Depolarizing channel

“With probability 1 − q: Each qubit is left alone. With equal probabilities q/3: X, Y, or Z affects the qubit.”

import Quantum_channel

def depolarize (q : ℝ) [0 ≤ q ≤ 1] (ρ) :=
(1-q) * ρ + q * I/2

theorem completely_positive_depolarize : 
completely_positive depolarize := sorry

theorem trace_preserving_depolarize :
tr(depolarize ρ) = tr(ρ) := sorry

theorem quantum_channel_depolarize :
quantum_channel depolarize :=
{
    completely_positive := completely_positive_depolarize,
    trace_preserving := trace_preserving_depolarize,
}

“Lower bound the achievable rate. Consider the rate for a random Stabilizer code. If we encode k qubits in n qubits using a random stabilizer S, the expected number of errors is nq. We need measure one of the errors to be distinguishable from each other.”

“However, the quantum Hamming bound only limits the efficiency of non-degenerate codes. The typical element of a random stabilizer will have weight 3n/4, which is much larger than pn for any p where the rate could possibly be nonzero.

  • So a random code will have a negligable number of degenerate errors. And the quantum Hamming bound still applies.

  • If the stabilizer has many degeneracies the quantum Hamming bound may be exceeded. Although existing codes only allow us to exceed the rate of a random code by a very small amount. Shor and Smolin showed that by concatenating a random code with a simple repetition code (|0〉 becomes the tensor product of |0〉’s and |1〉 becomes the tensor product of |1〉’s), the rate of the code is improved slightly near the zero-rate limit. The optimum block size for repetition turns out to be five.”

  • “We can set an upper bound on the efficiency of a degenerate Stabilizer code using similar arguments to those that gave us the capacity of a random stabilizer code. This upper bound does not necessarily apply to all codes, so it may not be a strict upper bound on the capacity. However, non-stabilizer codes are very difficult to work with, so it does provide a practical upper bound on the capacity.”

“To give this bound, assume that every element of S actually has weight xn. This bound is unlikely to be achievable, since the product of two operators of weight xn will only rarely have weight xn again. There are at least N(xn)/2ⁿ⁻ᵏ operators of weight n that commute with S, but 2ⁿ⁻ᵏ of them are in S. Therefore, in the best case, there are only N(xn)/2ⁿ⁻ᵏ − 2ⁿ⁻ᵏ operators that can potentially cause a problem. In the limit where n and k = rn are both large, either N(xn)/2ⁿ⁻ᵏ will dominate the number of troublesome operators, or N(xn)/2ⁿ⁻ᵏ ≪ 2ⁿ⁻ᵏ. In the first case, the calculation goes through as for a completely random stabilizer, giving us a capacity only at the quantum Hamming bound. In the second case,

N(xn) ≪ 2²ⁿ⁻²ᵏ

r = k/n < 1 − 1/{2n} log₂ N(xn) = 1 − x/2 log₂ 3 − 1/2 H(x).

Since x = 2q−4q²/3, this is higher than the quantum Hamming bound. This last equation gives an upper bound on the capacity of the depolarizing channel achievable using stabilizer codes.”

The Depolarizing channel has additive Holevo information;

import Degradable_channel

/--
"The depolarizing channel is an example 
of a quantum channel that is not degradable."
-/
theorem not_degradable_depolarize :
¬ degradable depolarize := sorry

/- 
"So we might expect it to exhibit some strange behavior with respect to its quantum capacity."
-/

import Channel_coherent_information

/--
Its coherent information is strictly superadditive 
when the channel becomes very noisy:
-/
theorem coh_info_depolarize_five_uses [depolarize 𝒩] : 
5 * coh_info(𝒩) < coh_info(𝒩^⊗5) := sorry

”

“The best strategy for achieving the quantum capacity of the depolarizing channel remains very poorly understood.”

“The rate in (24.1) is achievable by applying the direct part of the quantum capacity theorem to the channel 𝒩^⊗5.

Operationally, this strategy amounts to concatenating a random quantum code with a Five-qubit code.

The remarkable result: This concatenation strategy can beat the single-copy coherent information when the channel becomes very noisy. The concatenation strategy has positive coherent information even when the single-copy coherent information in (24.159) vanishes.

This demonstrates superadditivity of coherent information.”

/-- takes any input to the maximally mixed state -/
def completely_depolarizing [quantum_channel 𝒩] :=
∀ [quantum_state ρ], 𝒩 ρ = I

-- Don't forget to normalize.

Thermal state

References: