Collapse Miscellany

Abstract

An introduction to the CSL (Continuous Spontaneous Localization) theory of dynamical wave function collapse is provided, including a derivation of CSL from two postulates, a new result. There follows a review of applications to a free particle, or to a ‘small’ rigid cluster of free particles, in a single wave-packet and in interfering packets: the latter result is new. [Editors note: for a video of the talk given by Prof. Pearle at the Aharonov-80 conference in 2012 at Chapman University, see quantum.chapman.edu/talk-11.]

Appendices

Appendix A: Proof That R and S Must be Diagonal

We prove here that the real symmetric operators R and S in the Stratonovich Schrödinger equation for the un-normalized state vector,

$$ d|\phi,t\rangle=\bigl[RdB'+Sdt\bigr]|\phi,t\rangle $$

(9.77)

must be diagonal in the |a _n 〉 basis. This is in order that Eq. (9.77) give rise to the Itô gambler’s ruin condition Eq. (9.5),

$$ dx_{n}(t)=b_{n}({\bf x})dB(t). $$

(9.78)

After putting Eq. (9.4), dB′=dB+fdt, into Eq. (9.77), we convert that Stratonovich equation to an Itô equation, with the result

$$ d|\phi,t\rangle=[RdB+Vdt]|\phi,t\rangle\quad\hbox{where }V\equiv S+Rf+ \frac {\lambda}{2}R^{2}. $$

(9.79)

We note that V is also a real symmetric operator and, if we show R and V must be diagonal, then S must also be diagonal.

Using the rules for manipulating Itô equations, it is straightforward to find

$$\begin{aligned} d|\phi,t\rangle\langle\phi,t|={}&\bigl\{[RdB+Vdt],|\phi,t\rangle\langle \phi ,t|\bigr\} +\lambda dtR|\phi,t\rangle\langle\phi ,t|R, \end{aligned}$$

(9.80a)

$$\begin{aligned} d\langle\phi,t|\phi,t\rangle={}&2[ \langle\phi,t|R|\phi,t\rangle dB + \langle \phi,t|V|\phi,t\rangle dt ] +\lambda dt\langle\phi,t|R^{2}|\phi,t\rangle , \end{aligned}$$

(9.80b)

where {M,N}≡MN+NM. Defining the density matrix ρ(t)≡|ϕ,t〉〈ϕ,t|/〈ϕ,t|ϕ,t〉 and $\overline{M}\equiv\hbox{Trace} M\rho$, we obtain from Eqs. (9.80a)–(9.80b) and the Itô rules:

$$\begin{aligned} d\rho =&\bigl[\{R,\rho\}-2\rho\overline{R}\bigr]dB+dt\bigl[\{V,\rho \}-2\rho\overline {V}\bigr] \\ &{}+\lambda dt \bigl[\bigl[R\rho R-\rho\overline{R^{2}}\bigr]-2 \overline{R}\bigl[\{R,\rho\} -2\rho\overline{R}\bigr] \bigr]. \end{aligned}$$

(9.81)

Now, x _n (t)=〈a _n |ρ(t)|a _n 〉. Thus, in order that the diagonal elements of Eq. (9.81) agree with Eq. (9.78), we see that the diagonal elements of Eq. (9.81) which do not multiply dB must vanish for arbitrary ρ:

$$ 0=\bigl[\{V,\rho\}_{nn}-2\rho_{nn}\overline{V} \bigr]+\lambda \bigl[\bigl[(R\rho R)_{nn}-\rho_{nn} \overline{R^{2}}\bigr] -2\overline{R}\bigl[\{R,\rho\}_{nn}-2\rho _{nn}\overline{R}\bigr] \bigr], $$

(9.82)

where M _nm ≡〈a _n |M|a _m 〉

First, suppose that ρ _mm =1, where m≠n, and all other matrix elements of ρ vanish. It follows from Eq. (9.82) that

$$ 0=(R\rho R)_{nn}=(R_{nm})^{2}. $$

(9.83)

That is, all the off-diagonal elements of R vanish, so R is diagonal.

Second, choose a density matrix for which ρ _nn , ρ _mm =1−ρ _nn , ρ _nm do not vanish, but all other matrix elements of ρ do vanish. Then, using the diagonal nature of R, Eq. (9.82) may be written as

$$\begin{aligned} 0 =&2V_{nm}\rho_{nm}[1-2\rho_{nn}] \\ &{}+\rho_{nn}[1-\rho_{nn}] [2(V_{nn}-V_{mm}) \\ &{}+\lambda(R_{nn}-R_{mm})\bigl[(R_{nn}+R_{mm}) \\ &{}-4\bigl(R_{nn}\rho_{nn}+R_{mm}(1- \rho_{nn})\bigr) \bigr]. \end{aligned}$$

(9.84)

For fixed ρ _nn , a viable density matrix (non-negative eigenvalues which add up to 1) exists for $|\rho_{nm}|\leq\sqrt{\rho _{nn}(1-\rho_{nn})}$. But, as ρ _nm is varied, the first term in Eq. (9.84) varies while the rest of the terms remain fixed. Thus, the first term must vanish, and this means that V _nm =0 for n≠m, i.e., V is diagonal as well as R.

Appendix B: Transformation of Operators and White Noise

Consider the general CSL form for the evolution of the state vector, Eq. (9.33)

$$ |\phi,t\rangle=\mathcal{T}e^{-\int_{0}^{t}dt'\{iH(t')+\frac{1}{4\lambda }\sum_{\alpha}[w^{\alpha}(t')-2\lambda A^{\alpha}]^{2}\}}|\phi,0\rangle $$

(9.85)

We introduce a real orthonormal set of vectors $u_{\beta }^{\alpha}$, i.e., $\sum_{\alpha}u_{\beta}^{\alpha}u_{\beta'}^{\alpha }=\delta_{\beta\beta'}$, $\sum_{\beta}u_{\beta}^{\alpha}u_{\beta }^{\alpha'}=\delta_{\alpha\alpha'}$. Defining white noise functions v ^β(t) and complete commuting set of operators Z ^β by $w^{\alpha}(t)\equiv\sum_{\beta} u_{\beta}^{\alpha}v^{\beta}(t)$ and $A^{\alpha}(t)\equiv\sum_{\beta} u_{\beta}^{\alpha}Z^{\beta}(t)$, one readily sees that, in the exponent of Eq. (9.85),

$$ \sum_{\alpha}\bigl[w^{\alpha}(t)-2 \lambda A^{\alpha}\bigr]^{2}=\sum_{\beta } \bigl[v^{\beta}(t)-2\lambda Z^{\beta}\bigr]^{2}. $$

(9.86)

The Jacobian of the transformation from w’s to v’s is 1 so, in using the Probability Rule (9.32b), Dw=Dv.

We wish to apply such a transformation to Eqs. (9.41), (9.42) which, for simplicity, we limit to one-dimensional space:

$$\begin{aligned} |\phi,t\rangle&=\mathcal{T}e^{-i\int_{0}^{t}dt'\frac{1}{2MN}\hat {P}^{2}} e^{-\frac{1}{4\lambda}\int_{0}^{t}dt'\int dx'[w(x',t')-2\lambda NA(x')]^{2}}|\phi,0\rangle, \end{aligned}$$

(9.87a)

$$\begin{aligned} A\bigl(x'\bigr)&\equiv\frac{1}{(\pi a^{2})^{1/4}}e^{-\frac{1}{2a^{2}}[x'-\hat {X}]^{2}}. \end{aligned}$$

(9.87b)

We shall use as orthonormal functions the harmonic oscillator wave functions

$$ u_{n}(x)\equiv C_{n}H_{n}(x/a)e^{-\frac{1}{2a^{2}}x^{2}} \quad\mbox{where } C_{n}\equiv\frac{1}{\sqrt{\pi^{1/2}2^{n}n!a}}. $$

(9.88)

With the definitions v _n (t)≡∫dxw(x,t)u _n (x) and $\hat{Z}_{n}\equiv\int dx A(x)u_{n}(x)$, the exponent in Eq. (9.87a) may be written as

$$\begin{aligned} &-\frac{1}{4\lambda}\int_{0}^{t}dt' \int dx'\bigl[w\bigl(x',t'\bigr)-2\lambda NA\bigl(x'\bigr)\bigr]^{2} \\ &\quad=-\frac{1}{4\lambda}\sum_{n=0}^{\infty} \int_{0}^{t}dt'\bigl[v_{n} \bigl(t'\bigr)-2\lambda N\hat{Z}_{n}\bigr]^{2}. \end{aligned}$$

(9.89)

Thus, we see how a white-noise field gets converted to an equivalent sum of white noise functions.

Using the identity $\exp(-t^{2}+2tz)=\sum_{n=0}^{\infty }t^{n}H_{n}(z)/n!$, with $t\equiv\hat{X}/2a$, z≡x′/a, we find

$$\begin{aligned} \hat{Z}_{n} =&\int dx\frac{1}{(\pi a^{2})^{1/4}}e^{-\frac {1}{2a^{2}}[x'-\hat{X}]^{2}} u_{n}(x) \\ =&\frac{1}{C_{n}(\pi a^{2})^{1/4}}e^{-\frac{1}{4a^{2}}\hat{X}^{2}}\sum_{m=0}^{\infty} \frac{(\hat{X}/2a)^{m}}{m!}\int dx u_{m}(x)u_{n}(x) \\ =&e^{-\frac{1}{4a^{2}}\hat{X}^{2}}\frac{(\hat{X}/\sqrt{2}a)^{n}}{\sqrt{n!}}. \end{aligned}$$

(9.90)

This leads to the density matrix evolution equation

$$\begin{aligned} \frac{d\rho(t)}{d t} =&-i \biggl[\frac{\hat{P}^{2}}{2MN},\rho(t) \biggr] \\ &{} -\frac{\lambda N^{2}}{2}\sum_{n=0}^{\infty} \biggl[e^{-\frac{1}{4a^{2}}\hat {X}^{2}}\frac{(\hat{X}/\sqrt{2}a)^{n}}{\sqrt{n!}}, \biggl[e^{-\frac{1}{4a^{2}}\hat{X}^{2}} \frac{(\hat{X}/\sqrt{2}a)^{n}}{\sqrt {n!}}, \rho(t)\biggr]\biggr]. \end{aligned}$$

(9.91)

If we expand $\exp-\hat{X}^{2}/4a^{2}$, we see that the n=0 term goes as $(\hat{X}/a)^{4}$ and the rest of the terms go as $(\hat{X}/a)^{n}$ to lowest order. Therefore, the lowest order term comes from n=1. Upon neglect of the higher order terms, this gives the density matrix evolution equation

$$ \frac{d\rho(t)}{d t}=-i \biggl[\frac{\hat{P}^{2}}{2MN},\rho(t) \biggr] - \frac{\lambda N^{2}}{4a^{2}}\bigl[\hat{X},\bigl[\hat{X}, \rho(t)\bigr]\bigr], $$

(9.92)

which is identical to the one-dimensional version of Eq. (9.48).