The Casimir Effect

Overview

The Casimir effect is the attraction between two neutral, parallel conducting plates caused by the change in the quantum vacuum modes of the electromagnetic field. The plates are electrically neutral, so the force is not an ordinary electrostatic force. It appears because conducting boundaries restrict which field modes can exist between the plates. The zero-point energy of those allowed modes depends on the plate separation.

The main result for two ideal, perfectly conducting parallel plates of area $A$, separated by a distance $L$, is

$$E_0(L)=E_0(\infty)-\frac{\hbar c\pi^2 A}{720L^3}+O(L^{-4}),$$

and the corresponding pressure is

$$P=-\frac{1}{A}\frac{\partial E_0}{\partial L} =-\frac{\pi^2}{240}\frac{\hbar c}{L^4}.$$

The negative sign means the plates are pulled together.

The lesson is not only about one force formula. It is also a clean example of how quantum field theory handles divergent vacuum energies: isolate the part that changes with a physical parameter, regularize the calculation, and then extract the finite universal piece.

Learning objectives

By the end, you should be able to:

• explain why the vacuum of a quantum field is not classical emptiness;
• describe why conducting plates modify the electromagnetic vacuum modes;
• write the allowed standing-wave frequencies between and outside the plates;
• explain why only the $L$-dependent part of the vacuum energy can produce a force;
• understand why a high-frequency cutoff is physically justified;
• use the Euler--Maclaurin formula to approximate a sum by an integral plus correction terms;
• derive the leading Casimir energy and pressure for ideal parallel conducting plates;
• state where the idealized formula breaks down for real materials and non-planar geometries.

Prerequisites

You should be comfortable with:

• harmonic oscillator zero-point energy;
• normal modes of a wave equation;
• basic canonical quantization of fields;
• standing waves and boundary conditions;
• asymptotic expansions;
• elementary multivariable integrals.

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1. Vacuum states in the field representation

In ordinary quantum mechanics, the position basis is defined by

$$\hat q\,|q\rangle=q|q\rangle.$$

A field theory has an analogous field basis. For a scalar field, field-operator eigenstates satisfy

$$\hat\Phi_{\vec k}\,|\varphi_0\rangle = \Phi_{\vec k}\,|\varphi_0\rangle.$$

A quantum state can therefore be represented by a wave functional,

$$\Psi[\varphi_0]=\langle \varphi_0|\Psi\rangle.$$

This is the field-theory analogue of a position-space wavefunction $\psi(q)=\langle q|\psi\rangle$.

In this representation, the momentum conjugate to the field acts as a functional derivative:

$$\langle \varphi_0|\hat\Pi_{\vec k}|\Psi\rangle = -i\hbar\frac{\delta}{\delta\Phi_{\vec k}^{*}}\Psi[\varphi_0].$$

This mirrors ordinary quantum mechanics:

$$\hat p\rightarrow -i\hbar\frac{\partial}{\partial q}.$$

For a free Klein--Gordon field, the vacuum is defined by the annihilation condition

$$\hat a_{\vec k}|0\rangle=0.$$

Using the field and momentum operators, this condition can be written schematically as

$$\left(\omega_k\hat\Phi_{\vec k}+i\hat\Pi_{\vec k}\right)|0\rangle=0 \qquad \forall\,\vec k.$$

Solving this functional differential equation gives a Gaussian vacuum wave functional:

$$\Psi_0[\Phi] = Z\exp\left[-\frac{1}{\hbar}\int d^Dk\,|\Phi_k|^2\omega_k\right].$$

The exact normalization $Z$ is not important here. The important point is conceptual: the vacuum is a wave functional with nonzero width. It is centered around the zero field configuration, but it is not a delta-functional. The field still fluctuates around zero.

That spread is what people loosely call vacuum fluctuations.

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2. Vacuum fluctuations as constrained waves

The vacuum is the ground state of the field, not classical nothing. Each mode of a free field behaves like a harmonic oscillator, and a harmonic oscillator has ground-state energy

$$E_{\text{mode}}=\frac{1}{2}\hbar\omega.$$

For a field, there are many modes, so formally

$$E_0=\sum_{\text{all modes}}\frac{1}{2}\hbar\omega.$$

This raw expression usually diverges. That is not yet a physical prediction. A force appears only if the vacuum energy changes when a physical parameter changes. In the Casimir problem, the physical parameter is the plate separation $L$.

A useful intuition is this: waves outside two plates can have different allowed patterns from waves between them. If the allowed mode structure changes with separation, the vacuum energy changes with separation. The force follows from the derivative of that energy.

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3. Physical setup

Consider two large, parallel, rectangular conducting plates separated by distance $L$. Let the plate dimensions be $L_y$ and $L_z$, so that

$$A=L_yL_z.$$

The plates are electrically neutral. The electromagnetic field exists both between the plates and outside them.

For an ideal conductor, the tangential electric field vanishes at the surface:

$$\mathbf E_{\parallel}=0.$$

This boundary condition forces the allowed electromagnetic modes to become standing waves in the direction normal to the plates.

For the purpose of the mode-counting derivation, the electromagnetic field can be treated as two independent massless scalar-like sectors, one for each photon polarization. This is why a factor of two appears in the vacuum energy.

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4. Making the system finite

To avoid writing continuum expressions too early, enclose the entire system inside a much larger conducting box of dimensions

$$L_x\times L_y\times L_z.$$

The inner plate separation $L$ is variable. The other lengths are fixed and large. The two outside regions each have width

$$\tilde L=\frac{L_x-L}{2}.$$

At the end, one can take $L_x,L_y,L_z\rightarrow\infty$ in the appropriate way. The point of the box is to make the spectrum discrete while the calculation is being defined.

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5. Modes between the plates

For the region between the plates, a scalar-like standing-wave mode can be written as

$$\varphi(x,y,z) = \sum_{k\ell m}C_{k\ell m} \sin\left(\frac{k\pi x}{L}\right) \sin\left(\frac{\ell\pi y}{L_y}\right) \sin\left(\frac{m\pi z}{L_z}\right).$$

The corresponding angular frequencies are

$$\omega_{k\ell m} = c\pi \sqrt{ \frac{k^2}{L^2} + \frac{\ell^2}{L_y^2} + \frac{m^2}{L_z^2}} \qquad k,\ell,m=1,2,3,\dots$$

The factor $k\pi/L$ is the important one: it is the discretization in the direction normal to the plates. This is where the separation $L$ enters.

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6. Modes outside the plates

For each outside region, the width is

$$\tilde L=\frac{L_x-L}{2}.$$

The corresponding outside-mode frequencies are

$$\tilde\omega_{k\ell m} = c\pi \sqrt{ \frac{k^2}{\tilde L^2} + \frac{\ell^2}{L_y^2} + \frac{m^2}{L_z^2}}.$$

There are two outside regions, so their contribution is doubled.

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7. Total zero-point energy

Every mode contributes $\hbar\omega/2$. There are two polarizations. Therefore the total vacuum energy can be written as

$$E_0 = \sum_{\text{all modes}}\frac{\hbar\omega}{2}.$$

Including two polarizations, the inside region, and the two outside regions,

$$E_0 = 2\times\frac{\hbar}{2} \sum_{k=1}^{\infty}\sum_{\ell=1}^{\infty}\sum_{m=1}^{\infty} \left( \omega_{k\ell m}+2\tilde\omega_{k\ell m} \right).$$

Equivalently,

$$E_0 = \hbar \sum_{k\ell m} \left( \omega_{k\ell m}+2\tilde\omega_{k\ell m} \right).$$

The question is not whether this sum is infinite. It is. The question is:

$$\text{How does }E_0\text{ depend on }L?$$

If a term is independent of $L$, it produces no force when differentiated with respect to $L$. Even an infinite $L$-independent term is physically irrelevant for this force.

The pressure on the plates is obtained from

$$P=-\frac{1}{A}\frac{\partial E_0}{\partial L}.$$

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8. Why regularization is necessary

The ideal-conductor boundary condition cannot be correct for arbitrarily high frequencies. A real metal does not reflect arbitrarily short-wavelength radiation perfectly. At sufficiently high frequency, the mode no longer “sees” the plates as ideal reflecting boundaries.

So we introduce a smooth cutoff function $\Gamma(\omega)$ and replace

$$\sum_{k\ell m}\omega_{k\ell m} \longrightarrow \sum_{k\ell m}\omega_{k\ell m}\,\Gamma(\omega_{k\ell m}).$$

The cutoff satisfies

$$\lim_{\omega\to0}\Gamma(\omega)=1,$$

and

$$\lim_{\omega\to\infty}\Gamma(\omega)=0,$$

fast enough to make the sums converge.

The details of $\Gamma$ should not affect the final leading Casimir force. If the final answer depended strongly on the arbitrary cutoff shape, it would not be a universal physical prediction.

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9. Reducing the transverse sums to integrals

Define $\Xi(L)$ as the regularized dimensionless mode sum for a region of width $L$:

$$\Xi(L) := \lim_{L_y,L_z\to\infty} \frac{1}{L_yL_z} \sum_{k\ell m} \sqrt{ \frac{k^2}{L^2} + \frac{\ell^2}{L_y^2} + \frac{m^2}{L_z^2}} \, \Gamma\left( \sqrt{ \frac{k^2}{L^2} + \frac{\ell^2}{L_y^2} + \frac{m^2}{L_z^2}} \right).$$

In the limit of large transverse dimensions, the sums over $\ell$ and $m$ become integrals:

$$\Xi(L) = \sum_k \int_0^{\infty}d\lambda \int_0^{\infty}d\mu\, \sqrt{\frac{k^2}{L^2}+\lambda^2+\mu^2} \, \Gamma\left( \sqrt{\frac{k^2}{L^2}+\lambda^2+\mu^2} \right).$$

Use polar coordinates in the $(\lambda,\mu)$-plane:

$$\lambda=K\cos\theta, \qquad \mu=K\sin\theta.$$

The first quadrant gives $0\leq\theta\leq\pi/2$, so

$$\int_0^{\infty}d\lambda\int_0^{\infty}d\mu = \int_0^{\pi/2}d\theta\int_0^{\infty}dK\,K.$$

Therefore

$$\Xi(L) = \sum_{k=1}^{\infty} \int_0^{\pi/2}d\theta \int_0^{\infty}dK\,K \sqrt{\frac{k^2}{L^2}+K^2} \, \Gamma\left(\sqrt{\frac{k^2}{L^2}+K^2}\right).$$

Performing the angular integral gives

$$\Xi(L) = \frac{\pi}{2} \sum_{k=1}^{\infty} \int_0^{\infty}dK\,K \sqrt{\frac{k^2}{L^2}+K^2} \, \Gamma\left(\sqrt{\frac{k^2}{L^2}+K^2}\right).$$

Now set

$$\rho=K^2.$$

Since $d\rho=2K\,dK$, we get

$$\Xi(L) = \frac{\pi}{4} \sum_{k=1}^{\infty} \int_0^{\infty}d\rho\, \sqrt{\frac{k^2}{L^2}+\rho} \, \Gamma\left(\sqrt{\frac{k^2}{L^2}+\rho}\right).$$

If $L\to\infty$, the remaining sum over $k$ could also become an integral. But $L$ is finite. The finite discreteness of $k/L$ is exactly what produces the Casimir correction.

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10. The Euler--Maclaurin tool

The technical problem is now to approximate a sum of the form

$$\sum_{m=1}^{\infty}f\left(\frac{m}{\Lambda}\right)$$

by an integral plus correction terms.

For sufficiently well-behaved $f$, with derivatives vanishing at infinity,

$$\sum_{m=1}^{\infty}f\left(\frac{m}{\Lambda}\right) = \Lambda\int_0^{\infty}dx\,f(x) - \sum_{m=0}^{\infty} \frac{C_{m+1}}{\Lambda^m}f^{(m)}(0).$$

The first coefficients are

$$C_0=1, \qquad C_1=\frac{1}{2}, \qquad C_2=\frac{1}{12}, \qquad C_3=0, \qquad C_4=-\frac{1}{720}.$$

Keeping terms up to $C_4$, the expansion reads

$$\sum_{m=1}^{\infty}f\left(\frac{m}{\Lambda}\right) = \Lambda\int_0^{\infty}dx\,f(x) - \frac{1}{2}f(0) - \frac{1}{12\Lambda}f'(0) + \frac{1}{720\Lambda^3}f^{(3)}(0) + \cdots.$$

The signs follow from the convention used above. What matters physically is the universal $1/L^3$ correction obtained from $C_4f^{(3)}(0)$.

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11. Deriving the needed Euler--Maclaurin coefficients

For completeness, here is a compact derivation of the coefficients used above.

Start from the target expansion

$$P: \qquad \sum_{n=1}^{N}f(n\Delta) = \sum_{m=0}^{\infty} \Delta^{m-1}C_m \int_0^{N\Delta}dx\,f^{(m)}(x).$$

Introduce a sequence of propositions

$$P_M: \qquad \sum_{n=1}^{N}f(n\Delta) = \sum_{m=0}^{M} \Delta^{m-1}C_m \int_0^{N\Delta}dx\,f^{(m)}(x) + \sum_{m=M+1}^{\infty} D_{mM}\Delta^m \sum_{n=1}^{N}f^{(m)}(n\Delta).$$

For $M=0$, Taylor expand $f(x)$ around the upper endpoint $n\Delta$:

$$\sum_{n=1}^{N}f(n\Delta) = \Delta^{-1} \sum_{n=1}^{N} \int_{(n-1)\Delta}^{n\Delta}dx\,f(n\Delta),$$ $$= \Delta^{-1} \sum_{n=1}^{N} \int_{(n-1)\Delta}^{n\Delta}dx \left[ f(x)-\sum_{m=1}^{\infty} \frac{f^{(m)}(n\Delta)(x-n\Delta)^m}{m!} \right],$$

so

$$\sum_{n=1}^{N}f(n\Delta) = \Delta^{-1} \int_0^{N\Delta}dx\,f(x) - \sum_{m=1}^{\infty} \frac{(-\Delta)^m}{(m+1)!} \sum_{n=1}^{N}f^{(m)}(n\Delta).$$

Therefore

$$C_0=1, \qquad D_{m0}=\frac{(-1)^{m+1}}{(m+1)!}.$$

Iterating the same argument gives

$$D_{m0}=\frac{(-1)^{m+1}}{(m+1)!},$$ $$D_{m1}=D_{m0}+D_{10}\frac{(-1)^m}{m!} = \frac{(-1)^{m+1}}{(m+1)!} + \frac{(-1)^m}{2m!} = \frac{(-1)^m(m-1)}{2(m+1)!},$$ $$D_{m2} =D_{m1}+D_{21}\frac{(-1)^{m-1}}{(m-1)!} = \frac{(-1)^m(m-1)}{2(m+1)!} + \frac{(-1)^{m-1}}{12(m-1)!} = -\frac{(-1)^m}{12(m+1)!}(m-2)(m-3),$$

and

$$D_{m3}=D_{m2}+D_{32}\frac{(-1)^{m-2}}{(m-2)!}=D_{m2}.$$

The needed coefficients are therefore

$$C_0=1,$$ $$C_1=D_{10}=\frac{1}{2},$$ $$C_2=D_{21}=\frac{1}{12},$$ $$C_3=D_{32}=0,$$ $$C_4=D_{43}=-\frac{1}{720}.$$

These are the coefficients that generate the finite Casimir correction.

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12. Defining the function $f$

From the expression for $\Xi(L)$, define

$$\Xi(L)=\frac{\pi}{4}\sum_{k=1}^{\infty}f\left(\frac{k}{L}\right),$$

where

$$f(x) = \int_0^{\infty}d\rho\,\sqrt{x^2+\rho}\,\Gamma\left(\sqrt{x^2+\rho}\right).$$

Now set

$$u=\sqrt{x^2+\rho}.$$

Then

$$\rho=u^2-x^2, \qquad d\rho=2u\,du.$$

When $\rho=0$, $u=x$. When $\rho\to\infty$, $u\to\infty$. Hence

$$f(x) = \int_x^{\infty}2u^2\Gamma(u)\,du = 2\int_x^{\infty}du\,u^2\Gamma(u).$$

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13. Derivatives of $f$ at zero

From

$$f(x)=2\int_x^{\infty}du\,u^2\Gamma(u),$$

we get

$$f(0)=2\int_0^{\infty}du\,u^2\Gamma(u).$$

Differentiate once:

$$f'(x)=-2x^2\Gamma(x),$$

so

$$f'(0)=-2[x^2\Gamma(x)]_{x=0}=0.$$

Differentiate again:

$$f''(x)=-4x\Gamma(x)-2x^2\Gamma'(x),$$

so

$$f''(0)=\left[-4x\Gamma(x)-2x^2\Gamma'(x)\right]_{x=0}=0.$$

Differentiate a third time:

$$f^{(3)}(x) = -4\Gamma(x)-8x\Gamma'(x)-2x^2\Gamma''(x).$$

At zero, using $\Gamma(0)=1$,

$$f^{(3)}(0) = \left[-4\Gamma(x)-8x\Gamma'(x)-2x^2\Gamma''(x)\right]_{x=0} = -4.$$

That is the only derivative needed for the leading $1/L^3$ correction.

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14. Expanding $\Xi(L)$

Apply Euler--Maclaurin with $\Lambda=L$:

$$\Xi(L) = \frac{\pi}{4} \sum_{k=1}^{\infty}f\left(\frac{k}{L}\right).$$

Using the coefficients above,

$$\Xi(L) = \frac{\pi}{2}L \int_0^{\infty}dx\int_x^{\infty}dk\,k^2\Gamma(k) - \frac{\pi}{2}C_1 \int_0^{\infty}dk\,k^2\Gamma(k) + \frac{\pi C_4}{L^3} + O(L^{-4}).$$

The first double integral can be simplified by exchanging the order of integration:

$$\int_0^{\infty}dx\int_x^{\infty}dk\,k^2\Gamma(k) = \int_0^{\infty}dk\,k^2\Gamma(k) \int_0^k dx.$$

Since

$$\int_0^k dx=k,$$

we obtain

$$\int_0^{\infty}dx\int_x^{\infty}dk\,k^2\Gamma(k) = \int_0^{\infty}dk\,k^3\Gamma(k).$$

Therefore

$$\Xi(L) = \frac{\pi}{2}L \int_0^{\infty}dk\,k^3\Gamma(k) - \frac{\pi}{2}C_1 \int_0^{\infty}dk\,k^2\Gamma(k) + \frac{\pi C_4}{L^3} + O(L^{-4}).$$

Using

$$C_1=\frac{1}{2}, \qquad C_4=-\frac{1}{720},$$

this becomes

$$\Xi(L) = \frac{\pi}{2}L \int_0^{\infty}dk\,k^3\Gamma(k) - \frac{\pi}{4} \int_0^{\infty}dk\,k^2\Gamma(k) - \frac{\pi}{720L^3} + O(L^{-4}).$$

The first two terms depend on the cutoff. The $1/L^3$ term is universal.

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15. Total energy

The total regularized vacuum energy is

$$E_0 = \hbar c\pi L_yL_z \left[ \Xi(L)+2\Xi\left(\frac{L_x-L}{2}\right) \right].$$

Substituting the large-width expansion gives

$$E_0 = \frac{\hbar c\pi^2}{2}L_yL_z \int_0^{\infty}dk\,k^3\Gamma(k) \left[ L+2\frac{L_x-L}{2}\right] - \frac{3\hbar c\pi^2}{4}L_yL_z \int_0^{\infty}dk\,k^2\Gamma(k)$$ $$- \frac{\hbar c\pi^2L_yL_z}{720} \left[ L^{-3}+2\left(\frac{L_x-L}{2}\right)^{-3}\right] + \cdots.$$

The first bracket simplifies as

$$L+2\frac{L_x-L}{2}=L_x.$$

So the leading bulk term is independent of $L$:

$$\frac{\hbar c\pi^2}{2}L_yL_zL_x \int_0^{\infty}dk\,k^3\Gamma(k).$$

The next cutoff-dependent surface-like term is also independent of $L$:

$$- \frac{3\hbar c\pi^2}{4}L_yL_z \int_0^{\infty}dk\,k^2\Gamma(k).$$

Now take the outside box to be very large:

$$\frac{L_x-L}{2}\rightarrow\infty.$$

Then

$$2\left(\frac{L_x-L}{2}\right)^{-3}\rightarrow0.$$

The only leading $L$-dependent term left is

$$- \frac{\hbar c\pi^2L_yL_z}{720L^3}.$$

Since $A=L_yL_z$, the Casimir energy is

$$E_0(L) = E_0(\infty) - \frac{\hbar c\pi^2A}{720L^3} + O(L^{-4}).$$

This is the central energy result.

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16. Pressure between the plates

The pressure is force per area:

$$P=-\frac{1}{A}\frac{\partial E_0}{\partial L}.$$

Using

$$E_0(L)=E_0(\infty)-\frac{\hbar c\pi^2A}{720L^3},$$

we compute

\frac{\partial E_0}{\partial L} = -rac{\hbar c\pi^2A}{720}\frac{\partial}{\partial L}L^{-3}.

Since

$$\frac{\partial}{\partial L}L^{-3}=-3L^{-4},$$

we get

$$\frac{\partial E_0}{\partial L} = \frac{3\hbar c\pi^2A}{720L^4} = \frac{\hbar c\pi^2A}{240L^4}.$$

Therefore

$$P = -\frac{1}{A}\frac{\partial E_0}{\partial L} = -\frac{\pi^2}{240}\frac{\hbar c}{L^4}.$$

So

$$\boxed{P=-\frac{\pi^2\hbar c}{240L^4}}.$$

The pressure is negative, so the plates attract.

A useful numerical estimate is

$$P\approx-1.3\times10^{-3} \left(\frac{L}{\mu\mathrm m}\right)^{-4}\mathrm{Pa}.$$

At roughly

$$L\sim10\,\mathrm{nm},$$

this becomes comparable to atmospheric pressure in order of magnitude.

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17. What actually produced the force?

The force comes from the finite difference between a discrete sum and the corresponding continuum integral:

$$\sum_n f\left(\frac{n}{L}\right) - L\int_0^{\infty}dx\,f(x).$$

The integral term represents the continuum approximation. It contributes to large background terms, but these do not produce the plate force after the outside regions are included.

The nonzero finite correction is the $1/L^3$ term from Euler--Maclaurin:

\Delta E_0(L) = -rac{\hbar c\pi^2A}{720L^3}.

Differentiation turns that into the $1/L^4$ pressure:

$$P\propto -L^{-4}.$$

That is why the Casimir effect becomes important at very small separations.

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18. Why the cutoff disappears from the leading force

The cutoff-dependent terms are

$$\int_0^{\infty}dk\,k^3\Gamma(k)$$

and

$$\int_0^{\infty}dk\,k^2\Gamma(k).$$

These terms depend on the microscopic high-frequency behavior of the material model. However, in the parallel-plate calculation they combine into terms that are independent of $L$ after the outside regions are included.

The leading $L$-dependent term is instead controlled by

$$C_4=-\frac{1}{720}$$

and

$$f^{(3)}(0)=-4.$$

Those values only use the low-frequency condition

$$\Gamma(0)=1.$$

Therefore the leading ideal Casimir pressure does not depend on the detailed shape of the cutoff.

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19. Real materials

The ideal derivation assumes perfectly conducting, perfectly flat, infinitely large plates. Real plates are different.

Important microscopic scales include:

• the skin depth of the metal;
• the lattice spacing, often around

$$10^{-10}\,\mathrm m;$$

• material resonance frequencies, each associated with a length scale of order

$$\frac{c}{\omega_{\mathrm{res}}}.$$

For separations much larger than these microscopic scales, the ideal result is usually a good leading approximation. At very small separations, or for precision comparison with experiment, finite conductivity, temperature, roughness, and geometry corrections matter.

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20. Geometry matters

The parallel-plate formula is special. Casimir forces are sensitive to boundary shape. A simple universal formula does not exist for arbitrary surfaces.

For ideal parallel plates:

$$P<0,$$

so the force is attractive.

For other geometries, even the sign can change. For example, the stress associated with a hollow conducting spherical shell is not simply the same as the plate result with a different distance inserted. Geometry changes the spectrum, and the spectrum controls the vacuum energy.

The rough scaling rule for many small-gap configurations is:

$$\text{force density}\sim \frac{\hbar c}{d^4},$$

where $d$ is the relevant separation scale. But the coefficient and even the sign depend on the full geometry and material response.

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21. Common mistakes

Mistake 1: Treating the vacuum as classical nothing

The vacuum is the lowest-energy quantum state of the field. It has nonzero mode fluctuations and zero-point energy.

Mistake 2: Thinking the infinite vacuum energy itself is directly measured

The measurable quantity here is not the absolute value of the divergent vacuum energy. The measurable quantity is the change in vacuum energy with $L$.

Mistake 3: Ignoring outside modes

Only including the modes between the plates gives a misleading picture. The outside regions are needed to identify which divergent terms are independent of $L$.

Mistake 4: Forgetting the two polarizations

The electromagnetic field has two physical polarization states. This produces the factor of two in the mode sum.

Mistake 5: Assuming ideal metal behavior at all frequencies

Real metals stop behaving like perfect reflectors at sufficiently high frequencies. This is why a cutoff is physically sensible.

Mistake 6: Believing the force is always attractive

Attraction is the result for ideal parallel conducting plates. Other geometries can behave differently.

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22. Worked example: differentiating the energy

Start with

$$E_0(L)=E_0(\infty)-\frac{\hbar c\pi^2A}{720L^3}.$$

Differentiate:

$$\frac{dE_0}{dL} = -\frac{\hbar c\pi^2A}{720}\frac{d}{dL}L^{-3}.$$

Since

$$\frac{d}{dL}L^{-3}=-3L^{-4},$$

we get

$$\frac{dE_0}{dL} = \frac{\hbar c\pi^2A}{240L^4}.$$

Then

$$P=-\frac{1}{A}\frac{dE_0}{dL} =-\frac{\pi^2}{240}\frac{\hbar c}{L^4}.$$

The sign is negative, so the force pushes the plates toward smaller separation.

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23. Short summary

The Casimir effect arises because conducting plates change the allowed quantum electromagnetic modes. Each mode contributes zero-point energy $\hbar\omega/2$, so changing the mode spectrum changes the vacuum energy. The raw vacuum energy is divergent, so a physically motivated high-frequency cutoff is introduced. After converting transverse sums into integrals, the remaining finite-separation effect is the difference between a discrete sum and its continuum approximation. Euler--Maclaurin summation isolates the leading finite correction:

$$E_0(L)=E_0(\infty)-\frac{\hbar c\pi^2A}{720L^3}+O(L^{-4}).$$

Differentiating gives

$$P=-\frac{\pi^2}{240}\frac{\hbar c}{L^4}.$$

Thus ideal parallel conducting plates attract. The result is universal at leading order because the cutoff-dependent terms do not contribute to the $L$-dependent force. For real materials and nontrivial geometries, corrections can be important.

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Practice problems

1. Why does the vacuum wave functional have nonzero width even though the vacuum is the lowest-energy state?

2. Show that the standing-wave boundary condition between plates gives $k_x=k\pi/L$.

3. Explain why the transverse sums become integrals when $L_y,L_z\rightarrow\infty$.

4. Starting from

$$\Xi(L)=\frac{\pi}{4}\sum_{k=1}^{\infty}f\left(\frac{k}{L}\right),$$

use Euler--Maclaurin to identify which term produces the leading $1/L^3$ correction.

5. Verify that

$$f^{(3)}(0)=-4$$

for

$$f(x)=2\int_x^{\infty}du\,u^2\Gamma(u).$$

6. Derive the pressure from

$$E_0(L)=E_0(\infty)-\frac{\hbar c\pi^2A}{720L^3}.$$

7. Why do the cutoff-dependent terms not affect the final leading pressure?

8. Why should the ideal-conductor formula fail at sufficiently small separations?

9. Why is the statement “the Casimir force is always attractive” wrong?

10. Estimate how the pressure changes if the plate separation is reduced by a factor of two.

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References and further reading

• H. B. G. Casimir, “On the attraction between two perfectly conducting plates,” Proceedings of the Royal Netherlands Academy of Arts and Sciences, 1948.
• S. K. Lamoreaux, “The Casimir force: background, experiments, and applications,” Reports on Progress in Physics, 2005.
• M. Bordag, U. Mohideen, and V. M. Mostepanenko, “New developments in the Casimir effect,” Physics Reports, 2001.
• K. A. Milton, The Casimir Effect: Physical Manifestations of Zero-Point Energy, World Scientific, 2001.
• Wolfram MathWorld, “Euler--Maclaurin Integration Formulas.”
• J. D. Jackson, Classical Electrodynamics, for ideal-conductor electromagnetic boundary conditions.