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two-qubits [June 13, 2026 at 02:54] – Ivan Janevskitwo-qubits [June 13, 2026 at 03:23] (current) – Ivan Janevski
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 A general two-qubit state is a superposition over the four computational basis states, with the normalization condition saying the total probability must be 1: A general two-qubit state is a superposition over the four computational basis states, with the normalization condition saying the total probability must be 1:
  
-$$\lvert\psi\rangle = c_{00}\lvert 00\rangle + c_{01}\lvert 01\rangle + c_{10}\lvert 10\rangle + c_{11}\lvert 11\rangle, \qquad |c_{00}|^2 + |c_{01}|^2 + |c_{10}|^2 + |c_{11}|^2 = 1$$+$$\lvert\psi\rangle = \begin{pmatrix}c_{00}\\c_{01}\\c_{10}\\c_{11}\end{pmatrix}, \qquad \lvert\psi\rangle = c_{00}\underbrace{\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}_{\lvert 00\rangle} + c_{01}\underbrace{\begin{pmatrix}0\\1\\0\\0\end{pmatrix}}_{\lvert 01\rangle} + c_{10}\underbrace{\begin{pmatrix}0\\0\\1\\0\end{pmatrix}}_{\lvert 10\rangle} + c_{11}\underbrace{\begin{pmatrix}0\\0\\0\\1\end{pmatrix}}_{\lvert 11\rangle}, \qquad |c_{00}|^2 + |c_{01}|^2 + |c_{10}|^2 + |c_{11}|^2 = 1$$
  
 ## Product states ## Product states
  
-The simplest two-qubit states are **product states**: states where the two qubits are independent of each other. If qubit 0 is in $a\lvert 0\rangle + b\lvert 1\rangle$ and qubit 1 is in $c\lvert 0\rangle + d\lvert 1\rangle$, the combined state is just their tensor product:+The simplest two-qubit states are **product states**: states where the two qubits are independent of each other. If qubit 1 is in $a\lvert 0\rangle + b\lvert 1\rangle$ and qubit 0 is in $c\lvert 0\rangle + d\lvert 1\rangle$, the combined state is just their tensor product:
  
-$$\lvert\psi_0\rangle\otimes\lvert\psi_1\rangle = ac\lvert 00\rangle + ad\lvert 01\rangle + bc\lvert 10\rangle + bd\lvert 11\rangle$$+$$\lvert\psi_1\rangle\otimes\lvert\psi_0\rangle = ac\lvert 00\rangle + ad\lvert 01\rangle + bc\lvert 10\rangle + bd\lvert 11\rangle$$
  
-The shorthand $\lvert\psi_0\psi_1\rangle$ means $\lvert\psi_0\rangle\otimes\lvert\psi_1\rangle$. There are infinitely many product states. Any pair of single-qubit states gives you one. The ones worth naming come from picking a standard single-qubit basis and using it for both qubits.+The shorthand $\lvert\psi_1\psi_0\rangle$ means $\lvert\psi_1\rangle\otimes\lvert\psi_0\rangle$. There are infinitely many product states. Any pair of single-qubit states gives you one. The ones worth naming come from picking a standard single-qubit basis and using it for both qubits.
  
 ### Z-basis (Computational basis) ### Z-basis (Computational basis)
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 | [[ket-11|$\lvert 11\rangle$]] | Both qubits are $\lvert 1\rangle$. | | [[ket-11|$\lvert 11\rangle$]] | Both qubits are $\lvert 1\rangle$. |
  
-### X-basis+### X-basis (Hadamard basis)
  
 Pick [[ket-plus|$\lvert +\rangle$]] and [[ket-minus|$\lvert -\rangle$]] for each qubit instead. These states look "spread out" in the computational basis, because [[ket-plus|$\lvert +\rangle$]] $= (\lvert 0\rangle + \lvert 1\rangle)/\sqrt{2}$ mixes both. But in the X basis they are perfectly sharp, just like $\lvert 00\rangle$ is sharp in the Z basis. You get from the computational basis to this one by applying a Hadamard to every qubit. Pick [[ket-plus|$\lvert +\rangle$]] and [[ket-minus|$\lvert -\rangle$]] for each qubit instead. These states look "spread out" in the computational basis, because [[ket-plus|$\lvert +\rangle$]] $= (\lvert 0\rangle + \lvert 1\rangle)/\sqrt{2}$ mixes both. But in the X basis they are perfectly sharp, just like $\lvert 00\rangle$ is sharp in the Z basis. You get from the computational basis to this one by applying a Hadamard to every qubit.
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 ^ State ^ Expansion in computational basis ^ Description ^ ^ State ^ Expansion in computational basis ^ Description ^
 | $\lvert ++\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle + \lvert 01\rangle + \lvert 10\rangle + \lvert 11\rangle)$ | Both qubits in [[ket-plus|$\lvert +\rangle$]]. Uniform superposition over all four computational basis states; $H\otimes H\lvert 00\rangle = \lvert ++\rangle$. | | $\lvert ++\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle + \lvert 01\rangle + \lvert 10\rangle + \lvert 11\rangle)$ | Both qubits in [[ket-plus|$\lvert +\rangle$]]. Uniform superposition over all four computational basis states; $H\otimes H\lvert 00\rangle = \lvert ++\rangle$. |
-| $\lvert +-\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle - \lvert 01\rangle + \lvert 10\rangle - \lvert 11\rangle)$ | Qubit 0 in [[ket-plus|$\lvert +\rangle$]], qubit 1 in [[ket-minus|$\lvert -\rangle$]]. | +| $\lvert +-\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle - \lvert 01\rangle + \lvert 10\rangle - \lvert 11\rangle)$ | Qubit 1 in [[ket-plus|$\lvert +\rangle$]], qubit 0 in [[ket-minus|$\lvert -\rangle$]]. | 
-| $\lvert -+\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle + \lvert 01\rangle - \lvert 10\rangle - \lvert 11\rangle)$ | Qubit 0 in [[ket-minus|$\lvert -\rangle$]], qubit 1 in [[ket-plus|$\lvert +\rangle$]]. |+| $\lvert -+\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle + \lvert 01\rangle - \lvert 10\rangle - \lvert 11\rangle)$ | Qubit 1 in [[ket-minus|$\lvert -\rangle$]], qubit 0 in [[ket-plus|$\lvert +\rangle$]]. |
 | $\lvert --\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle - \lvert 01\rangle - \lvert 10\rangle + \lvert 11\rangle)$ | Both qubits in [[ket-minus|$\lvert -\rangle$]]. | | $\lvert --\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle - \lvert 01\rangle - \lvert 10\rangle + \lvert 11\rangle)$ | Both qubits in [[ket-minus|$\lvert -\rangle$]]. |
  
-### Y-basis+### Y-basis (Phase basis)
  
 Pick [[ket-plus-i|$\lvert +i\rangle$]] and [[ket-minus-i|$\lvert -i\rangle$]] for each qubit. These are the eigenstates of the Y gate. The coefficients are complex, so the Y-basis states are the first ones on this page with imaginary entries. You get from the computational basis to the Y basis by applying $SH$ to every qubit ($H$ first, then $S$). Pick [[ket-plus-i|$\lvert +i\rangle$]] and [[ket-minus-i|$\lvert -i\rangle$]] for each qubit. These are the eigenstates of the Y gate. The coefficients are complex, so the Y-basis states are the first ones on this page with imaginary entries. You get from the computational basis to the Y basis by applying $SH$ to every qubit ($H$ first, then $S$).
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 ^ State ^ Expansion in computational basis ^ Description ^ ^ State ^ Expansion in computational basis ^ Description ^
 | $\lvert +i,+i\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle + i\lvert 01\rangle + i\lvert 10\rangle - \lvert 11\rangle)$ | Both qubits in [[ket-plus-i|$\lvert +i\rangle$]]. | | $\lvert +i,+i\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle + i\lvert 01\rangle + i\lvert 10\rangle - \lvert 11\rangle)$ | Both qubits in [[ket-plus-i|$\lvert +i\rangle$]]. |
-| $\lvert +i,-i\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle - i\lvert 01\rangle + i\lvert 10\rangle + \lvert 11\rangle)$ | Qubit 0 in [[ket-plus-i|$\lvert +i\rangle$]], qubit 1 in [[ket-minus-i|$\lvert -i\rangle$]]. | +| $\lvert +i,-i\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle - i\lvert 01\rangle + i\lvert 10\rangle + \lvert 11\rangle)$ | Qubit 1 in [[ket-plus-i|$\lvert +i\rangle$]], qubit 0 in [[ket-minus-i|$\lvert -i\rangle$]]. | 
-| $\lvert -i,+i\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle + i\lvert 01\rangle - i\lvert 10\rangle + \lvert 11\rangle)$ | Qubit 0 in [[ket-minus-i|$\lvert -i\rangle$]], qubit 1 in [[ket-plus-i|$\lvert +i\rangle$]]. |+| $\lvert -i,+i\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle + i\lvert 01\rangle - i\lvert 10\rangle + \lvert 11\rangle)$ | Qubit 1 in [[ket-minus-i|$\lvert -i\rangle$]], qubit 0 in [[ket-plus-i|$\lvert +i\rangle$]]. |
 | $\lvert -i,-i\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle - i\lvert 01\rangle - i\lvert 10\rangle - \lvert 11\rangle)$ | Both qubits in [[ket-minus-i|$\lvert -i\rangle$]]. | | $\lvert -i,-i\rangle$ | $\tfrac{1}{2}(\lvert 00\rangle - i\lvert 01\rangle - i\lvert 10\rangle - \lvert 11\rangle)$ | Both qubits in [[ket-minus-i|$\lvert -i\rangle$]]. |
  
 ## Entangled states ## Entangled states
  
-Some two-qubit states cannot be written as $\lvert\psi_0\rangle\otimes\lvert\psi_1\rangle$ for any choice of single-qubit states. These are **entangled states**. The give-away is that the four amplitudes $c_{00}, c_{01}, c_{10}, c_{11}$ cannot be factored into two pairs: entanglement means $c_{00}c_{11} \neq c_{01}c_{10}$.+Some two-qubit states cannot be written as $\lvert\psi_1\rangle\otimes\lvert\psi_0\rangle$ for any choice of single-qubit states. These are **entangled states**. The give-away is that the four amplitudes $c_{00}, c_{01}, c_{10}, c_{11}$ cannot be factored into two pairs: entanglement means $c_{00}c_{11} \neq c_{01}c_{10}$.
  
 What does that mean physically? Measuring one qubit of an entangled state instantly determines the outcome of measuring the other, even though neither qubit has a definite value before measurement. The correlations are stronger than anything a classical system can produce. What does that mean physically? Measuring one qubit of an entangled state instantly determines the outcome of measuring the other, even though neither qubit has a definite value before measurement. The correlations are stronger than anything a classical system can produce.
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 ^ State ^ Definition ^ Correlations ^ ^ State ^ Definition ^ Correlations ^
-| [[ket-phi-plus|$\lvert\Phi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle + \lvert 11\rangle)$ | Same-value in Z and X; anti-correlated in Y. Prepared from $\lvert 00\rangle$ by $H\otimes I$ then CNOT. |+| [[ket-phi-plus|$\lvert\Phi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle + \lvert 11\rangle)$ | Same-value in Z and X; anti-correlated in Y. Prepared from $\lvert 00\rangle$ by $H\otimes I$ then CX. |
 | [[ket-phi-minus|$\lvert\Phi^-\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle - \lvert 11\rangle)$ | Same-value in Z; anti-correlated in X. The minus sign is invisible in Z-basis measurements. | | [[ket-phi-minus|$\lvert\Phi^-\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle - \lvert 11\rangle)$ | Same-value in Z; anti-correlated in X. The minus sign is invisible in Z-basis measurements. |
-| [[ket-psi-plus|$\lvert\Psi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 10\rangle)$ | Anti-correlated in Z; same-value in X. Prepared from $\lvert 01\rangle$ by $H\otimes I$ then CNOT. |+| [[ket-psi-plus|$\lvert\Psi^+\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 10\rangle)$ | Anti-correlated in Z; same-value in X. Prepared from $\lvert 01\rangle$ by $H\otimes I$ then CX. |
 | [[ket-psi-minus|$\lvert\Psi^-\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle - \lvert 10\rangle)$ | Anti-correlated in every basis. The only antisymmetric Bell state; also called the singlet. | | [[ket-psi-minus|$\lvert\Psi^-\rangle$]] | $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle - \lvert 10\rangle)$ | Anti-correlated in every basis. The only antisymmetric Bell state; also called the singlet. |
  
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 ## Gates ## Gates
  
-Two-qubit gates are $4\times 4$ unitary matrices acting on $\mathbb{C}^4$. The three most common ones are CX (CNOT), CZ, and SWAP. Row and column ordering follows the computational basis: $\lvert 00\rangle, \lvert 01\rangle, \lvert 10\rangle, \lvert 11\rangle$.+Two-qubit gates are $4\times 4$ unitary matrices acting on $\mathbb{C}^4$. The three most common ones are CX (CX), CZ, and SWAP. Row and column ordering follows the computational basis: $\lvert 00\rangle, \lvert 01\rangle, \lvert 10\rangle, \lvert 11\rangle$.
  
 $$CX = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix} $$CX = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}
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 qc = QuantumCircuit(2, 2) qc = QuantumCircuit(2, 2)
 qc.h(0)       # Hadamard on qubit 0 → superposition qc.h(0)       # Hadamard on qubit 0 → superposition
-qc.cx(0, 1)   # CNOT: control=qubit 0, target=qubit 1 → entanglement+qc.cx(0, 1)   # CX: control=qubit 0, target=qubit 1 → entanglement
 qc.measure([0, 1], [0, 1]) qc.measure([0, 1], [0, 1])
  
two-qubits.1781319293.txt.gz · Last modified: by Ivan Janevski