Quantum Teleportation and Grover’s Algorithm Without the Wavefunction

Abstract

In the same way as the quantum no-cloning theorem and quantum key distribution in two preceding papers, entanglement-assisted quantum teleportation and Grover’s search algorithm are generalized by transferring them to an abstract setting, including usual quantum mechanics as a special case. This again shows that a much more general and abstract access to these quantum mechanical features is possible than commonly thought. A non-classical extension of conditional probability and, particularly, a very special type of state-independent conditional probability are used instead of Hilbert spaces and wavefunctions.

Annex

Lemma 5

Suppose that the quantum logic E satisfies the Assumptions 1, 2 and 3 and that $\mathbb {P}(f|e) = p = \mathbb {P}(e|f)$ for some $e,f \in E$. Then, for $r = 1, 2, 3,\ldots$,

$$\begin{aligned} \mathbb {P}\Bigl (f|(S_e S_f)^{r}e\Bigr ) = \mathbb {P}\Bigl ((S_f S_e)^{r}f|e\Bigr ) = sin^{2}\Bigl ((2r+1)arcsin(\sqrt{p})\Bigr ). \end{aligned}$$

Proof

The first equality follows from the invariance of $\mathbb {P}(\ |\ )$ under the automorphism $(S_f S_e)^{r}$, which is its own inverse. For the proof of the second equality, consider the following four elements in the order-unit space A: $b_1 := e$, $b_2 := f$, $b_3 := U_{e^{\prime }} f$ and $b_4 := U_{f^{\prime }} e$. Note that Lemmas 2(a) and 4 are repeatedly applied in the following calculations. $S_f S_e b_1 = S_f S_e e = S_f e = 2 U_f e + 2 U_{f^{\prime }} e - e = 2 p f + 2 U_{f^{\prime }} e - e = - b_1 + 2 p b_2 + 2 b_4$

Then use the identity $S_e f = 2 U_e f + 2 U_{e^{\prime }} f - f = 2 p e + 2 U_{e^{\prime }} f - f$ to get

$$\begin{aligned} S_f S_e b_2= & {} S_f S_e f = 2 p S_f e + 2 S_f U_{e^{\prime }} f - S_f f \\= & {} 2 p (2 U_f e + 2 U_{f^{\prime }} e - e) + 2 (2 U_f U_{e^{\prime }} f + 2 U_{f^{\prime }} U_{e^{\prime }} f - U_{e^{\prime }} f) - f \\= & {} 2 p (2 p f + 2 U_{f^{\prime }} e - e) + 2 (2 (1-p)^{2} f + 2 U_{f^{\prime }} U_{e} f - U_{e^{\prime }} f) - f \\= & {} 2 p (2 p f + 2 U_{f^{\prime }} e - e) + 2 (2 (1-p)^{2} f + 2 p U_{f^{\prime }} e - U_{e^{\prime }} f) - f \\= & {} - 2 p e + (8p^{2} - 8p + 3) f - 2 U_{e^{\prime }} f + 8 p U_{f^{\prime }} e \\= & {} - 2 p b_1 + (8p^{2} - 8p + 3) b_2 - 2 b_3 + 8 p b_4 \end{aligned}$$

$$\begin{aligned} S_f S_e b_3= & {} S_f S_e U_{e^{\prime }} f = S_f U_{e^{\prime }} f = 2 U_f U_{e^{\prime }} f + 2 U_{f^{\prime }} U_{e^{\prime }} f - U_{e^{\prime }} f \\= & {} 2 (1-p)^{2} f + 2 U_{f^{\prime }} U_{e} f - U_{e^{\prime }} f = 2 (1-p)^{2} f + 2 p U_{f^{\prime }} e - U_{e^{\prime }} f \\= & {} 2 (1-p)^{2} b_2 - b_3 + 2 p b_4 \end{aligned}$$

$$\begin{aligned} S_f S_e b_4&= S_f S_e U_{f^{\prime }} e = 2 S_f U_e U_{f^{\prime }} e + 2 S_f U_{e^{\prime }} U_{f^{\prime }} e - S_f U_{f^{\prime }} e \\&= 2 (1 - p)^{2} S_f e + 2 S_f U_{e^{\prime }} U_{f} e - U_{f^{\prime }} e = 2 (1 - p)^{2} S_f e + 2 p S_f U_{e^{\prime }} f - U_{f^{\prime }} e \\&= 2 (1 - p)^{2} (2 U_f e + 2 U_{f^{\prime }}e - e) + 2 p ( 2 U_f U_{e^{\prime }} f + 2 U_{f^{\prime }} U_{e^{\prime }} f - U_{e^{\prime }} f ) - U_{f^{\prime }} e \\&= 2 (1 - p)^{2} (2 p f + 2 U_{f^{\prime }}e - e) + 2 p (2(1-p)^{2} f + 2 U_{f^{\prime }} U_{e} f - U_{e^{\prime }} f) - U_{f^{\prime }} e \\&= 2 (1 - p)^{2} (2 p f + 2 U_{f^{\prime }}e - e) + 2 p (2(1-p)^{2} f + 2 p U_{f^{\prime }} e - U_{e^{\prime }} f) - U_{f^{\prime }} e \\&= - 2 (1-p)^{2} e + 8 p (1-p)^{2} f - 2 p U_{e^{\prime }} f + (8 p^{2} - 8 p + 3) U_{f^{\prime }} e \\&= - 2 (1-p)^{2} b_1 + 8 p (1-p)^{2} b_2 - 2 p b_3 + (8 p^{2} - 8 p + 3) b_4 \end{aligned}$$

The linear subspace in A, generated by $b_1, b_2, b_3, b_4$, is invariant under $S_f S_e$, which follows from the above identities. With respect to this basis, the restriction of $S_f S_e$ to this subspace is represented by the following matrix:

$$\begin{aligned} M := \begin{pmatrix} -1 &{}\quad - 2 p &{}\quad 0 &{}\quad -2 (1-p)^{2}\\ \\ 2 p &{}\quad 8p^{2} - 8p + 3 &{}\quad 2 (1-p)^{2} &{}\quad 8 p(1-p)^{2} \\ \\ 0 &{}\quad -2 &{}\quad -1 &{}\quad -2p \\ \\ 2 &{}\quad 8p &{}\quad 2p &{}\quad 8p^{2} - 8p + 3 \end{pmatrix} \end{aligned}$$

The Jordan form of this $4\times 4$matrix is now computed in two steps, each one basically dealing with the better manageable $2\times 2$-matrices. First consider the following matrix $N_1$

$$\begin{aligned} N_1 = \begin{pmatrix} 1-p &{}\quad 0 &{}\quad 1-p &{}\quad 0 \\ \\ 0 &{}\quad 1-p &{}\quad 0 &{}\quad 1-p \\ \\ -1 &{}\quad 0 &{}\quad 1 &{}\quad 0 \\ \\ 0 &{}\quad -1 &{}\quad 0 &{}\quad 1 \\ \end{pmatrix} \end{aligned}$$

and its inverse

$$\begin{aligned} N_1^{-1} = \frac{1}{2(1-p)} \begin{pmatrix} 1 \ &{}\quad \ 0 \ &{}\quad p-1 &{}\quad 0 \\ \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad p-1 \\ \\ 1 &{}\quad 0 &{}\quad 1-p &{}\quad 0 \\ \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 1-p \\ \end{pmatrix} \end{aligned}$$

Then

$$\begin{aligned} N_1^{-1}MN_1 = \begin{pmatrix} -1 &{}\quad 2-4p &{}\quad 0 &{}\quad 0 \\ \\ -2 + 4p &{}\quad (3-4p)(1-4p) &{}\quad 0 &{}\quad 0 \\ \\ 0 &{}\quad 0 &{}\quad -1 &{}\quad -2 \\ \\ 0 &{}\quad 0 &{}\quad 2 &{}\quad 3 \\ \end{pmatrix} \end{aligned}$$

The Jordan forms of the two $2 \times 2$ submatrices top left and bottom right can be calculated separately. With

$$\begin{aligned} N_2 = \begin{pmatrix} 1 &{}\quad 1 &{}\quad 0 &{}\quad 0 \\ \\ 1 - 2p + 2 \sqrt{p(1-p)} \ i &{}\quad 1 - 2p - 2 \sqrt{p(1-p)} \ i &{}\quad 0 &{}\quad 0 \\ \\ 0 &{}\quad 0 &{}\quad 1 &{}\quad -2 \\ \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 2 \end{pmatrix} \end{aligned}$$

and

$$\begin{aligned} N_2^{-1} = \begin{pmatrix} \frac{1}{2} + \frac{1-2p}{4 \sqrt{p(1-p)}} i &{}\quad \frac{-1}{4\sqrt{p(1-p)}} i &{}\quad 0 &{}\quad 0 \\ \\ \frac{1}{2} - \frac{1-2p}{4 \sqrt{p(1-p)}} i &{}\quad \frac{1}{4\sqrt{p(1-p)}} i &{}\quad 0 &{}\quad 0 \\ \\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 1 \\ \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \frac{1}{2} \\ \end{pmatrix} \end{aligned}$$

the desired Jordan form of M is:

$$\begin{aligned} N_2^{-1} N_1^{-1} M N_1 N_2 = \begin{pmatrix} \alpha _1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ \\ 0 &{}\quad \alpha _2 &{}\quad 0 &{}\quad 0 \\ \\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 \\ \\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 1 \\ \end{pmatrix} \end{aligned}$$

where

$$\begin{aligned} \alpha _1 = 8p^{2} - 8p + 1 + 4(1-2p) \sqrt{p(1-p)} \ i, \end{aligned}$$

$$\begin{aligned} \alpha _2 = 8p^{2} - 8p + 1 - 4(1-2p) \sqrt{p(1-p)} \ i \end{aligned}$$

and 1 are the eigenvalues of M. This (almost diagonal) matrix can now easily be raised to the r-th power, and $M^{r}$ can be calculated:

$$\begin{aligned} M^{r}&= N_1 N_2 \begin{pmatrix} \alpha _1^{r} &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ \\ 0 &{}\quad \alpha _2^{r} &{}\quad 0 &{}\quad 0 \\ \\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 \\ \\ 0 &{}\quad 0 &{}\quad r &{}\quad 1 \\ \end{pmatrix} N_2^{-1} N_1^{-1}\\&= \begin{pmatrix} \cdots &{}\quad - r + \frac{1}{4\sqrt{p(1-p)}} Im (\alpha _1^{r}) &{}\quad \cdots &{}\quad \cdots \\ \\ \cdots &{}\quad \frac{1}{2}(2 r + 1 + Re(\alpha _1^{r}) + \frac{1-2p}{2\sqrt{p(1-p)}} Im(\alpha _1^{r})) &{}\quad \cdots &{}\quad \cdots \\ \\ \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots \\ \\ \cdots &{}\quad \frac{1}{2(1-p)} (2 r + 1 - Re(\alpha _1^{r}) - \frac{1-2p}{2\sqrt{p(1-p)}} Im(\alpha _1^{r})) &{}\quad \cdots &{}\quad \cdots \\ \end{pmatrix} \end{aligned}$$

Here, Re(z) [Im(z)] denotes the real [imaginary] part of the complex number z. Since $\alpha _2$ is the complex conjugate of $\alpha _1$, it does not anymore appear in this matrix. Note that only the second column is displayed, since only these entries will be used for the following calculation of $U_e (S_f S_e)^{r} b_2 = U_e (S_f S_e)^{r} f$. The third entry in this column is not needed, since $U_e b_3 = U_e U_{e^{\prime }} f = 0$. Moreover, recall that $U_e b_1 = U_e e = e$, $U_e b_2 = U_e f = p e$ and $U_e b_4 = U_e U_{f^{\prime }} e= (1 - p)^{2}e$.

$$\begin{aligned}&U_e (S_f S_e)^{r} f\\&\quad = \left( - r + \frac{1}{4\sqrt{p(1-p)}} Im (\alpha _1^{r}) \right) U_e(b_1) \\&\qquad + \frac{1}{2}\left( 2 r + 1 + Re(\alpha _1^{r}) + \frac{1-2p}{2\sqrt{p(1-p)}} Im(\alpha _1^{r})\right) U_e(b_2) \\&\qquad + \frac{1}{2(1-p)} \left( 2 r + 1 - Re(\alpha _1^{r}) - \frac{1-2p}{2\sqrt{p(1-p)}} Im(\alpha _1^{r})\right) U_e(b_4)\\&\quad = \left( - r + \frac{1}{4\sqrt{p(1-p)}} Im (\alpha _1^{r}) \right) e\\&\qquad + \frac{1}{2} \left( 2 r + 1 + Re(\alpha _1^{r}) + \frac{1-2p}{2\sqrt{p(1-p)}} Im(\alpha _1^{r}) \right) p e\\&\qquad + \frac{1}{2(1-p)} \left( 2 r + 1 - Re(\alpha _1^{r}) - \frac{1-2p}{2\sqrt{p(1-p)}} Im(\alpha _1^{r}) \right) (1 -p)^{2}e\\&\quad = \left( \frac{1}{2} - \frac{1-2p}{2} Re(\alpha _1^{r}) + \sqrt{p(1-p)} Im(\alpha _1^{r}) \right) e \end{aligned}$$

Therefore

$$\begin{aligned} \mathbb {P}((S_f S_e)^{r}f|e) = \frac{1}{2} - \frac{1-2p}{2} Re(\alpha _1^{r}) + \sqrt{p(1-p)} Im(\alpha _1^{r}) \end{aligned}$$

Since $\left| \alpha _1 \right| = 1$, $\alpha _1 = e^{it}$ with $t = arcsin (4(1-2p)\sqrt{p(1-p)})$. Furthermore, define $s := arcsin(2\sqrt{p(1-p)}$. Then $cos(s) = 1-2p$, since $(1-2p)^{2} + (2\sqrt{p(1-p)})^{2} = 1$, and

$$\begin{aligned} \mathbb {P}(f|(S_e S_f)^{r}e)&= \frac{1}{2} - \frac{1}{2} \Bigl ( cos(s) cos(rt) - sin(s) sin (rt) \Bigr ) \\&= \frac{1}{2} - \frac{1}{2} cos(s + rt) \\&= sin^{2} \Bigl ( \frac{s + rt}{2} \Bigr ) \\&= sin^{2} \Bigl ( (2r+1)arcsin(\sqrt{p}) \Bigr ). \end{aligned}$$

The second and the third equality follow from the trigonometric identities $cos(x) + cos(y) = cos(x) cos(y) - sin(x) sin(y)$ and $sin^{2}(\frac{x}{2}) = \frac{1 - cos(x)}{2}$. The last equality follows from the definitions of s and t and the following identity:

$$\begin{aligned} arcsin\left( 2\sqrt{x - x^{2}}\right) + r \ arcsin\left( 4(1-2x)\sqrt{x-x^{2}}\right) - (4r+2) arcsin\left( \sqrt{x}\right) = 0 \end{aligned}$$

by inserting $x = p$, which then gives $s+rt = (4r+2) arcsin(\sqrt{p})$. This identity can be proved by differentiation with respect to x: The derivative is constantly zero and the function thus constant; checking the function for $x=0$ yields that it is constantly zero. $\square$