# How can a quantum dot be used as a qubit?

Many people say that quantum dot is a potential physical representation of qubit. A qubit should have two distinguishable states which may carry quantum information. What are the two states of a quantum dot which could be regarded as the two states of a qubit?

• The last part "how do we build one" is a huge question and not really related to the rest, it might do better as its own question. Without that I think this is a nice concise question. I edited out that last sentence. If you want to reverse the edit see here. – Kyle Oman Apr 21 '16 at 8:01
• The quantum dots have spin states. Thesis on the subject – lemon Apr 21 '16 at 8:16

• There can be spin qubits realized within a quantum dot. Once we measure the charge stability diagram for a quantum dot, we can apply the corresponding bias voltages to make sure there is only one electron on that dot. Then the spin-up or spin-down state of that electron represents the logical $|0\rangle_L \equiv |\uparrow\rangle$ or $|1\rangle_L \equiv |\downarrow\rangle$ state. Alternatively, we can also make use of two adjacent quantum dots and drive the bias to the range where each dot has one single electron on it. If there is exchange interaction between the two electrons, they can be either in the spin-singlet or spin-triplet state $|0\rangle_L \equiv \frac{1}{\sqrt{2}}(|\uparrow \downarrow\rangle - |\downarrow \uparrow \rangle), |1\rangle_L \equiv \frac{1}{\sqrt{2}} (|\uparrow \downarrow \rangle + |\uparrow \downarrow \rangle)$ which serve as our two computational states.
• Charge qubits are also possible. For example, we again use two adjacent quantum dots to define one qubit. This time $|0\rangle_L$ is defined as having one electron in the left dot and no electron in the right dot, and $|1\rangle_L$ is defined as having no electron in the left and one electron in the right.