλ is the transverse propagator coupling the transverse degrees of freedom of the source. This result ties up with the previous analysis done in the canonical formalism in Sect. 3.6.1 and relates the transverse propagator to the interaction between the currents.

3.6.3 Casimir Effect

It is obvious that the success of our program, for quantizing a field and in-terpreting the states in the Hilbert space in terms of particles, is closely tied to decomposing the field into an infinite number of harmonic oscillators.

The facts: (i) the energy spectrum of the harmonic oscillator is equally spaced and (ii) its potential is quadratic, are the features which allow us to introduce quantum states labeled by a set of integers with the energy of the state changing byωk when n_k → nk+ 1. But if we take these oscil-lators seriously, we get into trouble with their zero point energy (1/2)ωk

which adds up to infinity. The conventional wisdom — which we have been faithfully advocating — is to simply throw this away by introducing normal ordering of operators like the Hamiltonian.

There is however one effect (called the Casimir effect, which is observed in the lab) for which the most natural explanation is provided in terms of the zero point energy of the oscillators. This is probably one of the most intriguing results in field theory and is worth understanding.

Consider two large, plane, perfect conductors of area L²kept at a dis-tance a apart in otherwise empty space. This is like two capacitor plates with L a (which allows us to ignore the edge effects) but we have not put any charges on them. Experiments show that these two plates exert a force of attraction on each other which varies as a⁻⁴. You need to explain

63The vanishing boundary condition is expected to mimic what happens in the case of electromagnetic field on the surfaces of the perfect conductors.

64This can be quite shocking to any-one innocent of the practices in high energy physics. One might have ex-pected that the sum, if not infinite, should at least be decent enough to be a positive integer rather than a nega-tive fraction! Such a conclusion, com-ing from physicists, would be quite sus-pect; but it was first obtained by a very respectable mathematician, Euler. In fact, most of our discussion will revolve around attributing meaning to diver-gent expressions in a systematic man-ner.

why two perfect conductors kept in the vacuum should exert such a force on each other!

The simplest explanation is the following: In the absence of the plates we are in the vacuum state of the electromagnetic field (made of an infinite number of oscillators) with corresponding mode functions varying in space as exp(ik·x) with all possible values for k. When you introduce two plates, you have to impose the perfect conductor boundary condition for the elec-tromagnetic field at the location of the plate. If the plates are located at z = 0 and z = a, this necessarily limits the z-component of the wave vector k to be discrete with the allowed values being k_z = nπ/a with integral n. The form of the total zero point energy for the oscillators will now be different (though, of course, still divergent) from the original form of the total zero point energy. The difference between the zero point energies calculated with and without the plates will involve subtracting one infinity from other. There are several ways of giving meaning to such an opera-tion, thereby extracting a finite result for the difference. This difference turns out to be negative and scales as (−a⁻³); this leads to an attractive force because reducing a decreases this energy. Thus, whether zero point energies exist or not, their differences seem to exist and seem to be finite!

Given the conceptual importance of this result, we will provide a fairly detailed discussion of its derivation. To understand the key issues which are involved, we will first consider a toy model of a scalar field in the (1+1) dimension and then study electromagnetic fields in (3+1).

Consider the quantum theory for a massless scalar field in the (1+1) dimension when the field is constrained to vanish⁶³ at x = 0 and x = L.

The allowed wave vectors for the modes are now given by k_n = (nπ/L).

The total zero-point energy of the vacuum in the presence of the plates is therefore given by

To give meaning to this expression, we need to evaluate the sum over all positive integers. It turns out that this sum⁶⁴ is equal to −(1/12) so that E = −(π/24L). If one imagines two point particles at x = 0 and x = L (which are the analogs of two perfect conductors in the case of electromagnetism), this energy will lead to an attractive force (∂E/∂L) = (π/24L²) between the particles. Let us now ask how we can obtain this result.

There are rigorous ways of defining sums of certain divergent series which could be used for this purpose. But before discussing this approach, let us consider a more physical way of approaching the question. We first note that the zero-point energy in the absence of plates is given by the expression

where we have used the substitution k = (πn/L) with n being a continuous variable. What we are really interested in is the energy difference given by

ΔE = π

To evaluate this difference, let us introduce a high-n cut-off by a function e^−λn into these expressions and take the limit of λ→ 0 at the end of the calculation. That is, we think of the energy difference as given by⁶⁵

ΔE = π

For finite λ, the energy difference can be evaluated as

ΔE =− π

We are interested in this expression near λ = 0 which can be computed by a Taylor series expansion. We see that:

 1

It follows, quite remarkably, that:

ΔE =− π

which is the same result we would have obtained by treating the sum over all positive integers as (−1/12) as we mentioned before!

The above procedure of introducing a cut-off function e^−λnand taking the limit of λ → 0 at the end of the calculation has one serious draw-back. The (−1/12) we obtained seems to have come from the Taylor series expansion of e^−λand hence it is not clear whether the same result will hold if we use some other cut-off function to regularize the sum and the integral.

To settle this question, we need to evaluate expressions like

Δf≡

where f (x) is a function which dies down rapidly for large values of the argument. It can be shown that (see Mathematical Supplement Sect. 3.8):

Δf =

where D = d/dx and B_k (called Bernoulli numbers) are defined through the series expansion

Consider now an arbitrary cut-off function F (λx) instead of exp(−λx) and let the modified function be f (x) = xF (λx) with the condition that F (0) = 1. Using Eq. (3.226) we can compute Δf with this cut-off function and take

65From a physical point of view, we could think of a factor like e^−λn as arising due to the finite conductivity of the metallic plates in the case of (1+3) electrodynamics which makes conduc-tors less than perfect at sufficiently high frequencies.

the limit λ→ 0. Equation (3.226) requires us to compute the derivatives of f (x) at the origin and take the limit λ→ 0. One can easily see that the first derivative evaluated at the origin is given by

[xF^(λx)λ + F (λx)]

x=0

= 1 (3.228)

and the second derivative vanishes

F^(λx)λ + xλ²F^(λx) + λF^(x) 

x=0→ 0 (3.229) when λ → 0. All further derivatives will introduce extra factors of λ and hence will vanish. We therefore find that

Δf =−B₂

2!f^(0) =−1

12 (3.230)

where we have used the fact that B2= (1/6). This shows that, for a wide class of cut-off functions, we recover the same result. Knowing this fact, we could have even taken f (n) = n in Eq. (3.226) and just retained the first derivative term (which is the only term that contributes) on the right hand side of Eq. (3.226).

Having done this warm-up exercise, let us turn our attention to the electromagnetic field in (1+3). We consider a region between two parallel conducting plates, each of area L× L, separated by a distance a. We will assume that L  a and will be interested in computing the force per unit area of the conducting plates by differentiating the corresponding expression for zero-point energy with respect to a. As in the previous case, we want to compute the zero-point energy in the presence of the plates and in their absence in the 3-dimensional volume L²a and compute their difference.

The energy contained in this region in the absence of plates is given by the integral

E0 = 2

 L²d²k (2π)²

 a dk3

2π 1

2 (

k²₁+ k²₂+ k₃² 

=

 L²d²k (2π)²

 _∞

0

dn



k₁²+ k₂²+

nπ a

2

(3.231) where the overall factor 2 in front in the first line takes into account two polarizations and we have set k₃ = (nπ/a) with a continuum variable n to obtain the second line. Writing the transverse component of the wave vector as k²_⊥ ≡ k1²+ k₂² ≡ (π/a)²μ, we can re-write this expression as an integral over μ and n in the form

E₀=π²L² 4a³

 _∞

0

dμ

 _∞

0

dn

μ + n²1/2

(3.232) Both the integrals, of course, are divergent — as to be expected.

Let us next consider the situation in the presence of the conducting plates. This will require replacing the integral over n by a summation over n when n= 0. When n = 0, the corresponding result has to be multiplied by a factor (1/2) because only one polarization state contributes. This comes about from the nature of the mode functions for the vector potential A in the presence of the plates. (See also Problem 4.) We will work in

the gauge with A⁰= 0,∇·A = 0 and decompose the vector potential into parallel and normal components A = A_+ A_⊥ where A_⊥ is along the z-axis while A_ is parallel to the plates in the x− y plane. If you impose the boundary condition, that the parallel component of the electric field and the normal component of the magnetic field should vanish on the plates, we get the conditions⁶⁶

∂A_ It is straightforward to determine the allowed modes in this case and you will find that there are two polarizations, each contributing the energy:

ωk,n= when n= 0, and one polarization contributing when n = 0. Therefore, the corresponding expression in the presence of the plates is given by

E₁

The difference between the energies per unit area is essentially determined by the combination

We will now use the result

∞

which can be easily obtained from Eq. (3.226) (see Mathematical Supple-ment Sect. 3.8). We now have to regularize the expression in Eq. (3.236) by multiplying f (n) by a cut-off function F (λn) and work out the derivatives at the origin and take the limit λ → 0. We will, however, cheat at this stage and will work with f (n) in Eq. (3.237) itself, because it leads to the same result. In this case, it is easy to see that

f^(n) with all further derivatives vanishing. Hence we get the final result to be

ΔE where we have used the fact that B₄=−(1/30). The corresponding force of attraction (per unit area of the plates) is given by

f =−∂E

∂a =− π²

240 a⁴ (3.241)

66Notation: The symbols and ⊥ do not mean curl-free and div-free parts in this context.

Exercise 3.21: Do this rigorously by introducing at cut-off function F (λx)

— which will ensure convergence — and prove that you get the same re-sult, independent of the choice of F .

68In the language of dimensional reg-ularization, a pet trick in high energy physics, this means that several power law divergences can be “regularized” to vanish.

69In the previous discussions, we in-terpreted the difference between two divergent expressions as finite. Using dimensional regularization, we are giv-ing meangiv-ing to the divergent E₀/L²in Eq. (3.235) directly without any sub-traction.

Exercise 3.22: For a massless scalar field in the (1+1) dimension, find the vacuum functionals with and without the vanishing boundary conditions at x = 0, L.

It is this force (which works out to 10⁻⁸ N for a = 1μm, L = 1 cm) that has been observed in the lab.⁶⁷

A more formal mathematical procedure for obtaining the Casimir effect in the case of the (1+1) scalar field or (1+3) electromagnetic field involves giving meaning to the zeta function

ζ(s) =

∞ n=1

1

n^s (3.242)

for negative values of s. We have already seen that, in the case of the scalar field in (1+1) dimension, we needed to evaluate the sum of all integers which is essentially ζ(−1). To see the corresponding result in the case of the electromagnetic field, consider the integral

 _∞

0

dμ (μ + n²)^−α= 1

(α− 1)n^−2(α−1) (3.243) which is well defined for sufficiently large α. We can therefore write

 _∞ Putting α =−1/2 in this relation, it can be shown that:

 _∞

This means we need not worry⁶⁸about the second integral within the square bracket in Eq. (3.235).

It follows that the quantity we needed to evaluate in Eq. (3.235) is given by

It is actually possible to define ζ(s) for negative integral values of s by analytically continuing a suitable integral representation of ζ(s) in the complex plane to these values.⁶⁹ Such an analysis actually shows that ζ(1− 2k) = −(B2k/2k) using which we can obtain the same results as above.

Finally we mention that the Casimir effect has a clearer intuitive expla-nation in the Schrodinger picture. The ground state wave functional which we computed for a scalar field in Eq. (3.128) explicitly depended on using the running modes exp(ik· x) to go back and forth between the field config-uration φ(x) and the oscillator variables q_k. If you introduce metal plates into the vacuum, thereby modifying the boundary conditions, the vacuum functional will change. In other words, the ground state wave functional in the presence of the metal plates is actually quite different from that in the absence of the plates. (This should be obvious from the fact that Ψ should now vanish for any field configuration which does not obey the boundary conditions.) Since the ground states are different, it should be no surprise that the energies are different too.

67So does it mean that zero point en-ergies are non-zero and normal order-ing is a nonsensical procedure? While the above derivation provides the most natural explanation for Casimir effect, it is possible to obtain it by comput-ing the direct Van der Waals like forces between the conductors. The fact that conductivity has a frequency de-pendence provides a natural high fre-quency cut-off. If you accept this al-ternative derivation (originally due to Liftshitz), then may be what you need to really understand is why zero point energies with dubious subtractions of infinities also give the same result. No-body knows for sure.

In document [Thanu Padmanabhan] Quantum Field Theory(BookZZ.org) (Page 132-138)