11 and introducing scaled variables as in Section III B with M replaced by N , we obtain the following expression for the Casimir force The evaluation of Z a f is also quite straightforward with this procedure; the analogue of (4.52) is: F cas f = − k$_{B}$T 2 π ∫ π − π γ ( ω ) dω 1 1 + [(1 + cos( δ $^{′}$( ω ))) / (1 − cos( δ $^{′}$( ω )))] exp[2 Nγ ( ω )] , (4.61) lim N $_{→∞}$N − $^{1}$Z a f = M 4 π ∫ − π dω γ ( ω ) + M 2 π ∫ π + iϵ − π + iϵ dω e iMω 1 − e $_{iMω}$γ ( ω ) (4.56) f free cas = − 2 πN 2 ∫ ∞ −∞ d Ω $^{√}$t 2 + Ω 2   (1 − t √ t $^{2}$+Ω$_{2}$) e − 4 √ t $^{2}$+Ω2 (1 + t √ t $^{2}$+Ω$_{2}$) + (1 − t √ t $^{2}$+Ω$_{2}$) e − 4 √ t $^{2}$+Ω2   (4.62) F a $_{cas}$( cyl. ) = − k$_{B}$T π ∫ π 0 du ln [1 − exp ( − M ˆ γ ( u ))] (4.57) M 2 ¯ f a $_{cas}$( cyl. ) → k$_{B}$T π ∫ ∞ ∞ dy ( x 2 + y $^{2}$)1 / 2 $^{[}$coth $^{(}$( x 2 + y $^{2}$)1 / 2 ) − 1 ] (4.58) lim K$_{0}$ $_{→∞}$exp [ K$_{0}$ M ∑ m =1 $^{(}$σ x $_{m}$σ x m +1 − 1 ) ] = | Φ 0 $_{+}$〉〈 Φ 0 $_{+}$| + | Φ 0 $_{−}$〉〈 Φ 0 $_{−}$| (4.63) which agrees with the expression (3.24) derived by other methods. An analogous argument allows us to evaluate the Casimir free energy for the strip with plus spins at both ends. The right hand side of (4.15) can be evaluated in a convenient form for calculations by relating it to the limit of V when K ∗ 1 → 0. Following this with the limit K$_{0}$ → ∞ then yields Note that this expression is also invariant under inter- change of K ∗ 1 and K$_{2}$ . The analogue of (4.55) for the Casimir force with the line of reversed bonds is: with an additional term sgn( T$_{c}$ − T ) γ (0), which will not appear in the Casimir term, since it is independent of M . Note that the only other difference is the sign in the denominator of the second integrand; this comes about because the sum leading to this term is over periodic wave numbers. The Casimir free energy in this case is denoted by F a $_{cas}$( cyl. ); it is given by: The integral for x = 0 has the value π $^{2}$/ 6. C. Free and fixed boundary conditions The states on the right hand side of (4.72) are defined by lim$_{K}$ ∗ $_{1}$→ 0 | Φ$_{±}$ 〉 = | Φ 0 $_{±}$〉 . Thus they are eigenvalues of V$_{2}$ ( ± ) written in a form which is useful for the calcula- tion. Then a straightforward calculation gives To calculate the Casimir force for free and fixed bound- ary conditions we consider the other limits, i.e. M → ∞ with N finite. This yields 〈 Φ 0 $_{±}$| ( V ( ± )) $^{N}$| Φ 0 $_{±}$〉 = ∏ k ∈ Ω$_{M}$( ± ) [cosh( Nγ ( k )) + sinh( Nγ ( k )) cos δ $^{∗}$( k )] 1 / 2 (4.64) lim M →∞ 1 M ln Z$_{f}$ = 1 4 π ∫ 2 π 0 ln$^{[}$ cosh( Nγ ( ω )) + sinh( Nγ ( ω )) cos δ $^{′}$( ω )$^{]}$ dω (4.59) f ++ cas = − 2 πN 2 ∫ ∞ −∞ d Ω $^{√}$t 2 + Ω 2   (1 + t √ t $^{2}$+Ω$_{2}$) e − 4 √ t $^{2}$+Ω2 (1 − t √ t $^{2}$+Ω$_{2}$) + (1 + t √ t $^{2}$+Ω$_{2}$) e − 4 √ t $^{2}$+Ω2   (4.65) F cas f = − k$_{B}$T 4 π ∫ π − π ln$^{[}$(1 + cos( δ $^{′}$( ω ))) +(1 − cos( δ $^{′}$( ω ))) exp[ − 2 Nγ ( ω )]$^{]}$ dω (4.60) where the angle δ ∗ is defined by (4.45) and (4.46) with K ∗ 1 and K$_{2}$ interchanged , which is just what we could have anticipated from duality, as explained above. Evaluating the analogue of (4.53) we find Extracting the bulk term we are left with an incremental part, denoted F cas f . This remainder is given by Taking the derivative of the above result with respect to N ,