Lecture 024 - Time Travel

Closed Timelike Curves (CTC): you could have two points in spacetime $(t_0, x_0, y_0, z_0), (t_1, x_1, y_1, z_1)$ such that the state of particles is the same.

Markov Chain

The following transition matrix

$\begin{bmatrix} 1-p & q\\ p & 1-q\\ \end{bmatrix}$

has steady state

$b = \begin{cases} 0 & \text{w.p. } \frac{q}{p+q}\\ 1 & \text{w.p. } \frac{p}{p+q}\\ \end{cases}$

Quantum Time Travel

Theorem: with 1 time-traveling bit, we can efficiently solve CIRCUIT-SAT (and therefore any NP-complete problems and coNP-complete probems)

NP, coNP $\subseteq$ BPP + 1 C.T.C. bit

Algorithm:

def SAT(b_0 qubit, C circuit):
  Flip n^2 coins
  if all head:
    b_1 = 0
  else:
    pick y in {0, 1}^n at random
    if C(y):
      b_1 = 1
    else:
      b_1 = b_0

When it is unsatisfiable, we have transition matrix

$\begin{bmatrix} 1 & 2^{-n^2}\\ 0 & 1-2^{-n^2}\\ \end{bmatrix}$

Therefore the steady state is:

$\begin{cases} 0 & \text{w.p. 100\%}\\ 1 & \text{w.p. 0\%}\\ \end{cases}$

When it is satisfiable, we have transition probability from $0$ to $1$ :

$\begin{align*} p \geq& Pr\{n^2 \text{ heads}\} Pr\{y \text{ is satisfying for } C\}\\ p =& 2^{-n} - 2^{-n^2}\\ \frac{p}{p+q} \geq& \frac{2^{-n} - 2^{-n^2}}{2^{-n}} \simeq 1 - 2^{-n^2}\\ \end{align*}$

Therefore the steady state is:

$\begin{cases} 1 & \text{w.p. } \simeq 1-2^{-n^2}\\ 0 & \text{w.p. } \simeq 2^{-n^2}\\ \end{cases}$

So we solved SAT with one-sided error.

Other Complexity from Time Travel

With CTC:

BPP + k CTC = BPP-path = BPP + CTC for any constant k (O'D-Say '14)
BQP + 1 CTC = PP (Aaronson '08)
PP != BPP-path unless polynomial hierarchy (PH) collapses
P or BPP or BQP or PSPACE + poly(n) CTC = PSPACE

Above result assume you can send bits back in $poly(n)$ time. If you assume you can send bits back in $O(1)$ time, then all decidable language can be decided in $O(1)$ expected time.

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