Circuit with memory bit

Fig. 4.4 shows a quantum circuit for the demon with a physical mem-ory bit. The design of this circuit follows the discussion in last subsection.

Initially, we prepared the two-level systems ρ^A and ρ^B as follows:

ρ^A(1) = P (ψ₁)|ψ1⟩ ⟨ψ1| + P (ψ2)|ψ2⟩ ⟨ψ2| , ρ^B(1) =|ψ1⟩ ⟨ψ1| . (4.1) For this initial state, the information in the chamber A is

S(A) =−kBT (P (ψ₁) ln P (ψ₁) + P (ψ₂) ln P (ψ₂)) , (4.2)

..

Figure 4.3: Use the C-NOT gate to mimic the semi-permeable membrane.

Left: the corresponding C-NOT gate for the OFF (opaque) case. Right:

the corresponding C-NOT gate for the ON (transparent) case. Flipping the target bit will increase the entropy of the target ensemble.

and in the chamber B is S(B) = 0. This preparation is similar to the initial state for the gas model⁷. Therefore, the goal of this circuit is to swap the information of A and B.

The design of this circuit follows the discussion in last subsection. In the first and the second parts of the circuit Fig. 4.4, the demon reads the states of the molecules and take the information down to its memory. For simplicity, we let the system ρ^A emits two bits at a time. The first bit of ρ^A is used to join the circuit operation, the second bit becomes the memory bit of demon instead. In the other words, ρ^A’hands over’ the memory bit to the demon. The arrow from ρ_A to M is the “hand” over this process. In the third part of circuit, then demon changes the entropy of each potential well.

...

Figure 4.4: The circuit for a quantum demon with a physical memory bit.

We set ground state as the ON state and labelled as ◦, set excited state as the ON state and labelled as •.

To avoid confusion, note that in the following discussion of the first

C-NOT operation, denoted by the path A, we will require the same shape of the potential wells for A and B so that they have the same eigen-spectrum, i.e., ˆHA = ˆHB = ∆A|1⟩ ⟨1|. However, for the remaining operations, i.e., the 2nd C-NOT and Control-Z denoted by the path B, we will deform the potential well of B so that ∆_B̸= ∆A.

First, we perform the thermodynamic path A in our circuit. Initially, the density matrix of the total system (including ρ^A, ρ^B and ρ^M) is given by

ρ^{AM B}(1) = P (ψ₁)|ψ1ψ₁ψ₁⟩ ⟨ψ1ψ₁ψ₁| + P (ψ2)|ψ2ψ₂ψ₁⟩ ⟨ψ2ψ₂ψ₁| . (4.3) Note that ρ^A hands over one of her bits to ρ^M, thus the reduced density matrix of ρ^A and ρ^M are the same. We also can say that ρ^A has the same information amount as ρ^M.

Next, we perform the C-NOT operation to ρ^B, and the state |ψ1⟩ of ρ^M is set to the ON state. After that, we arrive the new state as follows:

ρ^{AM B}(2) = P (ψ₁)|ψ1ψ₁ψ₂⟩ ⟨ψ1ψ₁ψ₂|

+ P (ψ₂)|ψ2ψ₂ψ₁⟩ ⟨ψ2ψ₂ψ₁| . (4.4) In the third step, the demon decreases the entropy of ρ^A also by the C-NOT operation, and arrive

ρ^{AM B}(3) = P (ψ₁)|ψ1ψ₁ψ₂⟩ ⟨ψ1ψ₁ψ₂|

+ P (ψ2)|ψ1ψ2ψ1⟩ ⟨ψ1ψ2ψ1| . (4.5) Then we get ρ(3)^A and ρ(3)^B by taking the partial trace, viz,

ρ^A:= Tr_B,M(ρ^{AM B}) and ρ^B:= Tr_A,M(ρ^{AM B}) (4.6)

⇒ ρ^A(3) =|ψ1⟩ ⟨ψ1| , (4.7)

ρ^B(3) = P (ψ₁)|ψ2⟩ ⟨ψ2| + P (ψ2)|ψ1⟩ ⟨ψ1| . (4.8) In the C-NOT operations of step 2 and step 3, the work done by ρ^A and ρ^B are

W_A= Tr(ρ^A(1) ˆH_A)− Tr(ρ^A(3) ˆH_A) = P (ψ₂)∆_A, (4.9) W_B = Tr(ρ^B(1) ˆH_B)− Tr(ρ^B(3) ˆH_B) =−P (ψ1)∆_A, (4.10) respectively. Thus, the total work extraction from ρ^A and ρ^B is

W_pathA = W_A+ W_B , (4.11)

= (P (ψ₂)− P (ψ1))∆_A . (4.12)

Then we perform the thermodynamic path B in the quantum circuit. The difference between the paths A and B is the choice of the operation basis.

The path B performs the operation with the basis {e1, e2}. The connection between the two basis set is by a θ-rotation:

|ψ2⟩ = sin θ |e1⟩ + cos θ |e2⟩ , (4.13)

|ψ1⟩ = cos θ |e1⟩ − sin θ |e2⟩ . (4.14) Now we start the operation of circuit for the path B. Since the initial entropy S(ρ^B(1)) = 0 so that ρ^B(1) = |e1⟩ ⟨e1|. Then the density matrix of the initial joint state is

ρ^{AM B}(1) = P (ψ₁)|ψ1ψ₁e₁⟩ ⟨ψ1ψ₁e₁| + P (ψ2)|ψ2ψ₂e₁⟩ ⟨ψ2ψ₂e₁| . (4.15) Since the demon executes the C-NOT operation with the basis{e1, e₂}, thus the outcome depends on the probabilities of projection ⟨ψi|ej⟩. There-fore we expand the state of memory bit |ψi⟩ by {e1, e₂} basis to yield

ρ^{AM B}(1) = P (ψ₁)[⟨ψ1|e1⟩ |ψ1e₁e₁⟩ + ⟨ψ1|e2⟩ |ψ1e₂e₁⟩]

[⟨ψ1e₁e₁| ⟨e1|ψ1⟩ + ⟨ψ1e₂e₁| ⟨e2|ψ1⟩]

+P (ψ₂)[⟨ψ2|e1⟩ |ψ2e₁e₁⟩ + ⟨ψ2|e2⟩ |ψ2e₂e₁⟩]

[⟨ψ2e₁e₁| ⟨e1|ψ2⟩ + ⟨ψ2e₂e₁| ⟨e2|ψ2⟩] . (4.16) We then perform the first C-NOT operation to ρ^B with |e1⟩ of ρ^M as ON. After that the density matrix of the joint state becomes

ρ^{AM B}(2) = P (ψ₁)[⟨ψ1|e1⟩ |ψ1e₁e₂⟩ + ⟨ψ1|e2⟩ |ψ1e₂e₁⟩]

[⟨ψ1e1e2| ⟨e1|ψ1⟩ + ⟨ψ1e2e1| ⟨e2|ψ1⟩]

+P (ψ2)[⟨ψ2|e1⟩ |ψ2e1e2⟩ + ⟨ψ2|e2⟩ |ψ2e2e1⟩]

[⟨ψ2e₁e₂| ⟨e1|ψ2⟩ + ⟨ψ2e₂e₁| ⟨e2|ψ2⟩] . (4.17) The next step is to decrease the entropy of ρ^A by acting with a sequential C-NOT and Control-Z operation. It turns out that the input state and the output state of this sequential operation differs at most only by a sign and a flip of the target bit. For example, consider the state |e2ψ₂⟩ under this

sequential operation:

|e2ψ₂⟩ = sin θ |e2e₁⟩ + cos θ |e2e₂⟩ (4.18) C-NOT operation =⇒ sin θ |e2e₂⟩ + cos θ |e2e₁⟩ (4.19) Control-Z operation =⇒ − sin θ |e2e₂⟩ + cos θ |e2e₁⟩ (4.20)

=|e2⟩ (cos θ |e1⟩ − sin θ |e2⟩) (4.21)

=|e2ψ₁⟩ (4.22)

By the similar process, we can derive the other maps for the above sequential operation:

|e2ψ₁⟩ −→ − |e2ψ₂⟩ , (4.23)

|e1ψ₂⟩ −→ |e1ψ₂⟩ , (4.24)

|e1ψ1⟩ −→ |e1ψ1⟩ . (4.25) Obviously, this sequential operation is a C-NOT operation with control and target bits in different basis.

After the above sequential C-NOT operation, by using the above maps we arrive the new state characterized by

ρ^{AM B}(3) = P (ψ₁)[⟨ψ1|e1⟩ |ψ1e₁e₂⟩ − ⟨ψ1|e2⟩ |ψ2e₂e₁⟩]

[⟨ψ1e₁e₂| ⟨e1|ψ1⟩ − ⟨ψ2e₂e₁| ⟨e2|ψ1⟩]

+P (ψ₂)[⟨ψ2|e1⟩ |ψ2e₁e₂⟩ + ⟨ψ2|e2⟩ |ψ1e₂e₁⟩]

[⟨ψ2e₁e₂| ⟨e1|ψ2⟩ + ⟨ψ1e₂e₁| ⟨e2|ψ2⟩] . (4.26) Based on (4.26), we can get the reduced density matrix of each subsystem by taking the partial traces:

ρ^A(3) = (P (ψ₁)P (e₁|ψ1) + P (ψ₂)P (e₂|ψ2))|ψ1⟩ ⟨ψ1|

+ (P (ψ₁)P (e₂|ψ1) + P (ψ₂)P (e₁|ψ2))|ψ2⟩ ⟨ψ2| , (4.27) ρ^M(3) = P (e₁)|e1⟩ ⟨e1| + P (e2)|e2⟩ ⟨e2| , (4.28) ρ^B(3) = P (e₂)|e1⟩ ⟨e1| + P (e1)|e2⟩ ⟨e2| . (4.29)

Now we can evaluate the work done by each subsystem as follows:

W_A= Tr_A(ρ^A(1) ˆH_A)− TrA(ρ^A(3) ˆH_A) (4.30)

= (P (ψ₂)P (e₂|ψ2)− P (ψ1)P (e₂|ψ1))∆_A (4.31) W_M = Tr_M(ρ^M(1) ˆH_M)− TrM(ρ^M(3) ˆH_M) (4.32)

= ∆M(P (e2)− P (ψ1)| ⟨e2|ψ1⟩ |² −P (ψ2)| ⟨e2|ψ2⟩ |²) = 0 (4.33) W_B = Tr_B(ρ^B(1) ˆH_B)− TrB(ρ^B(3) ˆH_B) =−P (e1)∆_B (4.34) Note that the work done by ρ^M is zero as there is no operation performed on ρ^M. Thus, the total work done in the path B is

W_pathB = W_A+ W_M + W_B (4.35)

= (P (ψ₂)P (e₂|ψ2)− P (ψ1)P (e₂|ψ1))∆_A− P (e1)∆_B . (4.36) Moreover, it is easy to see that the entropy S(ρ^A(3)) is not zero after the process of the path B. This is the same as in the gas model. In subsection 4.4 we will discuss the counterpart of the path C in the context of circuit model, and obtain the constraint from the 2nd law of thermodynamics, which is then used to test the Holevo bound.