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Quantum codes do not increase fidelity against isotropic errors

J. Lacalle, L.M. Pozo-Coronado,
A.L. Fonseca de Oliveira, R. Martín-Cuevas
Dep. de Matemática Aplicada a las TIC, ETS de Ingeniería de Sistemas Informáticos, Universidad Politécnica de Madrid, C/ Alan Turing s/n, 28031, Madrid, Spain ([email protected]).Dep. de Matemática Aplicada a las TIC, ETS de Ingeniería de Sistemas Informáticos, Universidad Politécnica de Madrid, C/ Alan Turing s/n, 28031, Madrid, Spain ([email protected]).Facultad de Ingeniería, Universidad ORT Uruguay, Montevideo, Uruguay ([email protected]).Programa de Doctorado en Ciencias y Tecnologías de la Computación para Smart Cities, ETS de Ingeniería de Sistemas Informáticos, Universidad Politécnica de Madrid, C/ Alan Turing s/n, 28031, Madrid, Spain ([email protected]).
(January 2022)
Abstract

Given an mm-qubit Φ0\Phi_{0} and an (n,m)(n,m)-quantum code 𝒞\mathcal{C}, let Φ\Phi be the nn-qubit that results from the 𝒞\mathcal{C}-encoding of Φ0\Phi_{0}. Suppose that the state Φ\Phi is affected by an isotropic error (decoherence), becoming Ψ\Psi, and that the corrector circuit of 𝒞\mathcal{C} is applied to Ψ\Psi, obtaining the quantum state Φ~\tilde{\Phi}. Alternatively, we analyze the effect of the isotropic error without using the quantum code 𝒞\mathcal{C}. In this case the error transforms Φ0\Phi_{0} into Ψ0\Psi_{0}. Assuming that the correction circuit does not introduce new errors and that it does not increase the execution time, we compare the fidelity of Ψ\Psi, Φ~\tilde{\Phi} and Ψ0\Psi_{0} with the aim of analyzing the power of quantum codes to control isotropic errors. We prove that F(Ψ0)F(Φ~)F(Ψ)F(\Psi_{0})\geq F(\tilde{\Phi})\geq F(\Psi). Therefore the best option to optimize fidelity against isotropic errors is not to use quantum codes.

Keywords: quantum error correcting codes, isotropic quantum computing errors, quantum computing error fidelity, quantum computing error variance

1 Introduction

Currently the biggest obstacle to the development of quantum computing continues to be control of quantum errors. Since the beginnings of quantum computing in the 90s of the last century one of the main research objectives was to solve this stumbling block. To address the problem, two fundamental tools were developed: quantum error correction codes [1, 2, 3, 4, 5, 6] in combination with fault tolerant quantum computing [7, 8, 9, 10, 11, 12, 13]. These studies culminated in the proof of the quantum threshold theorem, which reads as follows: a quantum computer with a physical error rate below a certain threshold can, through application of quantum error correction schemes, suppress the logical error rate to arbitrarily low levels. However, the proof of this theorem depends on the discretized treatment of quantum errors, inherited from the construction of quantum codes.

We believe that the quantum error model used for the proof of the quantum threshold theorem is not general and that the techniques developed to control quantum errors do not verify the golden rule of error control: correct all small errors exactly. For example, in the case of the coding of a qubit by means of the 5-qubit code [14, 15] it is argued, using error discretization and that this code exactly corrects errors in any of the qubits, that the error probability goes from pp to p2p^{2}, once the correction circuit has been applied. But, what is actually happening is that the probability of an error (small with high probability) in all qubits is 1 and that the code cannot correct these simultaneous errors. Then, an error occurs with probability 1 and, once the correction circuit is applied, it becomes undetectable.

Therefore it is necessary to make an analysis of quantum errors regardless of their discretization. The procedure indicated for this is to consider quantum errors as continuous random variables and characterize them by their corresponding density functions. In this article we analyze a specific type of error: isotropic quantum errors. An isotropic error of an nn-qubit Φ\Phi is one in which the probability of the error Ψ\Psi only depends on the distance between the two states, ΨΦ\|\Psi-\Phi\|, and not on the direction in which the imprecision Ψ\Psi occurs with respect to Φ\Phi. They are errors that are easy to analyze due to their central symmetry with respect to Φ\Phi.

In work [16] we have studied the ability of an arbitrary quantum code to correct these errors, using the variance as the error measure. If Φ\Phi is the nn-qubit without error state, Ψ\Psi the state resulting from a disturbance modeled by an isotropic quantum error and Φ~\tilde{\Phi} the result of applying the quantum code correction circuit, assuming that it does not introduce new errors, the result that we demonstrate in [16] is the following:

V(Φ~)V(Ψ),V(\tilde{\Phi})\geq V(\Psi),

where V(Φ~)=E[Φ~Φ2]V(\tilde{\Phi})=E[\|\tilde{\Phi}-\Phi\|^{2}] and V(Ψ)=E[ΨΦ2]V(\Psi)=E[\|\Psi-\Phi\|^{2}] are the variances of the corrected state Φ~\tilde{\Phi} and the disturbed state Ψ\Psi respectively. This means that no quantum code can handle isotropic errors, or even reduce their variance.

Now we are interested in analyzing the ability of quantum codes to increase fidelity against isotropic errors, since the fidelity allows to measure the quantum errors taking into account that the quantum states do not change if they are multiplied by a phase factor, while the variance used in [16] does not take this fact into account.

We represent nn-qubits as points of the unit real sphere of dimension 2d12d-1 being d=2nd=2^{n} [17], S2d1={x2d|x=1}S^{2d-1}=\{x\in\mathbb{R}^{2d}\ |\ \|x\|=1\}, taking coordinates with respect to the computational basis [|0,|1,,|2n1][|0\rangle,|1\rangle,\dots,|2^{n}-1\rangle],

Ψ=(x0+ix1,x2+ix3,,x2d2+ix2d1).\Psi=(x_{0}+ix_{1},x_{2}+ix_{3},\dots,x_{2d-2}+ix_{2d-1}). (1)

We consider quantum computing errors as random variables with density function defined on S2d1S^{2d-1}. In [16] it is mentioned that it is easy to relate this representation to the usual representation in quantum computing by density matrices and that the representation through random variables is more accurate.

We define the variance of a random variable XX as the mean of the quadratic deviation from the mean value μ\mu of XX, V(X)=E[Xμ2]V(X)=E[\|X-\mu\|^{2}]. In our case, since the random variable XX represents a quantum computing error, the mean value of XX is the nn-qubit Φ\Phi resulting from an errorless computation. Without loss of generality, we will assume that the mean value of every quantum computing error will always be Φ=|0\Phi=|0\rangle. To achieve this, it suffices to move Φ\Phi into |0|0\rangle through a unitary transformation. Therefore, using the pure quantum states given by Formula (1), the variance of XX will be:

V(X)=E[[ΨΦ2]=E[22x0]=22S2d1x0f(x)dx.V(X)=E\left[[\|\Psi-\Phi\|^{2}\right]=E[2-2x_{0}]=2-2\int_{S^{2d-1}}x_{0}f(x)dx. (2)

Obviously the variance satisfies V(X)[0,4]V(X)\in[0,4]. In [18] the variance of the sum of two independent errors on S2d1S^{2d-1} is presented for the first time. It is proved for isotropic errors and it is conjectured in general that:

V(X1+X2)=V(X1)+V(X2)V(X1)V(X2)2.V(X_{1}+X_{2})=V(X_{1})+V(X_{2})-\frac{V(X_{1})V(X_{2})}{2}. (3)

Considering the representation of errors through random variables, the definition of fidelity is very simple:

F2(X)=E[|Ψ|Φ|2]=E[x02+x12]=S2d1(x02+x12)f(x)𝑑x.F^{2}(X)=E\left[|\langle\Psi|\Phi\rangle|^{2}\right]=E\left[x_{0}^{2}+x_{1}^{2}\right]=\int_{S^{2d-1}}(x_{0}^{2}+x_{1}^{2})f(x)dx. (4)

Then, the problem we want to address is the following: Let Φ0\Phi_{0} be an mm-qubit and Φ\Phi the corresponding nn-qubit encoded by an (n,m)(n,m)-quantum code 𝒞\mathcal{C}. Suppose that the coded state Φ\Phi is changed by error, becoming the state Ψ\Psi. Now, to fix the error we apply the code correction circuit, obtaining the final state Φ~\tilde{\Phi}. While Φ\Phi is a pure state, Ψ\Psi and Φ~\tilde{\Phi} are random variables (mixed states).

We also want to study the possibility of not using quantum codes. In this case, we suppose that the initial state Φ0\Phi_{0} is changed by error, becoming the state Ψ0\Psi_{0}. State Ψ0\Psi_{0} is also a random variable. Then our goal is to compare the fidelities of Ψ\Psi, Φ~\tilde{\Phi} and Ψ0\Psi_{0}:

F(Ψ)=E[|Ψ|Φ|2],F(Φ~)=E[|Φ~|Φ|2]andF(Ψ0)=E[|Ψ0|Φ0|2].F(\Psi)=E\left[|\langle\Psi|\Phi\rangle|^{2}\right],\ F(\tilde{\Phi})=E\left[|\langle\tilde{\Phi}|\Phi\rangle|^{2}\right]\ \text{and}\ F(\Psi_{0})=E\left[|\langle\Psi_{0}|\Phi_{0}\rangle|^{2}\right].

In order to compare the fidelities we will assume that the corrector circuit of 𝒞\mathcal{C} does not introduce new errors and it does not increase the execution time. In other words, we are going to estimate the theoretical capacity of the code to correct quantum computing errors.

In the case of isotropic errors we shall prove that:

F(Ψ0)F(Φ~)F(Ψ).F(\Psi_{0})\geq F(\tilde{\Phi})\geq F(\Psi). (5)

This result leads us to the conclusion that the best option to optimize fidelity against isotropic errors is not to use quantum codes. This result goes in the same direction as that obtained in [16], which indicates that quantum codes do not reduce the variance against isotropic errors.

However, the most widely used model of errors in quantum computing is qubit independent errors. The study of this type of quantum error is much more complex than that of isotropic errors, because it does not have the same symmetry. Despite this technical difficulty, we have proved in [19] that the 55-qubit code [14, 15] is not able to reduce the variance against qubit independent errors. This result, together with those obtained in [16] and in this article, clearly reveals the difficulty of the quantum error control challenge and clearly shows that the continuous nature of quantum errors cannot be ignored.

There are many works related to the control of quantum computing errors, in addition to those already mentioned above. General studies and surveys on the subject [20, 21, 22, 23, 24, 25, 26, 27], about the quantum computation threshold theorem [28, 29, 30, 31], quantum error correction codes [32, 33, 34, 35], concatenated quantum error correction codes [36, 37] and articles related to topological quantum codes [38, 39]. Lately, quantum computing error control has focused on both coherent errors [40, 41] and cross-talk errors [42, 43]. Finally, we cannot forget the hardest error to control in quantum computing, the quantum decoherence [44]. As we have commented above, these quantum computing errors can be analyzed in the framework of random variables that has been set in [16, 18]. In the conclusions we analyze in more detail the characteristics of the different types of error from the point of view of their control and in view of the result obtained in this paper.

The outline of the article is as follows: in section 2 we study the fidelity of the quantum stages Ψ\Psi, Ψ0\Psi_{0} and Φ~\tilde{\Phi}; in section 3 we prove the relationship between them given by Formula (5); finally, in section 4 we analyze the conclusions that can be obtained from the proved result.

2 Analysis of fidelity

Associated with the (n,m)(n,m)-quantum code 𝒞\mathcal{C}, the following parameters are defined: d=2nd=2^{n} is the dimension of 𝒞\mathcal{C}, d=2md^{\,\prime}=2^{m} and d′′d^{\,\prime\prime} is the number of discrete errors that 𝒞\mathcal{C} corrects.

First we are going to study how we can compare the fidelity of the quantum states Ψ\Psi and Φ~\tilde{\Phi}, which are nn-qubits encoded with the quantum code 𝒞\mathcal{C}, and the fidelity of the state Ψ0\Psi_{0}, which is an unencoded mm-qubit state. The working scheme in these two scenarios is illustrated in Figure 1. We assume that the 𝒞\mathcal{C} correction circuit, which is applied after each quantum gate in the coded algorithm, does not introduce new errors and is ideally applied for a time t=0t=0. In this way we study the theoretical capacity of 𝒞\mathcal{C} to control isotropic errors, that is, its capacity to increase the fidelity of the final state Φ~\tilde{\Phi} with respect to Ψ\Psi, and we can compare it with the fidelity of the final state Ψ0\Psi_{0} in the scheme without the quantum code 𝒞\mathcal{C}.

Refer to caption
Figure 1: Uncoded/coded work scheme.

We analyze the isotropic error as a decoherence error over a unit of time, which corresponds to the time it takes to apply a quantum gate in the coded algorithm. To compare it with the uncoded algorithm we have to bear in mind that the unit of time in this case will be at most the nn-th part of the unit of time in the coded algorithm. To relate the probability distributions in both cases we use the following equality of variances:

V(E)=V(E1+E2++En),V(E)=V(E_{1}+E_{2}+\cdots+E_{n}),

where EE is the decoherence error during a unit of time in the coded algorithm and E1E_{1}, E2E_{2}, …EnE_{n} are independent decoherence errors corresponding to a unit of time in the uncoded algorithm. Using the following generalization of Formula (3) demonstrated in [18]:

V(E1+E2++En)=22(1vu2)n,V(E_{1}+E_{2}+\cdots+E_{n})=2-2\left(1-\frac{v_{u}}{2}\right)^{n}, (6)

where vuv_{u} is the variance of each of the independent errors, we obtain the following relation of vuv_{u} with the variance vcv_{c} of the error EE:

vc=22(1vu2)nvu=22(2vc2)1/n.v_{c}=2-2\left(1-\frac{v_{u}}{2}\right)^{n}\quad\Leftrightarrow\quad v_{u}=2-2\left(\frac{2-v_{c}}{2}\right)^{1/n}. (7)

In the case of the normal probability distribution defined in [16, 18], with the following density function:

fn(σ,θ0)=(2d2)!!(2π)d(1σ2)(1+σ22σcos(θ0))d,f_{n}(\sigma,\theta_{0})=\frac{(2d-2)!!}{(2\pi)^{d}}\frac{(1-\sigma^{2})}{(1+\sigma^{2}-2\sigma\cos(\theta_{0}))^{d}}, (8)

where the parameter σ\sigma belongs to the interval [0,1)[0,1), the above variances have a very simple expression and are independent of the dimension: vc=2(1σc)v_{c}=2(1-\sigma_{c}) and vu=2(1σu)v_{u}=2(1-\sigma_{u}). The relationship between them given in Formula (7) translates into a very simple relationship between the corresponding sigma parameters:

σc=σunσu=σc1/n.\sigma_{c}=\sigma_{u}^{n}\quad\Rightarrow\quad\sigma_{u}=\sigma_{c}^{1/n}. (9)

From now on we are going to follow the scheme proposed in [16] to calculate the variances of states Ψ\Psi and Φ~\tilde{\Phi}, but to calculate the fidelities of these states and of state Ψ0\Psi_{0}.

2.1 Fidelity of Ψ\Psi and Ψ0\Psi_{0}

The state Ψ\Psi, described in Cartesian coordinates in Formula (1) is represented in spherical coordinates as follows:

Ψ=(θ0,θ1,,θ2d2){0θ0,,θ2d3π0θ2d22π,xj=sin(θ0)sin(θj1)cos(θj)for all0j2d2,x2d1=sin(θ0)sin(θ2d2).\begin{array}[]{l}\Psi=(\theta_{0},\,\theta_{1},\,\dots,\,\theta_{2d-2})\quad\left\{\begin{array}[]{l}\vrule height=8.0pt,depth=8.0pt,width=0.0pt0\leq\theta_{0},\,\dots,\,\theta_{2d-3}\leq\pi\\ \vrule height=12.0pt,depth=2.0pt,width=0.0pt0\leq\theta_{2d-2}\leq 2\pi\end{array}\right.,\\ \vrule height=16.0pt,depth=8.0pt,width=0.0ptx_{j}=\sin(\theta_{0})\,\cdots\,\sin(\theta_{j-1})\,\cos(\theta_{j})\quad\text{for all}\quad 0\leq j\leq 2d-2,\\ \vrule height=12.0pt,depth=8.0pt,width=0.0ptx_{2d-1}=\sin(\theta_{0})\,\cdots\,\sin(\theta_{2d-2}).\end{array}

Using this representation of Ψ\Psi, the fidelity entered in Formula (4) is as follows:

F2(X)=E[cos2(θ0)+sin2(θ0)cos2(θ1)]=1E[sin2(θ0)sin2(θ1)].F^{2}(X)=E\left[\cos^{2}(\theta_{0})+\sin^{2}(\theta_{0})\cos^{2}(\theta_{1})\right]=1-E\left[\sin^{2}(\theta_{0})\sin^{2}(\theta_{1})\right]. (10)
Theorem 1.

The fidelity of the isotropic random variable Ψ\Psi with density function f(θ0)f(\theta_{0}) is equal to:

F2(Ψ)=14(2π)d1(2d1)!!(d1)E¯[sin2d(θ0)],F^{2}(\Psi)=1-4\,\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,(d-1)\,{\bar{E}}\left[\sin^{2d}(\theta_{0})\right], (11)

where E¯[sin2d(θ0)]=0πf(θ0)sin2d(θ0)𝑑θ0\displaystyle{\bar{E}}\left[\sin^{2d}(\theta_{0})\right]=\int_{0}^{\pi}f(\theta_{0})\sin^{2d}(\theta_{0})d\theta_{0}.

Proof.

We have to calculate the expected value of an expression that depends only on angles θ0\theta_{0} and θ1\theta_{1} and the isotropic density function depends only on angle θ0\theta_{0}. Therefore, using Formula (10):

F2(Ψ)=1|S2d3|E¯[sin2d(θ0)]0πsin2d1(θ1)𝑑θ1=1(2π)d1(2d4)!! 2(2d2)!!(2d1)!!E¯[sin2d(θ0)]=14(2π)d1(2d1)!!(d1)E¯[sin2d(θ0)].\begin{array}[]{lll}\vrule height=14.0pt,depth=12.0pt,width=0.0ptF^{2}(\Psi)&=&\displaystyle 1-|S^{2d-3}|\,{\bar{E}}[\sin^{2d}(\theta_{0})]\,\int_{0}^{\pi}\sin^{2d-1}(\theta_{1})d\theta_{1}\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-\dfrac{(2\pi)^{d-1}}{(2d-4)!!}\,2\dfrac{(2d-2)!!}{(2d-1)!!}\,{\bar{E}}[\sin^{2d}(\theta_{0})]\\ \vrule height=18.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-4\,\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,(d-1)\,{\bar{E}}[\sin^{2d}(\theta_{0})].\\ \end{array}

We have used equalities from the Appendix. ∎

Corollary 1.

The fidelity of the isotropic random variable Ψ\Psi with normal distribution fn(σc,θ0)f_{n}(\sigma_{c},\theta_{0}) is equal to:

F2(Ψ)=1+(d1)σc2d.F^{2}(\Psi)=\dfrac{1+(d-1)\sigma_{c}^{2}}{d}. (12)
Proof.

Using the definition of the normal distribution given in Formula (8) and the Appendix:

F2(Ψ)=14(2π)d1(2d1)!!(d1)E¯[sin2d(θ0)]=14(2π)d1(2d1)!!(d1)(2d2)!!(2π)d(1σc2)(2d1)!!(2d)!!π=1d1d(1σc2)=1+(d1)σc2d.\begin{array}[]{lll}\vrule height=14.0pt,depth=12.0pt,width=0.0ptF^{2}(\Psi)&=&\displaystyle 1-4\,\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,(d-1)\,{\bar{E}}[\sin^{2d}(\theta_{0})]\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-4\,\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,(d-1)\,\dfrac{(2d-2)!!}{(2\pi)^{d}}\,(1-\sigma_{c}^{2})\,\dfrac{(2d-1)!!}{(2d)!!}\pi\\ \vrule height=18.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-\dfrac{d-1}{d}\,(1-\sigma_{c}^{2})=\dfrac{1+(d-1)\sigma_{c}^{2}}{d}.\\ \end{array}

Theorem 1 and Corollary 1 also apply to state Ψ0\Psi_{0}, changing the parameter dd to dd^{\prime}.

Corollary 2.

The fidelity of the isotropic random variable Ψ0\Psi_{0} with density function f(θ0)f(\theta_{0}) is equal to:

F2(Ψ0)=14(2π)d1(2d1)!!(d1)E¯[sin2d(θ0)],F^{2}(\Psi_{0})=1-4\,\dfrac{(2\pi)^{d^{\prime}-1}}{(2d^{\prime}-1)!!}\,(d^{\prime}-1)\,{\bar{E}}\left[\sin^{2d^{\prime}}(\theta_{0})\right], (13)

where E¯[sin2d(θ0)]=0πf(θ0)sin2d(θ0)𝑑θ0\displaystyle{\bar{E}}\left[\sin^{2d^{\prime}}(\theta_{0})\right]=\int_{0}^{\pi}f(\theta_{0})\sin^{2d^{\prime}}(\theta_{0})d\theta_{0}. And, if the probability distribution of Ψ0\Psi_{0} is normal with density function fn(σu,θ0)f_{n}(\sigma_{u},\theta_{0}), the fidelity is equal to:

F2(Ψ0)=1+(d1)σu2d.F^{2}(\Psi_{0})=\dfrac{1+(d^{\prime}-1)\sigma_{u}^{2}}{d^{\prime}}. (14)

To compare the fidelities of Ψ0\Psi_{0} and Φ~\tilde{\Phi} we need to obtain their values as a function of their variances vuv_{u} and vcv_{c} respectively. The relationship between these variances obtained in Formula (7) will allow us to relate the fidelities of these states.

Theorem 2.

The fidelity of the isotropic random variable Ψ0\Psi_{0} with density function f(θ0)f(\theta_{0}) satisfy:

F2(Ψ0)12d22d1(vu(vu2)2).F^{2}(\Psi_{0})\geq 1-\dfrac{2d^{\prime}-2}{2d^{\prime}-1}\left(v_{u}-\left(\dfrac{v_{u}}{2}\right)^{2}\right). (15)
Proof.

First we prove, similar to the proof of Theorem 1, the following:

F2(Ψ0)=1|S2d3|E¯[sin2d(θ0)]0πsin2d1(θ1)𝑑θ1=1|S2d3|E¯[sin2d(θ0)]0πsin2d3(θ1)𝑑θ10πsin2d1(θ1)𝑑θ10πsin2d3(θ1)𝑑θ1=1S2d1f(θ0)sin2(θ0)0πsin2d1(θ1)𝑑θ10πsin2d3(θ1)𝑑θ1\begin{array}[]{lll}\vrule height=14.0pt,depth=12.0pt,width=0.0ptF^{2}(\Psi_{0})&=&\displaystyle 1-|S^{2d^{\prime}-3}|\,{\bar{E}}[\sin^{2d^{\prime}}(\theta_{0})]\int_{0}^{\pi}\sin^{2d^{\prime}-1}(\theta_{1})d\theta_{1}\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-|S^{2d^{\prime}-3}|\,{\bar{E}}[\sin^{2d^{\prime}}(\theta_{0})]\int_{0}^{\pi}\sin^{2d^{\prime}-3}(\theta_{1})d\theta_{1}\,\dfrac{\displaystyle\int_{0}^{\pi}\sin^{2d^{\prime}-1}(\theta_{1})d\theta_{1}}{\displaystyle\int_{0}^{\pi}\sin^{2d^{\prime}-3}(\theta_{1})d\theta_{1}}\\ \vrule height=18.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-\int_{S^{2d^{\prime}-1}}f(\theta_{0})\sin^{2}(\theta_{0})\,\dfrac{\displaystyle\int_{0}^{\pi}\sin^{2d^{\prime}-1}(\theta_{1})d\theta_{1}}{\displaystyle\int_{0}^{\pi}\sin^{2d^{\prime}-3}(\theta_{1})d\theta_{1}}\\ \end{array}

And, using the formulas in the Appendix, we obtain:

F2(Ψ0)=1E[sin2(θ0)]2d22d1.F^{2}(\Psi_{0})=\displaystyle 1-E[\sin^{2}(\theta_{0})]\,\dfrac{2d^{\prime}-2}{2d^{\prime}-1}.

Using Jensen’s inequality we obtain a lower bound for E[sin2(θ0)]E[\sin^{2}(\theta_{0})]:

(E[1cos(θ0)])2E[(1cos(θ0))2]=E[1+cos2(θ0)2cos(θ0)]=E[22cos(θ0)sin2(θ0)]=vuE[sin2(θ0)].\begin{array}[]{lll}\vrule height=14.0pt,depth=12.0pt,width=0.0pt(E[1-\cos(\theta_{0})])^{2}&\leq&E[(1-\cos(\theta_{0}))^{2}]\\ \vrule height=14.0pt,depth=12.0pt,width=0.0pt&=&E[1+\cos^{2}(\theta_{0})-2\cos(\theta_{0})]\\ \vrule height=14.0pt,depth=12.0pt,width=0.0pt&=&E[2-2\cos(\theta_{0})-\sin^{2}(\theta_{0})]\\ \vrule height=14.0pt,depth=12.0pt,width=0.0pt&=&v_{u}-E[\sin^{2}(\theta_{0})].\\ \end{array}

And then:

E[sin2(θ0)]vu(E[1cos(θ0)])2=vu(vu2)2.E[\sin^{2}(\theta_{0})]\leq v_{u}-(E[1-\cos(\theta_{0})])^{2}=v_{u}-\left(\dfrac{v_{u}}{2}\right)^{2}.

Substituting in the formula of F2(Ψ0)F^{2}(\Psi_{0}) the previous lower bound of E[sin2(θ0)]E[\sin^{2}(\theta_{0})], the proof is concluded:

F2(Ψ0)12d22d1(vu(vu2)2).F^{2}(\Psi_{0})\geq 1-\dfrac{2d^{\prime}-2}{2d^{\prime}-1}\left(v_{u}-\left(\dfrac{v_{u}}{2}\right)^{2}\right).

2.2 Fidelity of Φ~\tilde{\Phi}

The formula for the fidelity of the state Φ~\tilde{\Phi} is very similar to that of the state Ψ\Psi, Formula (11), although the proof is more complex because the quantum code 𝒞\mathcal{C} is involved.

Theorem 3.

The fidelity of the isotropic random variable Φ~\tilde{\Phi} with density function f(θ0)f(\theta_{0}) is equal to:

F2(Φ~)=14(2π)d1(2d1)!!(dd′′)E¯[sin2d(θ0)],F^{2}(\tilde{\Phi})=1-4\,\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,(d-d^{\prime\prime})\,{\bar{E}}\left[\sin^{2d}(\theta_{0})\right], (16)

where E¯[sin2d(θ0)]=0πf(θ0)sin2d(θ0)𝑑θ0\displaystyle{\bar{E}}\left[\sin^{2d}(\theta_{0})\right]=\int_{0}^{\pi}f(\theta_{0})\sin^{2d}(\theta_{0})d\theta_{0}.

Proof.

Taking into account Theorem 3 and Corollary 1 of [16] the fidelity of Φ~\tilde{\Phi} is the following:

F2(Φ~)=E[P0|Φ|Π0Ψ|2]+(d′′1)E[P1|E1Φ|Π1Ψ|2].F^{2}(\tilde{\Phi})=E\left[P_{0}|\langle\Phi|\Pi_{0}\Psi\rangle|^{2}\right]+(d^{\prime\prime}-1)\,E\left[P_{1}|\langle E_{1}\Phi|\Pi_{1}\Psi\rangle|^{2}\right].

where P0P_{0} and P1P_{1} are the probabilities of measuring the syndromes 0 and 11 respectively, Π0\Pi_{0} and Π1\Pi_{1} the (normalized) projectors corresponding to the discrete errors E0=IE_{0}=I and E1E_{1} associated with the aforementioned syndromes and E1Φ=E1|0=|2dE_{1}\Phi=E_{1}|0\rangle=|2d^{\prime}\rangle.

The first expected value in the above expression is equal to F2(Ψ)F^{2}(\Psi) by the Formula (10) and, using Theorem 1, it is obtained:

E[P0|Φ|Π0Ψ|2]=E[1sin2(θ0)sin2(θ1)]=14(2π)d1(2d1)!!(d1)E¯[sin2d(θ0)].\begin{array}[]{lll}\vrule height=14.0pt,depth=8.0pt,width=0.0ptE\left[P_{0}|\langle\Phi|\Pi_{0}\Psi\rangle|^{2}\right]&=&E\left[1-\sin^{2}(\theta_{0})\sin^{2}(\theta_{1})\right]\\ \vrule height=16.0pt,depth=12.0pt,width=0.0pt&=&1-4\,\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,(d-1)\,{\bar{E}}\left[\sin^{2d}(\theta_{0})\right].\\ \end{array}

And the second is the following:

E[P1|E1Φ|Π1Ψ|2]=E[sin2(θ0)sin2(θ2d1)(1sin2(θ2d)sin2(θ2d+1))].E\left[P_{1}|\langle E_{1}\Phi|\Pi_{1}\Psi\rangle|^{2}\right]=E\left[\sin^{2}(\theta_{0})\cdots\sin^{2}(\theta_{2d^{\prime}-1})\left(1-\sin^{2}(\theta_{2d^{\prime}})\sin^{2}(\theta_{2d^{\prime}+1})\right)\right].\\

Using the Appendix the following is obtained:

E[sin2(θ0)sin2(θ2d1)]=E¯[sin2d(θ0)]0πsin2d1(θ0)dθ10πsin2d2d+1(θ2d1)dθ2d1|S2d2d1|=E¯[sin2d(θ0)] 2(2d2)!!(2d1)!!π(2d3)!!(2d2)!!2(2d2d)!!(2d2d+1)!!(2π)dd(2d2d2)!!=E¯[sin2d(θ0)] 4(2π)d1(2d1)!!(dd).\begin{array}[]{l}\vrule height=14.0pt,depth=8.0pt,width=0.0pt\displaystyle E\left[\sin^{2}(\theta_{0})\dots\sin^{2}(\theta_{2d^{\prime}-1})\right]={\bar{E}}[\sin^{2d}(\theta_{0})]\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt\displaystyle\cdot\int_{0}^{\pi}\sin^{2d-1}(\theta_{0})d\theta_{1}\cdots\int_{0}^{\pi}\sin^{2d-2d^{\prime}+1}(\theta_{2d^{\prime}-1})d\theta_{2d^{\prime}-1}|S_{2d-2d^{\prime}-1}|\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt\displaystyle={\bar{E}}[\sin^{2d}(\theta_{0})]\,2\dfrac{(2d-2)!!}{(2d-1)!!}\,\pi\dfrac{(2d-3)!!}{(2d-2)!!}\cdots 2\dfrac{(2d-2d^{\prime})!!}{(2d-2d^{\prime}+1)!!}\,\dfrac{(2\pi)^{d-d^{\prime}}}{(2d-2d^{\prime}-2)!!}\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt\displaystyle={\bar{E}}[\sin^{2d}(\theta_{0})]\,4\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,(d-d^{\prime}).\end{array}

Similarly we obtain:

E[sin2(θ0)sin2(θ2d+1)]=E¯[sin2d(θ0)] 4(2π)d1(2d1)!!(dd1).E\left[\sin^{2}(\theta_{0})\dots\sin^{2}(\theta_{2d^{\prime}+1})\right]={\bar{E}}[\sin^{2d}(\theta_{0})]\,4\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,(d-d^{\prime}-1).

With the last two results the following is obtained:

E[P1|E1Φ|Π1Ψ|2]=E¯[sin2d(θ0)] 4(2π)d1(2d1)!!.E\left[P_{1}|\langle E_{1}\Phi|\Pi_{1}\Psi\rangle|^{2}\right]={\bar{E}}[\sin^{2d}(\theta_{0})]\,4\dfrac{(2\pi)^{d-1}}{(2d-1)!!}.\\

Finally we get the result we are looking for:

F2(Φ~)=E[P0|Φ|Π0Ψ|2]+(d′′1)E[P1|E1Φ|Π1Ψ|2]=1E¯[sin2d(θ0)] 4(2π)d1(2d1)!!(d1(d′′1))=1E¯[sin2d(θ0)] 4(2π)d1(2d1)!!(dd′′).\begin{array}[]{lll}\vrule height=14.0pt,depth=8.0pt,width=0.0pt\displaystyle F^{2}(\tilde{\Phi})&=&\displaystyle E\left[P_{0}|\langle\Phi|\Pi_{0}\Psi\rangle|^{2}\right]+(d^{\prime\prime}-1)E\left[P_{1}|\langle E_{1}\Phi|\Pi_{1}\Psi\rangle|^{2}\right]\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-{\bar{E}}[\sin^{2d}(\theta_{0})]\,4\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\left(d-1-(d^{\prime\prime}-1)\right)\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-{\bar{E}}[\sin^{2d}(\theta_{0})]\,4\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\left(d-d^{\prime\prime}\right).\\ \end{array}

If the probability distribution of Ψ\Psi is normal the fidelity of Φ~\tilde{\Phi} is much simpler.

Corollary 3.

If Ψ\Psi has a normal probability distribution with parameter σc\sigma_{c} the fidelity of Φ~\tilde{\Phi} satisfies:

F2(Φ~)=1+(d1)σc2d.F^{2}(\tilde{\Phi})=\dfrac{1+(d^{\prime}-1)\sigma_{c}^{2}}{d^{\prime}}. (17)
Proof.

To prove the result, it is enough to substitute in Theorem 3 the value of the integral E¯[sin2d(θ0)]{\bar{E}}[\sin^{2d}(\theta_{0})] from the Appendix and consider that d=dd′′d=d^{\prime}d^{\prime\prime}. ∎

To compare the fidelities of Ψ0\Psi_{0} and Φ~\tilde{\Phi} we need to obtain F2(Φ~)F^{2}(\tilde{\Phi}) as a function of the variances vcv_{c} of the state Ψ\Psi.

Theorem 4.

If the state Ψ\Psi has an isotropic distribution with density function f(θ0)f(\theta_{0}) such that:

0π(1cos(θ0))cos(θ0)f(θ0)0,\int_{0}^{\pi}(1-\cos(\theta_{0}))\cos(\theta_{0})f(\theta_{0})\geq 0, (18)

the fidelity of Φ~\tilde{\Phi} satisfies:

F2(Φ~)1dd′′2d1vc.F^{2}(\tilde{\Phi})\leq 1-\dfrac{d-d^{\prime\prime}}{2d^{\prime}-1}\,v_{c}. (19)
Proof.

First we prove, similar to the proofs of Theorems 1 and 2, the following:

F2(Φ~)=14(2π)d1(2d1)!!(dd′′)E¯[sin2d(θ0)]=14(2π)d1(2d1)!!dd′′|S2d2|E¯[sin2d(θ0)]|S2d2|=14(2π)d1(2d1)!!(2d3)!!2(2π)d1(dd′′)E[sin2(θ0)]=12dd′′2d1E[sin2(θ0)].\begin{array}[]{lll}\vrule height=14.0pt,depth=12.0pt,width=0.0ptF^{2}(\tilde{\Phi})&=&\displaystyle 1-4\,\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,(d-d^{\prime\prime})\,{\bar{E}}\left[\sin^{2d}(\theta_{0})\right]\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-4\,\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,\dfrac{d-d^{\prime\prime}}{|S^{2d-2}|}\,{\bar{E}}\left[\sin^{2d}(\theta_{0})\right]\,|S^{2d-2}|\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-4\,\dfrac{(2\pi)^{d-1}}{(2d-1)!!}\,\dfrac{(2d-3)!!}{2(2\pi)^{d-1}}\,(d-d^{\prime\prime})\,E\left[\sin^{2}(\theta_{0})\right]\\ \vrule height=20.0pt,depth=14.0pt,width=0.0pt&=&\displaystyle 1-2\,\dfrac{d-d^{\prime\prime}}{2d-1}\,E\left[\sin^{2}(\theta_{0})\right].\\ \end{array}

Now, using Formula (18), we obtain the following lower bound:

E[sin2(θ0)]=E[(1cos(θ0))(1+cos(θ0))]E[1cos(θ0)]=vc2.\begin{array}[]{lll}\vrule height=12.0pt,depth=8.0pt,width=0.0ptE\left[\sin^{2}(\theta_{0})\right]&=&\displaystyle E\left[(1-\cos(\theta_{0}))(1+\cos(\theta_{0}))\right]\\ \vrule height=18.0pt,depth=12.0pt,width=0.0pt&\geq&\displaystyle E\left[1-\cos(\theta_{0})\right]=\dfrac{v_{c}}{2}.\\ \end{array}

The proof is concluded by introducing the previous lower bound in the expression obtained for F2(Φ~)F^{2}(\tilde{\Phi}). ∎

3 Relationship between the fidelity of the states Ψ0\Psi_{0}, Φ~\tilde{\Phi} and Ψ\Psi

The results obtained in the previous section allow us to easily prove the following theorem.

Theorem 5.

If the state Ψ\Psi has an isotropic distribution the following relationship between the fidelities of Φ~\tilde{\Phi} and Ψ\Psi holds:

F2(Φ~)F2(Ψ).F^{2}(\tilde{\Phi})\geq F^{2}(\Psi). (20)
Proof.

Theorems 1 and 3 allow us to prove the result directly, considering that d1dd′′d-1\geq d-d^{\prime\prime}. ∎

To compare the fidelities of states Ψ0\Psi_{0} and Φ~\tilde{\Phi} we need to use Theorems 2 and 4. However, we must establish a previous result in order to establish the relationship between these states.

Lemma 1.

Given nn\in\mathbb{N}, n2n\geq 2, and xx\in\mathbb{R}, 0x40\leq x\leq 4, the following is satisfied:

g(n,x)=22(1x2)n(x(x2)2)0.g(n,x)=2-2\left(1-\dfrac{x}{2}\right)^{n}-\left(x-\left(\dfrac{x}{2}\right)^{2}\right)\geq 0.
Proof.

The change of variable y=(1x2)\displaystyle y=\left(1-\dfrac{x}{2}\right) allows us to better analyze the function:

g(n,y)=1+y22ynandx[0,4]y[1,1].g(n,y)=1+y^{2}-2y^{n}\quad\text{and}\quad x\in[0,4]\ \Leftrightarrow\ y\in[-1,1].

Property 1,y2|yn|1,\ y^{2}\geq|y^{n}| for all y[1,1]y\in[-1,1] allows us to conclude that g(n,y)0g(n,y)\geq 0 for all y[1,1]y\in[-1,1] and this shows that:

g(n,x)0for allx[0,4].g(n,x)\geq 0\quad\text{for all}\quad x\in[0,4].

The previous lemma allows us to obtain the main result of this article.

Theorem 6.

If states Ψ0\Psi_{0} and Ψ\Psi have isotropic distributions with variances vuv_{u} and vcv_{c} respectively and the density function of Ψ\Psi satisfies Formula (18), the following relationship between the fidelities of Ψ0\Psi_{0} and Φ~\tilde{\Phi} holds:

F2(Ψ0)F2(Φ~).F^{2}(\Psi_{0})\geq F^{2}(\tilde{\Phi}). (21)
Proof.

Theorems 2 and 4 allow us to prove the result, just establishing that the following inequality holds:

dd′′2d1vc2d22d1(vu(vu2)2).\dfrac{d-d^{\prime\prime}}{2d^{\prime}-1}\,v_{c}\geq\dfrac{2d^{\prime}-2}{2d^{\prime}-1}\left(v_{u}-\left(\dfrac{v_{u}}{2}\right)^{2}\right).

Taking into account that d=dd′′d=d^{\prime}d^{\prime\prime} the above inequality is equivalent to the following:

vc2d′′(vu(vu2)2).v_{c}\geq\dfrac{2}{d^{\prime\prime}}\left(v_{u}-\left(\dfrac{v_{u}}{2}\right)^{2}\right).

Using the fact that d′′2d^{\prime\prime}\geq 2 would suffice to prove the first of the following two inequalities:

vcvu(vu2)22d′′(vu(vu2)2).v_{c}\geq v_{u}-\left(\dfrac{v_{u}}{2}\right)^{2}\geq\dfrac{2}{d^{\prime\prime}}\left(v_{u}-\left(\dfrac{v_{u}}{2}\right)^{2}\right).

Substituting the value of vcv_{c} given in Formula (7) and using the function g(n,x)g(n,x) of Lemma 1 we have:

vcvu(vu2)2g(n,vu)0.v_{c}\geq v_{u}-\left(\dfrac{v_{u}}{2}\right)^{2}\quad\Leftrightarrow\quad g(n,v_{u})\geq 0.

Finally, Lemma 1 allows us to conclude the proof, using the fact that the variance vu[0,4]v_{u}\in[0,4]. ∎

If the isotropic distributions of Ψ\Psi and Ψ0\Psi_{0} are normal the condition given in Formula (18) for Theorems 4 and 6 is not necessary. Indeed, Corollaries 1, 2 and 3 clearly imply that:

F(Ψ0)F(Φ~)F(Ψ).F(\Psi_{0})\geq F(\tilde{\Phi})\geq F(\Psi). (22)

On the other hand, the condition given by Formula (18) for Theorems 4 and 6 is a sufficient condition. However, it is not necessary because it has been obtained by underestimating the fidelity of Ψ0\Psi_{0} and overestimating that of Φ~\tilde{\Phi}. It is verified for very general isotropic distributions, such as for density functions f(θ0)f(\theta_{0}) that satisfy the following:

f(θ0)=0for allθ0(π2,π].f(\theta_{0})=0\quad\text{for all}\quad\theta_{0}\in\left(\dfrac{\pi}{2},\pi\right].
Refer to caption
Refer to caption
Figure 2: Representation of fidelities as a function of σ\sigma.

Figure 2 shows the curves of F2(Ψ0)F^{2}(\Psi_{0}), F2(Φ~)F^{2}(\tilde{\Phi}) and F2(Ψ)F^{2}(\Psi) for normal isotropic distributions and n=5n=5 (d=32d=32), in the extreme cases d=16d^{\prime}=16 (d′′=2d^{\prime\prime}=2) and d=2d^{\prime}=2 (d′′=16d^{\prime\prime}=16).

The conclusion of the study carried out in this article, in view of the results summarized in Formula (22), is that the best option to obtain the highest fidelity against isotropic errors is not to use quantum codes. On the other hand, the improvement of the fidelity of Φ~\tilde{\Phi} versus that of Ψ\Psi seems to be closely related to the dimension of the subspaces to which these states belong: dd^{\prime} for Φ~\tilde{\Phi} versus dd for Ψ\Psi. See Theorems 1 and 3 and Corollaries 1 and 3.

4 Conclusions

In this article we have analyzed the ability of quantum codes to increase fidelity of quantum states affected by isotropic decoherence errors. The results obtained, despite being those expected for this type of quantum errors, are not good from the point of view of controlling errors in quantum computing. The ability of quantum codes to reduce errors does not compensate the multiplication of the number of gates that they require. This fact implies that the best option against isotropic errors is not to use quantum codes. This result is similar to that obtained in [16]: quantum codes do not reduce the variance of isotropic errors; and in [19]: the 55-qubit quantum code do not reduce the variance of qubit independent errors. The last result is more worrying since it negatively affects the standard model of error in quantum computing. For this reason, it would be important to study the behavior of fidelity in this case.

These results indicate that continuous errors must be taken into account, since it is not possible to ensure that the golden rule of error control “correct all small errors exactly” is fulfilled. Therefore, the study of the stochastic model of quantum errors, focused on discrete errors, must be extended to continuous errors.

For future research, we believe that the continuous quantum computing error model should be further developed. The results on the ability of quantum codes to increase the fidelity or to reduce the variance of quantum errors should be extended to other types of error. It is also important to develop models of the behavior of quantum errors in highly entangled quantum systems. We need to know better the behavior of errors in this type of systems so important for quantum computing. Finally, all these approaches should allow a reformulation of fault-tolerant quantum computing for continuous errors.

5 Appendix

The values of the integrals that have been used throughout the article are included in this Appendix.


0πsink(θ)𝑑θ={2(k1)!!k!!k=1, 3, 5,π(k1)!!k!!k=2, 4, 6,\displaystyle\int_{0}^{\pi}\sin^{k}(\theta)d\theta=\left\{\begin{array}[]{l}\displaystyle 2\,\frac{(k-1)!!}{k!!}\quad\mbox{}\quad k=1,\,3,\,5,\dots\\ \\ \displaystyle\pi\,\frac{(k-1)!!}{k!!}\quad\mbox{}\quad k=2,\,4,\,6,\dots\end{array}\right.


0πsin2d2(θ0)(1+σ22σcos(θ0))d𝑑θ0=(2d3)!!(2d2)!!π(1σ2)d=1, 2, 3,\displaystyle\int_{0}^{\pi}\frac{\sin^{2d-2}(\theta_{0})}{(1+\sigma^{2}-2\sigma\cos(\theta_{0}))^{d}}d\theta_{0}=\frac{(2d-3)!!}{(2d-2)!!}\frac{\pi}{(1-\sigma^{2})}\quad\mbox{}\quad d=1,\,2,\,3,\,\,\dots


0πcos(θ0)sin2d2(θ0)(1+σ22σcos(θ0))d𝑑θ0=(2d3)!!(2d2)!!σ(1σ2)πd=1, 2, 3,\displaystyle\int_{0}^{\pi}\frac{\cos(\theta_{0})\sin^{2d-2}(\theta_{0})}{(1+\sigma^{2}-2\sigma\cos(\theta_{0}))^{d}}d\theta_{0}=\frac{(2d-3)!!}{(2d-2)!!}\frac{\sigma}{(1-\sigma^{2})}\pi\quad\mbox{}\quad d=1,\,2,\,3,\,\,\dots


0πsin2d(θ0)(1+σ22σcos(θ0))d𝑑θ0=(2d1)!!(2d)!!πd=0, 1, 2,\displaystyle\int_{0}^{\pi}\frac{\sin^{2d}(\theta_{0})}{(1+\sigma^{2}-2\sigma\cos(\theta_{0}))^{d}}d\theta_{0}=\frac{(2d-1)!!}{(2d)!!}\pi\quad\mbox{}\quad d=0,\,1,\,2,\,\,\dots


Starting from the first integral, the surface of a unit sphere of arbitrary even (2d)(2d) or odd (2d1)(2d-1) dimension can be calculated.


|𝒮2d|=0π0π02πsin2d1(θ0)sin1(θ2d2)𝑑θ0𝑑θ2d2𝑑θ2d1=2(2d2)!!(2d1)!!(2d3)!!(2d2)!!π 2(2d4)!!(2d3)!!(21)!!2!!π 2(11)!!1!! 2π=2(2π)d(2d1)!!\begin{array}[]{ccl}|{\cal S}_{2d}|&=&\displaystyle\int_{0}^{\pi}\cdots\int_{0}^{\pi}\int_{0}^{2\pi}\sin^{2d-1}(\theta_{0})\,\cdots\,\sin^{1}(\theta_{2d-2})\ d\theta_{0}\,\cdots\,d\theta_{2d-2}d\theta_{2d-1}\\ \\ &=&\displaystyle 2\frac{(2d-2)!!}{(2d-1)!!}\ \frac{(2d-3)!!}{(2d-2)!!}\pi\ 2\frac{(2d-4)!!}{(2d-3)!!}\ \cdots\ \frac{(2-1)!!}{2!!}\pi\ 2\frac{(1-1)!!}{1!!}\ 2\pi\\ \\ &=&\displaystyle\frac{2(2\pi)^{d}}{(2d-1)!!}\\ \end{array}


|𝒮2d1|=0π0π02πsin2d2(θ0)sin1(θ2d3)𝑑θ0𝑑θ2d3𝑑θ2d2=(2d3)!!(2d2)!!π 2(2d4)!!(2d3)!!(21)!!2!!π 2(11)!!1!! 2π=(2π)d(2d2)!!\begin{array}[]{ccl}|{\cal S}_{2d-1}|&=&\displaystyle\int_{0}^{\pi}\cdots\int_{0}^{\pi}\int_{0}^{2\pi}\sin^{2d-2}(\theta_{0})\,\cdots\,\sin^{1}(\theta_{2d-3})\ d\theta_{0}\,\cdots\,d\theta_{2d-3}d\theta_{2d-2}\\ \\ &=&\displaystyle\frac{(2d-3)!!}{(2d-2)!!}\pi\ 2\frac{(2d-4)!!}{(2d-3)!!}\ \cdots\ \frac{(2-1)!!}{2!!}\pi\ 2\frac{(1-1)!!}{1!!}\ 2\pi\\ \\ &=&\displaystyle\frac{(2\pi)^{d}}{(2d-2)!!}\\ \end{array}

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