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system of N particles in a partitioned box

The system in the picture has a box filled with $N$ classical particles which can either be on the left or the right side. Let this system have a Hamiltonian $$H= \frac{\mathbf p^2}{\sum 2m_i} + V(x, t) $$ where V becomes infinite at $\pm \frac{L}{2} $. Let it be represented by a point in $2N$-dimensional phase space $$ (x_1, p_1, ... , x_N, p_N) $$ If we say that a positive x coordinate corresponds to 1, and negative to 0, then this system will encode a string of bits $N$ digits long.

Suppose I want a quantum computer to track the information in the string of bits above. The computer will contain $M$ qubits where $M = \log N$. Assuming this is a perfect quantum computer with no errors, the general state of the computer will be a density matrix in $2^M$-dimensional Hilbert space. $$ \rho = \begin{bmatrix} c_{1,1} & c_{1,2} & \dots & c_{1,N} \\ c_{2,1} & c_{2,2} \\ \vdots & & \ddots \\ c_{N,1} & & & c_{N,N} \end{bmatrix}$$

Now we will quantify the amount of information in both systems.

In order to quantify how much information the box of particles contains, we can use the Shannon Entropy. For a random variable $P_i$ representing the probability of finding a one of a zero in the string at position $i$, the information entropy is$$ S = - \sum_i P_i \log P_i $$ If we assume the particles are sufficiently mixed up in the box, then every $P_i$ will have a probability $\frac{1}{2}$, and then total entropy will be $\frac{N}{2} \log 2$

Next we turn the quantum computer, where the measure of the amount of information is Von Neumann Entropy. Given a density matrix $\rho$ for the quantum computer, the amount of information is given by $$ S = - Tr(\rho \log \rho) $$

Question 1: Does there exist a density matrix that contains the same amount of information as the classical system?

Question 2: Does there exist a Hamiltonian to control the evolution of the quantum computer so that the information contained in both systems are always equal?

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    $\begingroup$ Actually, the classical phase space grows exponentially with N as well (assuming constant amount of free energy per particle.) Can you rephrase your question more precisely? $\endgroup$ – TotallyRhombus Nov 5 '17 at 18:23
  • $\begingroup$ Really? I thought there was a point in 2N-dimensional phase space for N particles. Why does it grow exponentially due to free energy? $\endgroup$ – psitae Nov 5 '17 at 18:26
  • $\begingroup$ Because as the dimension increases linearly, the 'number' (or volume) of configurations grows exponentially. $\endgroup$ – TotallyRhombus Nov 5 '17 at 23:22
  • $\begingroup$ That's right. But couldn't you say the same thing about the vector in Hilbert space, that it grows super exponentially in M, because the dimensions grow exponentially in M? $\endgroup$ – psitae Nov 6 '17 at 1:46
  • $\begingroup$ Not exactly. There are way more configurations of the wave function than there are states of the system from a specific choice of measurement basis. In a sense this is because the wave function describes the state of the system according to an arbitrary choice of measurement basis. There is no classical analogue to the wave function, and there is no simple classical analogue of the discrete configuration space of q-bits. The dimension of the quantum system configuration space increases linearly with $M$ in the sense that the dimension of $(\mathbb Z_2)^M$ increases linearly with $M$. $\endgroup$ – TotallyRhombus Nov 6 '17 at 2:06
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What do you mean by "information"?

There is classical information, which lives in binary strings, and there is quantum information, which lives in Hilbert spaces. They aren't the same thing.

Holevo's theorem says that the most classical information you can extract from a system of $t$ qubits is $t$ classical bits. However, if you want to approximately describe an arbitrary quantum state on $t$ qubits, you need $c^t$ classical bits to do so ($c$ is a constant that depends on how good you want the approximation to be).

So for your question, the amount of information you need to prepare the system is the same: $N$ bits, and the amount of information you can extract from the quantum system is only $\log N$ bits. Which of these is the real, bona fide information? That depends on your definition of information.

The standard definition of information in quantum information theory is von Neumann information, which for a system with density matrix $\rho$ is $-\mathrm{Tr} \,\rho \log \rho$. When you extend the theorems of classical information theory to quantum mechanics, this is generally the correct definition. By this measure, the quantum system has $\log N$ bits of information. So by the standard definition, they're different.

But as for how much "information" there really is in the system — this seems like a philosophy question to me.

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  • $\begingroup$ Hi Peter. By simulate I mean that the dynamics of the quantum system correspond to the dynamics of the classical system, so that, assuming that the phase space and Hilbert same have the same dimension, you could infer the position of one from the other. $\endgroup$ – psitae Nov 6 '17 at 2:52
  • $\begingroup$ Also, what if I'm not worried about extracting the information. Is the fact that it's there, even if it's not recoverable, physically meaningful? $\endgroup$ – psitae Nov 6 '17 at 2:53
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    $\begingroup$ Most $n$-bit classical systems need $n$ qubits to simulate them in any meaningful sense, because if you want to do it with fewer, the dynamics of the classical system are very constrained. If someone can identify a system where you can do it with less, then your question gets very interesting. $\endgroup$ – Peter Shor Nov 6 '17 at 2:56
  • $\begingroup$ I will update my question to show the specific correspondence between the two spaces. $\endgroup$ – psitae Nov 6 '17 at 2:58
  • $\begingroup$ I've edited my question again. It seems to me that both systems have information proportional no N log N. $\endgroup$ – psitae Nov 12 '17 at 6:58
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In general, the entropy of a system is written $$ S_\text{tot}=S_{EE}+S_\text{thermal} $$ where EE denotes the entanglement entropy. In a thermal classical system, you have no ignorance associated with the possibility of states being entangled. However, a quantum computer simulates the classical system using some quantum circuit that has non-zero entanglement entropy so clearly the information content of the two systems are different.

Ignoring the quantum computer itself, the information pertaining to the classical simulation the quantum computer performs contains the same amount of information as the classical system being simulated.

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  • $\begingroup$ For the purposes of my question, let's assume that we a have a perfect, error-free quantum computer. In that case, even though there will exist entanglement between qubits within the computer, the wavefunction of the entire computer will be a pure state, and hence the entanglement entropy will be zero. For reference, quantiki.org/wiki/entropy-entanglement-0 $\endgroup$ – psitae Nov 6 '17 at 2:27

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