The de Broglie–Bohm Theory

Abstract

The de Broglie-Bohm theory, which is a deterministic theory of matter in motion, is explained in this chapter. We first show what the trajectories look like and, then, how to derive the usual quantum predictions, both for the positions of particles and for other observables, using spin and momentum as examples. This derivation rests on the assumption of quantum equilibrium, which we shall explain and discuss. We also see how nonlocality appears in the de Broglie-Bohm theory and we shall present and answer several objections to that theory.

Appendices

Appendices

5.A The Continuity Equation for the $|\Psi |^2$ Distribution

Let us consider Schrödinger’s equation (5.1.1.1) with H given by (5.1.1.2). We will first prove the following continuity equation:

$$\begin{aligned} \frac{\partial |\Psi ({\mathbf{x}}, t)|^2}{ \partial t} + \nabla \cdot {\mathbf{j}}({\mathbf{x}}, t)= 0\;, \end{aligned}$$

(5.A.1)

where $\nabla = (\partial / \partial x_1, \ldots , \partial / \partial x_N)$ is the gradient, the dot denotes the scalar product, and ${\mathbf{j}}({\mathbf{x}},t)= (j_k ({\mathbf{x}}, t))_{k=1}^N$ (we index the components here by k instead of j) is the “probability current” or the “quantum flux” defined by

$$\begin{aligned} j_k ({\mathbf{x}}, t)&= \frac{1}{ 2i m_k } \left[ \Psi ^*({\mathbf{x}}, t) \frac{\partial }{ \partial x_k} \Psi ({\mathbf{x}}, t) - \Psi ({\mathbf{x}}, t)\frac{\partial }{ \partial x_{k}} \Psi ^*({\mathbf{x}}, t)\right] \\&= \frac{1}{m_k } \mathrm{Im}\left[ \Psi ^*({\mathbf{x}}, t)\frac{\partial }{ \partial x_k}\Psi ({\mathbf{x}}, t)\right] \;. \end{aligned}$$

(5.A.2)

To prove (5.A.1), write

$$\begin{aligned} \frac{\partial |\Psi ({\mathbf{x}}, t)|^2}{ \partial t} = \Psi ^*({\mathbf{x}}, t)\frac{\partial \Psi ({\mathbf{x}}, t)}{ \partial t} + \Psi ({\mathbf{x}}, t) \frac{\partial \Psi ^*({\mathbf{x}}, t)}{ \partial t}\;, \end{aligned}$$

(5.A.3)

take the complex conjugate of (5.1.1.1), viz.,

$$\begin{aligned} -i \frac{\partial \Psi ^*}{ \partial t} ({\mathbf{x}}, t) = H \Psi ^*({\mathbf{x}}, t)\;, \end{aligned}$$

(5.A.4)

and insert this and (5.1.1.1) into (5.A.3). The potential term cancels out and we get

$$\begin{aligned} \frac{\partial |\Psi ({\mathbf{x}}, t)|^2}{ \partial t} = -\frac{1}{ i} \sum ^N_{k=1} \frac{1}{ 2 m_k} \left[ \Psi ^*({\mathbf{x}}, t) \frac{\partial ^2 }{ \partial x_k^2} \Psi ({\mathbf{x}}, t) - \Psi ({\mathbf{x}}, t) \frac{\partial ^2 }{ \partial x_k^2} \Psi ^*({\mathbf{x}}, t)\right] \;. \end{aligned}$$

(5.A.5)

It is enough to observe that

$$\begin{aligned}&\frac{\partial }{ \partial x_k} \left[ \Psi ^*({\mathbf{x}}, t) \frac{\partial }{ \partial x_k} \Psi ({\mathbf{x}}, t) - \Psi ({\mathbf{x}}, t) \frac{\partial }{ \partial x_k} \Psi ^*({\mathbf{x}}, t)\right] =\Psi ^*({\mathbf{x}}, t) \frac{\partial ^2 }{ \partial x_k^2} \Psi ({\mathbf{x}}, t) \\&\quad - \Psi ({\mathbf{x}}, t)\frac{\partial ^2 }{ \partial x_k^2} \Psi ^*({\mathbf{x}}, t) \end{aligned}$$

to obtain (5.A.1).

Now, if we consider (5.1.1.4), we see that the function ${\mathbf{j}}({\mathbf{x}},t)= (j_k ({\mathbf{x}}, t))_{k=1}^N$ [whose arguments are the generic variables ${\mathbf{x}}$ rather than the actual positions of the particles as in (5.1.1.4)] is of the form

$$\begin{aligned} {\mathbf{j}}({\mathbf{x}})={\mathbf{V}}_{\Psi } ({\mathbf{x}}, t) \ |\Psi ({\mathbf{x}}, t)|^2\;, \end{aligned}$$

(5.A.6)

and we can write (5.A.1) as

$$\begin{aligned} \frac{\partial |\Psi ({\mathbf{x}},t)|^2}{ \partial t} + \nabla \cdot \Big [{\mathbf{V}}_{\Psi }({\mathbf{x}}, t) |\Psi ({\mathbf{x}},t) |^2\Big ]=0\;, \end{aligned}$$

(5.A.7)

which is the continuity equation⁹⁷ one would obtain for a “fluid” of density $\rho ({\mathbf{x}},t)=|\Psi ({\mathbf{x}},t)|^2$, with local velocity ${\mathbf{V}}_{\Psi } ({\mathbf{x}},t)$. This will be useful when we come to prove the equivariance of the $|\Psi ({\mathbf{x}},t)|^2$ distribution in Appendix 5.C.

5.B A Second Order Dynamics and the Quantum Potential

Here we derive (5.1.1.5) from (5.1.1.1)–(5.1.1.3). We insert $\Psi ({\mathbf{x}},t)=R({\mathbf{x}},t)e^{iS({\mathbf{x}},t)}$ into Schrödinger’s equation (5.1.1.1), where ${\mathbf{x}}= (x_1, \dots , x_N)$. Suppressing here the arguments $({\mathbf{x}}, t)$, we get

$$\begin{aligned} \frac{i \partial R}{ \partial t} e^{iS} -\frac{\partial S}{\partial t} \Psi = - \sum _{j=1}^N \frac{1}{ 2 m_j}\left[ \frac{\partial ^2 R}{ \partial x_j^2} - R \left( \frac{\partial S}{ \partial x_j}\right) ^2 +2i \frac{\partial R}{ \partial x_j} \frac{\partial S}{ \partial x_j} + iR\frac{\partial ^2 S}{ \partial x_j^2} \right] e^{iS} + V\Psi \;. \end{aligned}$$

(5.B.1)

Dividing both sides by $e^{iS}$ and separating the real and imaginary parts give rise to two equations:

$$\begin{aligned}&\frac{R\partial S}{ \partial t} = \sum _{j=1}^N \frac{1}{ 2 m_j} \left[ \frac{\partial ^2R}{ \partial x_j^2} - R\left( \frac{\partial S }{ \partial x_j}\right) ^2 \right] - VR \;,\end{aligned}$$

(5.B.2)

$$\begin{aligned}&\frac{\partial R}{ \partial t} + \sum _{j=1}^N \frac{1}{ 2 m_j} \left[ 2 \frac{\partial R}{ \partial x_j} \frac{\partial S}{ \partial x_j} + R\frac{\partial ^2 S}{ \partial x_j^2} \right] =0\;. \end{aligned}$$

(5.B.3)

Dividing (5.B.2) by R gives

$$\begin{aligned} \frac{\partial S}{ \partial t} = \sum _{j=1}^N \frac{1}{ 2 m_j} \left[ \frac{1}{ R} \frac{\partial ^2 R}{ \partial x_j^2} - \left( \frac{\partial S}{ \partial x_j}\right) ^2\right] - V\;. \end{aligned}$$

(5.B.4)

Multiplying (5.B.3) by 2R gives

$$ \frac{\partial R^2}{ \partial t} + \sum _{j=1}^N \frac{1}{ m_j} \frac{\partial }{ \partial x_j} \left( R^2 \frac{\partial S}{ \partial x_j} \right) =0\;, $$

which, with $R^2=|\Psi |^2$ and using (5.A.6), is the same as (5.A.1).

Now, if we consider (5.1.1.3) and take its time derivative, remembering that S depends on time through $\big (X_1(t),\ldots ,X_N(t)\big )$ and t, we get

$$\begin{aligned} m_k\frac{d^2}{ dt^2} X_k (t) = \sum _{j=1}^N \frac{\partial ^2 S}{ \partial x_j \partial x_k} \dot{X}_j + \frac{\partial ^2S}{\partial x_k \partial t}\;. \end{aligned}$$

(5.B.5)

If we use (5.1.1.3) to replace $\dot{X}_j$ by $(1/m_j)(\partial S/\partial x_j)$ in (5.B.5), and insert (5.B.4) in the second term of (5.B.5), evaluating the functions at $\mathbf{X}$ instead of $\mathbf{x}$, the terms

$$ \sum _{j=1}^N \frac{1}{ m_j}\frac{\partial ^2 S}{ \partial x_j \partial x_k} \frac{\partial }{\partial x_j} S \quad \text{ and }\quad -\frac{\partial }{\partial x_k} \sum _{j=1}^N \frac{1}{2 m_k} \left( \frac{\partial S}{\partial x_j}\right) ^2 $$

cancel each other and we get

$$\begin{aligned} m_k\frac{d^2}{ dt^2} X_k (t) =-\frac{\partial }{\partial x_k} (Q+V)\;, \end{aligned}$$

where the quantum potential Q is given by

$$\begin{aligned} Q=- \sum _{j=1}^N \frac{1}{ 2 m_j} \frac{1}{R} \frac{\partial ^2 R}{\partial x_j^2}\;. \end{aligned}$$

(5.B.6)

This proves (5.1.1.5).

5.C Proof of the Equivariance of the $|\Psi ({\mathbf{x}},t)|^2$ Distribution

5.1.1 5.C.1 Equivariance in $\mathbb {R}^N$

We will first prove⁹⁸ that, for a system in $\mathbb {R}^N$ and $\forall A \subset \mathbb {R}^N$,

$$\begin{aligned} P\big ({\mathbf{X}}(t)\in A\big )=\int _A |\Psi ({\mathbf{x}},t)|^2 d{\mathbf{x}}\;, \end{aligned}$$

(5.C.1.1)

provided that, at $t=0$, $\forall A \subset \mathbb {R}^N$,

$$\begin{aligned} P\big ({\mathbf{X}}(0)\in A\big ) = \int _A |\Psi ({\mathbf{x}},0)|^2 d{\mathbf{x}}\;. \end{aligned}$$

(5.C.1.2)

In (5.C.1.1), the left-hand side involves ${\mathbf{X}}(t)$, the solution of (5.1.1.4), while on the right-hand side, $\Psi ({\mathbf{x}},t)$ is the solution of Schrödinger’s equation (5.1.1.1).

Consider first a general differential equation

$$\begin{aligned} \frac{d{\mathbf{X}}(t)}{ dt} = {\mathbf{V}}\big ({\mathbf{X}}(t),t\big )\;, \end{aligned}$$

(5.C.1.3)

with ${\mathbf{X}}:\mathbb {R}\rightarrow \mathbb {R}^N$. We assume that ${\mathbf{V}}$ is such that (5.C.1.3) has solutions for all times. Let $\phi ^t({\mathbf{x}})$ denote the solution at time t of (5.C.1.3) with initial condition ${\mathbf{X}}(0)= {\mathbf{x}}$ at time 0. Then, given any probability density $\rho ({\mathbf{x}}, 0)$ on the initial conditions ${\mathbf{X}}(0)={\mathbf{x}}$, define⁹⁹ $\rho ({\mathbf{x}},t)$, $\forall A \subset \mathbb {R}^N$, by

$$\begin{aligned} \int _A \rho ({\mathbf{x}},t) d{\mathbf{x}}= P\big ({\mathbf{X}}(t)\in A\big ) = \int {\mathbbm {1}}_A \big (\phi ^t ({\mathbf{x}})\big )\rho ({\mathbf{x}}, 0)d{\mathbf{x}}\;, \end{aligned}$$

(5.C.1.4)

where ${\mathbbm {1}}_A$ is the indicator function of the set A.

Then we have

$$\begin{aligned} \frac{\partial \rho ({\mathbf{x}},t)}{ \partial t} + \mathbbm {\nabla } \cdot \Big [{\mathbf{V}}({\mathbf{x}},t)\rho ({\mathbf{x}},t)\Big ]=0\;. \end{aligned}$$

(5.C.1.5)

To prove (5.C.1.5), observe that, by approximating integrals by sums, for any smooth function with compact support, we get from (5.C.1.4):

$$\begin{aligned} \int F({\mathbf{x}}) \rho ({\mathbf{x}},t)d{\mathbf{x}}= \int F \big (\phi ^t({\mathbf{x}})\big ) \rho ({\mathbf{x}}, 0)d{\mathbf{x}}\;. \end{aligned}$$

(5.C.1.6)

Then, differentiating both sides of (5.C.1.6) with respect to t, and using the fact that (5.C.1.3) implies $\partial \phi ^t ({\mathbf{x}})/\partial t = {\mathbf{V}}\big (\phi ^t ({\mathbf{x}}), t\big )$, we have

$$\begin{aligned} \int F({\mathbf{x}}) \frac{\partial \rho ({\mathbf{x}},t)}{ \partial t} d{\mathbf{x}}= \int \mathbbm {\nabla } F \big (\phi ^t ({\mathbf{x}})\big ) \cdot {\mathbf{V}}\big (\phi ^t ({\mathbf{x}}),t\big ) \rho ({\mathbf{x}}, 0)d{\mathbf{x}}\;. \end{aligned}$$

Using (5.C.1.6) with $F (\cdot )$ replaced by ${\mathbbm {\nabla }} F (\cdot ) \cdot {\mathbf{V}}(\cdot ,t)$, the right-hand side of this equals

$$\begin{aligned} \int {\mathbbm {\nabla }} F ({\mathbf{x}}) \cdot {\mathbf{V}}({\mathbf{x}},t) \rho ({\mathbf{x}},t)d{\mathbf{x}}\;. \end{aligned}$$

Integrating by parts, this is

$$\begin{aligned} - \int F({\mathbf{x}}) \nabla \cdot \Big [{\mathbf{V}}({\mathbf{x}},t)\rho ({\mathbf{x}},t)\Big ] d{\mathbf{x}}\;. \end{aligned}$$

(5.C.1.7)

Since (5.C.1.7) is valid for any F, it implies (5.C.1.5).

Now, if we consider the differential equation (5.C.1.3) with ${\mathbf{V}}({\mathbf{X}},t)={\mathbf{V}}_{\Psi } ({\mathbf{X}}, t)$, with ${\mathbf{V}}_{\Psi } ({\mathbf{X}}, t)$ given by (5.1.1.4), and $\rho ({\mathbf{x}},t)=|\Psi ({\mathbf{x}},t)|^2$, we get, using (5.C.1.4), (5.C.1.5), and (5.A.7),

$$\begin{aligned} \frac{d}{ dt} P\big ({\mathbf{X}}(t)\in A\big )&= \int _A \frac{\partial \rho ({\mathbf{x}},t)}{\partial t} d{\mathbf{x}} \nonumber \\&= - \int _A \nabla \cdot \big [{\mathbf{V}}_{\Psi } ({\mathbf{x}}, t)\rho ({\mathbf{x}},t)\big ] d{\mathbf{x}} \nonumber \\&= \frac{d}{dt} \int _A |\Psi ({\mathbf{x}},t)|^2 d{\mathbf{x}}\;, \end{aligned}$$

(5.C.1.8)

which proves (5.C.1.1) if we assume that (5.C.1.1) holds at $t=0$, which is (5.C.1.2).

5.1.2 5.C.2 Proof of Equivariance of the Empirical Distributions

To finish the proof of (5.1.3.2), consider a large number N of copies of non-interacting identical systems,¹⁰⁰ whose initial wave function¹⁰¹ is

$$\begin{aligned} \Psi ({\mathbf{x}}, 0)=\Psi (x_1, \dots , x_N, 0)= \prod _{i=1}^N \psi (x_i, 0) \;. \end{aligned}$$

(5.C.2.1)

The density of the associated probability distribution is

$$\begin{aligned} |\Psi (x_1, \dots , x_N, 0)|^2= \prod _{i=1}^N |\psi (x_i, 0)|^2\;, \end{aligned}$$

(5.C.2.2)

which means that the variables $x_i$ are independent identically distributed random variables with distribution $|\psi (x, 0)|^2$. This implies, by the law of large numbers (see Appendix 3.A), that the empirical density distribution $\rho (x,0)$ of the N particles will be given, as $N\rightarrow \infty$, by $\rho (x,0)= |\psi (x, 0)|^2$.

By (5.C.1.1), the probability distribution density of $(x_1, \dots , x_N)$ at time $t>0$ will be

$$\begin{aligned} |\Psi (x_1, \dots , x_N, t)|^2= \prod _{i=1}^N |\psi (x_i, t)|^2\;, \end{aligned}$$

(5.C.2.3)

and the empirical density distribution $\rho (x,t)$ of the N particles at time t will be given, as $N\rightarrow \infty$, by $\rho (x,t)= |\psi (x, t)|^2$. This completes the proof of (5.1.3.2).

5.D Examples of de Broglie–Bohm Dynamics

5.1.1 5.D.1 The Gaussian Wave Function

Consider the time evolution of a Gaussian wave function (2.A.2.23), with initial condition $\Psi (x, 0)=\pi ^{-1/4}\exp (- x^2/2)$, which we copy here¹⁰²:

$$\begin{aligned} \Psi (x,t) = \frac{1}{(1+it/m)^{1/2}} \frac{1}{\pi ^{1/4}}\exp \left[ - \frac{x^2}{2 (1+it/m)}\right] \;. \end{aligned}$$

(5.D.1.1)

It is easy to see that, if we write $\Psi (x,t)= R (x,t) \exp \big [iS (x,t)\big ]$, we have (up to a constant term)

$$ S(x,t)= \frac{t x^2}{2m (1+t^2/m^2)}\;, $$

so that (5.1.1.3) gives rise to the equation of motion

$$\begin{aligned} \frac{d}{dt} X(t)= \frac{t X(t)}{m^2+t^2}\;, \end{aligned}$$

(5.D.1.2)

which can be easily integrated by a separation of variables:

$$\begin{aligned} X(t)=\frac{X(0)}{m} \sqrt{m^2+t^2}= X(0) \sqrt{1+\left( \frac{t}{m}\right) ^2}\;. \end{aligned}$$

(5.D.1.3)

This gives the explicit dependence of the position at time t on the initial condition and shows that the particle does not move if it is initially at $X(0)=0$, but moves asymptotically, when $t\rightarrow \infty$, according to $X(t)\sim X(0)t/m$.

Equation (2.A.3.3) can be checked explicitly here. Indeed, putting $m=1$ as in Appendix 2.A.3, we get from (2.A.2.24)

$$\begin{aligned} t|\Psi ( pt, t)|^2= t\frac{1}{\sqrt{\pi (1+t^2)} }\exp \left( - \frac{p^2t^2}{ 1+t^2}\right) \;, \end{aligned}$$

(5.D.1.4)

whose limit as $t \rightarrow \infty$ is $\pi ^{-1/2}\exp (- p^2) =|\widehat{\Psi }(p,0)|^2$.

It is also easy to verify explicitly the property (5.1.3.2) of equivariance. From (5.D.1.3), we get $\phi ^t(x)= x \sqrt{1+(t/m)^2}$. If we make the change of variable $y= \phi ^t(x)$ in the integral on the right-hand side of (5.1.3.1), we get, with the initial distribution $\rho (x,0)= | \Psi (x,0)|^2= \pi ^{-1/2}\exp (- x^2)$,

$$\begin{aligned} \rho (y,t)= \frac{1}{\sqrt{\pi \big [1+(t/m)^2\big ]} }\exp \left[ - \frac{y^2}{ 1+(t/m)^2}\right] \;. \end{aligned}$$

(5.D.1.5)

Changing the y variable back to x, this coincides with $| \Psi (x,t)|^2$ given by (2.A.2.24), with the prefactor $1/\sqrt{1+(t/m)^2}$ coming from $dx= dy/\sqrt{1+(t/m)^2}$.

5.1.2 5.D.2 The Particle in a Box

Consider a particle in a box,¹⁰³ in one dimension, of size L, chosen to be the interval [0, L]. The particle is free in the box, but constrained to remain in it. Let us look for solutions of the Schrödinger equation (5.1.1.1), with $N=1$, of the form $\exp (-iEt) \Psi (x)$, whose associated probability distribution $|\exp (-iEt) \Psi (x)|^2= | \Psi (x)|^2$ is independent of time.¹⁰⁴

We model the presence of the box by setting the potential in (5.1.1.2) such that $V=0$ inside the box and $V=\infty$ outside the box, which means that $\Psi (x)$ vanishes outside V. So we have to solve

$$\begin{aligned} \frac{1}{2} \frac{d^2}{dx^2} \Psi (x) =- E \Psi (x)\;, \end{aligned}$$

(5.D.2.1)

with $\Psi (0) = \Psi (L)=0$. We have set the mass $m=1$.

The general solution of (5.D.2.1) is

$$\begin{aligned} A \exp \big (i \sqrt{2E} x\big ) + B \exp \big (-i \sqrt{2E} x\big )\;, \end{aligned}$$

(5.D.2.2)

and the conditions $\Psi (0)=\Psi (L)=0$ imply $A=-B$ and $E_n=n^2\pi ^2/ L^2$, for $n\in \mathbbm {Z}$, $n \ne 0$. From (5.D.2.2), we get the eigenvectors

$$\begin{aligned} \Psi _n (x) = \sqrt{\frac{2}{L}} \sin (k_n x) {\mathbbm {1}}_{[0, L]} (x)\;, \end{aligned}$$

(5.D.2.3)

where ${\mathbbm {1}}_{[0, L]}$ is the indicator function of [0, L], and $k_n = n\pi /L$. The sine function being odd, we can restrict ourselves to $n=1,2,\ldots , \infty$, since the solution should be nonzero and is defined up to a multiplicative factor which does not affect the motion of the particle, as can be seen from (5.1.1.4).¹⁰⁵ The factor $\sqrt{2/L}$ ensures that $\int _{\mathbbm {R}} |\Psi _n(x)|^2 dx =1$.

If we take a function $\Psi _n(x)$ as initial condition in (5.1.1.1), then the time-dependent solution is

$$\begin{aligned} \Psi _n (x,t) = \sqrt{\frac{2}{L}} e^{-iE_nt} \sin (k_n x) {\mathbbm {1}}_{[0, L]} (x)\;. \end{aligned}$$

(5.D.2.4)

Now comes the apparent paradox: the phase $S_n$ of $\Psi _n (x,t)$ is constant in x, so $\partial S_n / \partial x=0$, and by (5.1.1.3), the particle is at rest in the box. Let us see how to solve that paradox. As we said, one method to measure the momentum is to remove the walls of the box and to detect the position x of the particle after a certain time t, and give x / t as a “measure” of momentum.

To see what happens when the walls of the box are removed, we have to compute the time evolution of the wave function with initial condition (5.D.2.3). Since the walls are removed, the time evolution is governed by the free Schrödinger equation (5.1.1.1). The solution is therefore obtained from (2.A.2.21) and we need to compute $\widehat{\Psi }(p,0)$ from (5.D.2.3), leaving out the index n on $\Psi$. Thus,

$$\begin{aligned} \widehat{\Psi }(p,0) =&\frac{1}{ \sqrt{2\pi }} \int _{{\mathbbm {R}}} e^{-ipx} \sqrt{\frac{2}{ L}} \sin (k_n x) {\mathbbm {1}}_{[0, L]} (x) dx\\ {}&= \frac{1}{ \sqrt{2\pi }} \int ^L_0 e^{-ipx} \sqrt{\frac{2}{ L}} \sin ( k_n x) dx\;. \end{aligned}$$

(5.D.2.5)

Writing $\sin (k_n x) = (e^{ik_n x}-e^{-ik_n x})/2i$, making the change of variable $x=x'+L/ 2$, and integrating over $x'\in [-L/ 2,L/ 2]$, we obtain

$$\begin{aligned} \widehat{\Psi }(p,0) = \frac{-i}{\sqrt{\pi L}} \left[ e^{i(k_n-p)L/ 2}\frac{\sin \big [(p-k_n)L/ 2\big ]}{ p-k_n} - e^{-i(k_n+p)L/ 2}\frac{\sin \big [(p+k_n)L/ 2\big ]}{ p+k_n}\right] \;. \end{aligned}$$

(5.D.2.6)

This can be written as:

$$\begin{aligned} \widehat{\Psi }(p,0)&= \frac{-i}{\sqrt{\pi L}} e^{-ip L/ 2} e^{ik_nL/ 2} \left[ \frac{\sin \big [(p-k_n)L/ 2\big ]}{ p-k_n} - e^{-ik_nL}\frac{\sin \big [(p+k_n)L/ 2\big ]}{ p+k_n}\right] \end{aligned}$$

$$\begin{aligned}&= \frac{-i}{\sqrt{\pi L}} e^{-ip L/ 2} e^{ik_nL/ 2} \left[ \frac{\sin \big [(p-k_n)L/ 2\big ]}{ p-k_n} +(-1)^{n+1}\frac{\sin \big [(p+k_n)L/ 2\big ]}{ p+k_n}\right] \;, \end{aligned}$$

(5.D.2.7)

since $e^{-ik_nL}= e^{-in\pi }=(-1)^n$. From (2.A.2.21) we get, using the variable $x'=x - L/2$,

$$\begin{aligned}&\Psi (x',t) = \frac{-i}{\pi \sqrt{2 L}}e^{ik_nL/ 2}\int _{{\mathbbm {R}}} \exp \left( -it \frac{p^2}{2}+ipx'\right) \left[ \frac{\sin \big [(p-k_n)L/ 2\big ]}{ p-k_n}\right. \\&\quad \left. +\,(-1)^{n+1}\frac{\sin \big [(p+k_n)L/ 2\big ]}{ p+k_n}\right] dp\;. \end{aligned}$$

(5.D.2.8)

From Plancherel’s theorem [see also (2.A.2.26)], we know that

$$ \int _{{\mathbbm {R}}} |\Psi (x',t) |^2 dx'= \int _{{\mathbbm {R}}} |\widehat{\Psi }(p,t) |^2 dp= \int _{\mathbbm {R}} |\widehat{\Psi }(p,0) |^2 dp= \int _{\mathbbm {R}} |\Psi (x,0) |^2 dx=1\;. $$

Now, for $t\ne 0$, if we write $\Psi (x',t)=R (x',t) e^{iS(x',t)}$, there is no reason why we should have $\partial S(x',t)/\partial x'=0$ and an analysis of (5.D.2.8) shows that we do indeed have $\partial S(x',t)/ \partial x'\ne 0$ for $t\ne 0$. In fact, we can see this indirectly using (2.A.3.2) and (2.A.3.3), which imply

$$\begin{aligned} \lim _{t\rightarrow \infty } \int _{At} |\Psi (x',t)|^2 dx'= \int _A |\widehat{\Psi }(p,0)|^2 dp\;. \end{aligned}$$

(5.D.2.9)

This means that the support of $\Psi (x',t)$ spreads as $t \rightarrow \infty$ (away from $x'=0$, i.e., away from the center of the box $x=L/ 2$), and by equivariance (5.1.3.1), this means that the particles will move towards either $+\infty$ or $-\infty$.

To illustrate this, consider what happens in the limit of large n, L (mathematically, $n, L \rightarrow \infty$) with $k_n = n\pi / L$ fixed.¹⁰⁶ It is well known (from the theory of Fourier transforms) that

$$\begin{aligned} \frac{\sin (Ly/2)}{ y}=\frac{1}{2} \int _{-L/2}^{L/2} e^{isy} ds \;\longrightarrow \; \pi \delta (y)\;, \end{aligned}$$

(5.D.2.10)

as $L\rightarrow \infty$, in the sense of distributions. So in that limit, we may change variables $y=p-k_n$ or $y=p+k_n$ in (5.D.2.8) and, to a good approximation, replace the integral in (5.D.2.8) by the value of the factor $\exp (it p^2/2+ipx)$ at $p=\pm k_n$. We then obtain

$$\begin{aligned} \Psi (x',t) \approx \frac{-i e^{ik_nL/2}}{ \sqrt{2 L}} \left[ e^{-it k^2_n/ 2+ik_n x'} +(-1)^{n+1} e^{-it k^2_n/ 2-ik_n x'} \right] \;, \end{aligned}$$

(5.D.2.11)

which represents two plane waves propagating in opposite directions with absolute velocity $k_n$, since $\partial (\pm k_n x')/\partial x'= \pm k_n$.

Of course, one has to be a bit careful with this limit, since we have $L\rightarrow \infty$ and $1/\sqrt{ L}$ in (5.D.2.11). But the function in square brackets in (5.D.2.11) is not square integrable over $\mathbbm {R}$ (while we know that $\int _{\mathbbm {R}} |\Psi (x',t) |^2 dx'=1$ for all t), and moreover, its phase is constant in $x'$. So what the limit means is that, for large t, $\Psi (x',t)$ is the sum of two square integrable functions, one whose Fourier transform is localized near $p=k_n$, and the other whose Fourier transform is localized around $p=-k_n$. And by (5.D.2.8) and equivariance, we find that the particles that are in the support of one of those two wave functions move with a speed approximately equal to $\pm k_n$.

The result (5.D.2.11) is exactly what Einstein asked for, when he thought of the classical limit as being made of a motion going back and forth between the walls of the box,¹⁰⁷ except that here this only happens once we remove the walls of the box.¹⁰⁸

5.E The Effective Collapse of the Quantum State and Decoherence

Let us describe the measurement process of Appendix 2.D in more detail (here we follow [70, Sect. 6.1]). When the particle is deflected by the magnetic field , either up or down, its presence is detected by a macroscopic device. One example of such a device would be, for a charged particle, an ion chamber. A cascade of ionization of atoms will eventually produce a pulse that can be observed, at a macroscopic level, by a galvanometer.

Consider two measuring devices, one detecting the particle if it goes up, and one detecting it if it goes down. Let the initial wave function for the two devices be

$$\begin{aligned} \Phi _0 (x_1,\ldots ,x_N,y_1,\ldots ,y_N)= \Phi _0^{\text {{up}}} (x_1,\ldots ,x_N) \Phi _0^{\text {{down}}} (y_1,\ldots ,y_N)\;, \end{aligned}$$

(5.E.1)

where $x_1 \ldots x_N$ are the coordinates of the particles making up the first device and $y_1 \ldots y_N$ are those for the second device, while N is of the order of the Avogadro number $(N\sim 10^{23})$ . Let $\Phi ^{\text {{up}}}_\mathrm{f} (x_1,\ldots ,x_N)$ and $\Phi _\mathrm{f}^{\text {{down}}} (y_1,\ldots ,y_N)$ (f for “final”) denote the wave functions of the detector after the particle has been detected. Note that these wave functions have a support that is disjoint from the support of $\Phi _0^{\text {{up}}}$, $\Phi _0^{\text {{down}}}$ for each variable $x_i$, $y_i$. Indeed, if we think of an ion chamber, then the electrons lost by the ions will move towards an electric plate in $\Phi _\mathrm{f}^{\text {{up}}}$, while they will remain bound to their atoms in $\Phi _0^{\text {{up}}}$ (and the same holds for $\Phi _0^{\text {{down}}}$, $\Phi _\mathrm{f}^{\text {{down}}}$). But this can occur only if the supports of the wave functions are disjoint, for each electron .

If the initial quantum state of the particle is given by (5.1.4.7), then the combined initial quantum state of the particle and the ion chamber is

$$ \Phi _0 = \Phi _0 (x_1,\ldots ,x_N,y_1,\ldots ,y_N) \Psi (z) \big (c_1 |1 \uparrow \rangle + c_2 |1 \downarrow \rangle \big )\;. $$

Then, the final quantum state [see (5.1.4.8)] will be:

$$\begin{aligned} \Phi ^{\text {{up}}}_\mathrm{f} \Phi ^{\text {{down}}}_0 \Psi (z-t) c_1 |1 \uparrow \rangle + \Phi ^{\text {{up}}}_0 \Phi ^{\text {{down}}}_\mathrm{f} \Psi (z+t) c_2 |1 \downarrow \rangle \;, \end{aligned}$$

(5.E.2)

where we have left out the arguments of the wave functions $\Phi ^{\text {{up}}}_\mathrm{f} \Phi ^{\text {{down}}}_0$ and $\Phi ^{\text {{up}}}_0 \Phi ^{\text {{down}}}_\mathrm{f}$. This means that, for the $\Psi (z-t) c_1 |1 \uparrow \rangle$ part of the quantum state, the up detector will be activated and the down part will remain in its original state, and vice versa for the $\Psi (z+t) c_2 |1 \downarrow \rangle$ part.

Now, since the two parts in (5.E.2) have disjoint supports, the original particle can only be in one of them, and so will be guided only by that part of the quantum state in whose support it happens to lie. However, we cannot generally neglect the part of the wave function in the support of which the particle is not located, because the two parts could subsequently be brought together, in such a way that their supports overlap . This is exactly what causes the interference phenomena, exemplified by the Mach–Zehnder interferometer , as discussed in Sect. 2.2. In our example, the quantum state of the particle, viz.,

$$ \Psi (z) \big (c_1 |1 \uparrow \rangle + c_2 |1 \downarrow \rangle \big )\;, $$

evolves into

$$\begin{aligned} \Psi (z-t) c_1 |1 \uparrow \rangle +\Psi (z+t) c_2 |1 \downarrow \rangle \;, \end{aligned}$$

(5.E.3)

where $\Psi (z-t)$ and $\Psi (z+t)$ have disjoint supports (for t not too small). In a quantum state like (5.E.3), the particle is guided by $\Psi (z-t)$ or $\Psi (z+t)$, depending on where it is, but we cannot “reduce” the quantum state and keep only that part, because $\Psi (z-t)$ and $\Psi (z+t)$ may later evolve into wave functions with overlapping supports.

But how can one do that with a quantum state like (5.E.2) involving of the order of $10^{23}$ particles? It is not enough to make $\Psi (z-t)$ and $\Psi (z+t)$ overlap, we need to do that also with $\Phi ^{\text {{up}}}_\mathrm{f}$ $\Phi ^{\text {{down}}}_0$ and $\Phi ^{\text {{up}}}_0$ $\Phi ^{\text {{down}}}_\mathrm{f}$. But this is impossible in practice. Let us see why.

Remember that these wave functions represent a collection of $N\sim 10^{23}$ particles, all of which have different locations, some moving more or less freely and some being bound to their atoms. Hence, the wave functions must have different supports for each particle. Consider, for simplicity, a wave function that factorizes :

$$\begin{aligned} \Phi ^{\text {{up}}}_0(x_1, \ldots , x_N) = \prod ^N_{i=1} \Phi _{0 i} (x_i)\;, \end{aligned}$$

(5.E.4)

and similarly for $\Phi ^{\text {{up}}}_\mathrm{f}$, $\Phi ^{\text {{down}}}_0$, and $\Phi ^{\text {{down}}}_\mathrm{f}$. The support of $\Phi _{0 i}$ is disjoint from the support of $\Phi _{\mathrm{f} i}$ for every particle since, in the support of $\Phi _{0 i}$, the particle is bound to the atom and, in the support of $\Phi _{\mathrm{f} i}$, it is moving.

For the two wave functions to interfere again, we need the wave functions $\Phi _{0 i}(x_i)$ of all the particles to overlap in the future. Indeed, $\Phi ^{\text {{up}}}_0$ and $\Phi ^{\text {{up}}}_\mathrm{f}$ have disjoint supports if

$$\begin{aligned} \Phi ^{\text {{up}}}_0(x_1, \ldots , x_N)\Phi ^{\text {{up}}}_\mathrm{f}(x_1, \ldots , x_N)=0\;, \end{aligned}$$

(5.E.5)

$\forall (x_1, \ldots , x_N) \in {{\mathbb {R}}}^N$. Of course, the same equation also holds for $\Phi ^{\text {{down}}}_0(x_1, \ldots , x_N)\Phi ^{\text {{down}}}_\mathrm{f}(x_1, \ldots , x_N)$. But for (5.E.5) to be true, it is enough to have $\Phi _{0 i} (x_i) \Phi _{\mathrm{f} i}(x_i)=0$, $\forall x_i \in {{\mathbb {R}}}$, for a single value of i. Hence, for interference to occur, we need to get the wave functions of every particle in (say) the ion chamber to overlap again.

It is clear that this is, in practice, impossible. The same is true for the particles constituting the live cat and the dead cat, the pointer pointing up or down, the exploded and unexploded bomb, etc. Now, of course, once the position of the original particle is in the support of one of the two terms in (5.E.2), we can simply reduce the quantum state to that term, since we can be sure that the other term will never again interfere with it.

But we can know in which of the supports of the two terms the particle happens to be located simply by looking at the (by definition) macroscopic result. Hence, in some sense, we do “collapse ” the quantum state after we look at the result of an experiment. However, this is an entirely practical matter. We can still consider, if we wish, that the true quantum state is and remains forever given by the time evolution of (5.E.2). Put simply, there is one of the terms that does not guide the motion of the particle, either now or ever again.

5.F Proof of the Factorization Formula

We now prove (5.1.7.5). Note that (5.1.7.4) implies that

$$\begin{aligned} \Psi (x_1,\ldots ,x_M,{\mathbf{Y}})=\psi (x_i) \Phi _i(\hat{x}_i, {\mathbf{Y}})\;, \end{aligned}$$

(5.F.1)

where $\hat{x}_i = (x_1,\ldots ,x_M)$ with $x_i$ missing and we have put in $\Phi _i(\hat{x}_i, {\mathbf{Y}})$ the arguments that were not indicated in (5.1.7.4).

To prove (5.1.7.5), consider $M=2$. By (5.F.1), we have

$$\begin{aligned} \Psi (x_1,x_2,{\mathbf{Y}})&= \psi (x_1) \Phi _1 (x_2,{\mathbf{Y}}) \end{aligned}$$

$$\begin{aligned}&=\psi (x_2) \Phi _2 (x_1,{\mathbf{Y}})\;. \end{aligned}$$

(5.F.2)

Dividing by $\psi (x_1)\psi (x_2)$, we get

$$\begin{aligned} \frac{\Phi _1 (x_2,{\mathbf{Y}})}{ \psi (x_2)} = \frac{\Phi _2(x_1,{\mathbf{Y}})}{\psi (x_1)}\;, \end{aligned}$$

(5.F.3)

which means that both sides are functions of ${\mathbf{Y}}$ alone, which we can write $\Phi ({\mathbf{Y}})$. Thus,

$$\begin{aligned} \frac{\Psi (x_1,x_2,{\mathbf{Y}})}{ \psi (x_1)\psi (x_2)}=\Phi ({\mathbf{Y}})\;, \end{aligned}$$

which is (5.1.7.5) for $M=2$. The argument easily extends to all M.