EPR-Bell-Schrödinger Proof of Nonlocality Using Position and Momentum

Abstract

Based on his extension of the classical argument of Einstein, Podolsky and Rosen, Schrödinger observed that, in certain quantum states associated with pairs of particles that can be far away from one another, the result of the measurement of an observable associated with one particle is perfectly correlated with the result of the measurement of another observable associated with the other particle. Combining this with the assumption of locality and some “no hidden variables” theorems, we showed in a previous paper [11] that this yields a contradiction. This means that the assumption of locality is false, and thus provides us with another demonstration of quantum nonlocality that does not involve Bell’s (or any other) inequalities. In [11] we introduced only “spin-like” observables acting on finite dimensional Hilbert spaces. Here we will give a similar argument using the variables originally used by Einstein, Podolsky and Rosen, namely position and momentum.

We dedicate this paper to the memory of Giancarlo Ghirardi, who devoted his life to understanding quantum mechanics. He was a friend of John Bell, who was inspired by Giancarlo’s work. He was also a friend of two of us (J.B. and S.G.)

To appear in: Do wave functions jump? Perspectives on the work of G. C. Ghirardi. Editors: V. Allori, A. Bassi, D. Dürr, N. Zanghì.

Appendix 1: Proof of Clifton’s Theorem 4.1

The proof we give here is taken from a paper by Myrvold [29], which is a simplified version of the result of Clifton [14] and is similar to proofs of Mermin [28], and to Peres [31, 32] in the case of spins. We note that the same proof would apply to the regularized EPR state of Sect. 3.6.

Proof of Theorem 4.1

We will need the operators $U_j (b)= \exp (-i b Q_j)$, $V_j (c)= \exp (-i c P_j)$, $j=1, 2$, with $Q_j$, $P_j$ defined by formulas (4.3), (4.4), but acting in $L^2(\mathbb R^2)$ instead of $L^2(\mathbb R^4)$, and $b, c \in \mathbb R$. They act as

$$\begin{aligned} U_j (b)\Psi ( x_1, x_2)=\exp (-i b x_j) \Psi ( x_1, x_2)\;,\quad j=1,2\;, \end{aligned}$$

(A.1)

which follows trivially from (4.3), and

$$\begin{aligned} V_1 (c) \Psi ( x_1, x_2)= \Psi ( x_1-c, x_2)\;, \end{aligned}$$

(A.2)

and similarly for $V_2 (c)$. Equation (A.2) follows from (4.4) by expanding both sides in a Taylor series, for functions $\Psi$ such that the series converges, and by extending the unitary operator $V_2 (b)$ to more general functions $\Psi$ (see, e.g., [33, Chap. 8] for an explanation of that extension).

We choose now the following functions of the operators $Q_i$, $P_i\,$:

$$\begin{aligned} A_1 = \cos (a Q_1)\;,\quad A_2 = \cos (a Q_2) \;,\quad B_1 = \cos \frac{\pi P_1}{a}\;,\quad B_2 = \cos \frac{\pi P_2}{a}\;,\qquad \quad \end{aligned}$$

(A.3)

where a is an arbitrary constant, and the functions are defined by (A.1), (A.2), and the Euler relations:

$$\begin{aligned} \begin{array}{l} \displaystyle \cos (a Q_j) =\frac{ \exp (i a Q_j)+ \exp (-i a Q_j)}{2}\;,\\ \displaystyle \cos \frac{\pi P_j}{ a} =\frac{ \exp ( i\pi P_j/ a)+ \exp (- i\pi P_j/ a)}{2}\;, \end{array} \end{aligned}$$

(A.4)

for $j=1, 2$. Note that $A_1,A_2, B_1,B_2$ are self-adjoint. By applying (4.8) several times to pairs of commuting operators made of products of such operators, we will derive a contradiction.

We have the relations

$$\begin{aligned}{}[A_1,A_2]= [B_1,B_2]=[A_1,B_2]=[A_2,B_1]=0\;, \end{aligned}$$

(A.5)

since the relevant operators act on different variables.

We can also prove:

$$\begin{aligned} A_1B_1 = -B_1A_1\;,\qquad A_2B_2 = -B_2A_2\;. \end{aligned}$$

(A.6)

To show (A.6), note that, from (A.1) and (A.2), one gets

$$\begin{aligned} U_j (b) V_j (c)= \exp (-ibc) V_j (c)U_j (b)\;, \end{aligned}$$

(A.7)

for $j=1,2$, which, for $bc=\pm \pi$, means

$$\begin{aligned} U_j (b) V_j (c)= - V_j (c)U_j (b)\;. \end{aligned}$$

(A.8)

Now use (A.4) to expand the product $\cos (a Q_j) \cos (\pi P_j/ a)$, for $j=1,2$, into a sum of four terms; each term will have the form of the left-hand side of (A.7) with $b=\pm a$, $c= \pm \pi / a$, whence $bc=\pm \pi$. Then applying (A.8) to each term proves (A.6).

The relations (A.5) and (A.6) imply that

$$\begin{aligned}{}[A_1A_2, B_2 B_1]= 0 \end{aligned}$$

(A.9)

since two anticommutations (A.6) suffice to move the B’s to the left of the A’s. Similarly we have that

$$\begin{aligned}{}[A_1B_2, A_2 B_1]= 0. \end{aligned}$$

(A.10)

We also have, using (A.6) once, that

$$\begin{aligned} A_1A_2 B_2 B_1=-A_1 B_2 A_2 B_1. \end{aligned}$$

(A.11)

Thus, with $C= (A_1 A_2)(B_2 B_1)$ and $D= (A_1 B_2)(A_2 B_1)$, we have that

$$\begin{aligned} C=-D. \end{aligned}$$

(A.12)

Now suppose there is a value map v as described in Theorem 4.1. Then, from (4.7) with $f(x)=-x$, we have that

$$\begin{aligned} v(C)=-v(D). \end{aligned}$$

(A.13)

But by (A.5), (A.9) and (A.10), we also have, by (4.8), that

$$\begin{aligned} v(C)=v(A_1 A_2) v(B_2 B_1)= v(A_1)v(A_2)v(B_2)v(B_1) \end{aligned}$$

(A.14)

and

$$\begin{aligned} v(D)=v(A_1 B_2) v(A_2 B_1)= v(A_1)v(B_2)v(A_2)v(B_1). \end{aligned}$$

(A.15)

Thus $v(C)= v(D)$. This is a contradiction unless $v(C)=0$, i.e. unless at least one of $v(A_i)$, $v(B_i)$, $i=1,2$ vanishes. But, by (4.7),

$$\begin{aligned} v(A_i) =\cos (a v(Q_i)) \end{aligned}$$

and

$$\begin{aligned} v(B_i)=\cos \left( \frac{\pi }{a} v(P_i)\right) , \end{aligned}$$

and thus a can be so chosen that $v(A_i)$ and $v(B_i)$ are all nonvanishing. $\blacksquare$