¿Cómo afecta la medición de un qubit a los otros?

Para representar el estado de una computadora cuántica, todos los qubits contribuyen a un vector de estado (esta es una de las principales diferencias entre la computación cuántica y la clásica según tengo entendido). Entiendo que es posible medir solo un qubit de un sistema de múltiples qubits. ¿Cómo afecta la medición de ese qubit a todo el sistema (específicamente, cómo afecta al vector de estado)?

Hay muchas formas diferentes de ver qubits, y el formalismo vectorial de estado es solo una de ellas. En un sentido lineal-algebraico general, una medición es una proyección sobre una base. Aquí proporcionaré información con un ejemplo desde el punto de vista observable de Pauli, que es el modelo de circuito habitual de control de calidad.

En primer lugar, es interesante en qué base se proporciona el vector de estado: cada operador de medición viene con un conjunto de estados propios y cualquier medida que observe (por ejemplo , $X,Y,Z, XX, XZ$ , etc.) determine la base que podría ser mejor para usted para escribir el vector de estado. La manera más fácil de responder a su pregunta es si sabe qué base le interesa y, lo que es más importante, si conmuta con la medición que acaba de hacer .

Entonces, por simplicidad, digamos que comienzas con dos qubits acoplados en un estado arbitrario escrito en la base $Z$ para ambos qubits:

$$| \psi \rangle = a | 0_{Z} \rangle \otimes | 0_{Z} \rangle +b | 0_{Z} \rangle \otimes | 1_{Z} \rangle + c | 1_{Z} \rangle \otimes | 0_{Z} \rangle + d | 1_{Z} \rangle \otimes | 1_{Z} \rangle$$

Las medidas más simples posibles que podría hacer serían , que es el operador Z en el primer qubit, seguido de Z 2 , el operador Z en el segundo qubit. ¿Qué hace la medición? Proyecta el estado en uno de los estados propios. Puede pensar en esto como eliminar todas las respuestas posibles que sean inconsistentes con la que acabamos de medir. Por ejemplo, digamos que medimos Z 1 y obtenemos el resultado 1 , entonces el estado resultante que tendríamos sería:$Z_{1}$$Z$$Z_{2}$$Z$$Z_{1}$$1$

$$| \psi \rangle = \frac{1}{\sqrt{|c|^{2} +|d|^{2}}} \left(c | 1_{Z} \rangle \otimes | 0_{Z} \rangle + d | 1_{Z} \rangle \otimes | 1_{Z} \rangle \right)$$

Tenga en cuenta que el coeficiente en el frente es solo para la renormalización. Entonces nuestra probabilidad de medir es $Z_{2}=0$. Tenga en cuenta que esto es diferente de la probabilidad que teníamos en el estado inicial, que era| a| 2$\frac{1}{|c|^{2} +|d|^{2}} |c^{2}|$$|a|^{2}+|c|^{2}$.

Suppose the next measurement you make does not commute with the previous one, however. This is trickier because you have to implement a change of basis on the state vector in order to understand the probabilities. With Pauli measurements, though, it tends to be easy since the eigenbases relate in a nice way, that is:

$$| 0_{Z} \rangle = \frac{1}{\sqrt{2}} (|0_{X}\rangle + |1_{X} \rangle )$$

$$| 1_{Z} \rangle = \frac{1}{\sqrt{2}} (|0_{X}\rangle - |1_{X} \rangle )$$

A good way to check your understanding: What is the probability of measuring $X= +1$ after the $Z_{1}=1$ measurement above? What is the probability if we have not made the $Z_{1}$ measurement? Then a more complicated question is to look at product operators that act on both qubits at once, for instance, how does a measurement of $Z_{1}Z_{2}=+1$ affect the initial state? Here $Z_{1}Z_{2}$ measures the product of the two operators.

Suppose that, prior to measurement, your $n$-qubit system is in some state $\lvert \psi \rangle \in \mathcal H_2^{\otimes n}$, where $\mathcal H_2 \cong \mathbb C^2$ is the Hilbert space of a single qubit. Write

$$\lvert \psi \rangle = \sum_{x \in \{0,1\}^n} u_x \lvert x \rangle$$ for some coefficients $u_x \in \mathbb C$ such that $\sum_x \lvert u_x \rvert^2 = 1$.

• If you are measuring the first qubit in the standard basis, define

  $$\begin{aligned} \lvert \varphi_0 \rangle &= \!\!\!\!\!\sum_{x' \in \{0,1\}^{n-1}}\!\!\!\!\!\! u_{0x'} \,\lvert0\rangle \lvert x' \rangle, \\ \lvert \varphi_1 \rangle &= \!\!\!\!\!\sum_{x' \in \{0,1\}^{n-1}}\!\!\!\!\!\! u_{1x'} \,\lvert1\rangle \lvert x' \rangle,\end{aligned}$$ and let $\lvert \psi_0 \rangle = \lvert \varphi_0 \rangle \big/\! \sqrt{\langle \varphi_0 \vert \varphi_0 \rangle}\,$ and $\,\lvert \psi_1 \rangle = \lvert \varphi_1 \rangle \big/\! \sqrt{\langle \varphi_1 \vert \varphi_1 \rangle}\,$. It is not too difficult to show that, if you measure the first qubit and obtain the state $\lvert 0 \rangle$, the state of the entire system "collapses" to $\lvert \psi_0 \rangle$, and if you obtain $\lvert 1 \rangle$ what you obtain is $\lvert \psi_1 \rangle$.

  This is broadly analogous to the idea of conditional probability distributions: you might think of $\lvert \psi_0 \rangle$ as the state of the system conditioned on the first qubit being $\lvert 0 \rangle$, and $\lvert \psi_1 \rangle$ as the state of the system conditioned on the first qubit being $\lvert 1 \rangle$ (except of course that the story is a bit more complicated, on account of the fact that the first qubit is not "secretly" in either the state $0$ or $1$).

• The above is not strongly dependent on measuring the first qubit: we can define $\lvert \varphi_0 \rangle$ and $\lvert \varphi_1 \rangle$ in terms of fixing any particular bit in the bit string $x$ to either $0$ or $1$, summing over only those components which are consistent with either the choice $0$ or $1$, and proceeding as above.

• The above is also not strongly dependent on measuring in the standard basis, as Emily indicates. If we wish to consider measuring the first qubit in the basis $\lvert \alpha \rangle, \lvert \beta \rangle$, where $\lvert \alpha \rangle = \alpha_0 \lvert 0 \rangle + \alpha_1 \lvert 1 \rangle$ and $\lvert \beta \rangle = \beta_0 \lvert 0 \rangle + \beta_1 \lvert 1 \rangle$, we define

  $$\begin{aligned} \lvert \varphi_0 \rangle &= \Bigl(\lvert \alpha \rangle\!\langle \alpha \lvert \otimes I^{\otimes n-1}\Bigr)\lvert \psi\rangle = \!\!\!\!\!\sum_{x' \in \{0,1\}^{n-1}}\!\!\!\!\!\! \bigl(\alpha_0^\ast u_{0x'} + \alpha_1^\ast u_{1x'}\bigr) \,\lvert\alpha\rangle \lvert x' \rangle\,, \\ \lvert \varphi_1 \rangle &= \Bigl(\lvert \beta\rangle\!\langle \beta \lvert \otimes I^{\otimes n-1}\Bigr)\lvert \psi\rangle = \!\!\!\!\!\sum_{x' \in \{0,1\}^{n-1}}\!\!\!\!\!\! \bigl(\beta_0^\ast u_{0x'} + \beta_1^\ast u_{1x'}\bigr) \,\lvert\beta\rangle \lvert x' \rangle\,, \end{aligned}$$ and then proceeding as above.

Less formally-stated than the other answers, but for beginners I like the intuitive method outlined by Prof. Vazirani in this video.

Suppose you have a general two-qbit state:

$|\psi\rangle = \begin{bmatrix} \alpha_{00} \\ \alpha_{01} \\ \alpha_{10} \\ \alpha_{11} \end{bmatrix} = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle$

Now suppose you measure the most-significant (leftmost) qbit in the computational basis (as in, collapse it to either $|0\rangle$ or $|1\rangle$). There are two questions we might ask:

1. What is the probability that the measured qbit collapses to $|0\rangle$? What about $|1\rangle$?
2. What is the state of the 2-qbit system after measurement?

For the first question, the intuitive answer is this: take the sum of squares of all amplitudes associated with the value for which you want to find the probability of collapse. So, if you want to know the probability of the measured qbit collapsing to $|0\rangle$, you'd look at the amplitudes associated with cases $|00\rangle$ and $|01\rangle$, because those are the cases where the measured qbit is $|0\rangle$. Thus:

$P[|0\rangle] = |\alpha_{00}|^2 + |\alpha_{01}|^2$

Similarly, for $|1\rangle$ you look at the amplitudes associated with cases $|10\rangle$ and $|11\rangle$, so:

$P[|1\rangle] = |\alpha_{10}|^2 + |\alpha_{11}|^2$

As for the state of the 2-qbit system after measurement, what you do is cross out all the components of the superposition which are inconsistent with the answer you got. So, if you measured $|0\rangle$, then the state after measurement is:

$\require{cancel} |\psi\rangle = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \cancel{\alpha_{10}|10\rangle} + \cancel{\alpha_{11}|11\rangle} = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle$

However, this state is not normalized - the sum of squares does not add up to 1, and so you have to normalize it:

$|\psi\rangle = \frac{\alpha_{00}|00\rangle + \alpha_{01}|01\rangle}{\sqrt{|\alpha_{00}|^2 + |\alpha_{01}|^2}}$

Similarly, if you measured $|1\rangle$ then you'd get:

$\require{cancel} |\psi\rangle = \cancel{\alpha_{00}|00\rangle} + \cancel{\alpha_{01}|01\rangle} + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle = \alpha_{10}|10\rangle + \alpha_{11}|11\rangle$

Normalized:

$|\psi\rangle = \frac{\alpha_{10}|10\rangle + \alpha_{11}|11\rangle}{\sqrt{|\alpha_{10}|^2 + |\alpha_{11}|^2}}$

And that's how you calculate the action of measuring one qbit in a multi-qbit state, in the simplest case!