91影视

Now there are following four equations that satisfied the amplitudes:

• $|唯鉄�={伪}_0|0鉄�+{尾}_0|1鉄�$
• $|唯鉄�={伪}_0|0鉄�+{尾}_0|1鉄�$
• $|x_1鉄�={伪}_0|0鉄�+{伪}_1|1鉄�$
• $|x_2鉄�={尾}_0|0鉄�+{尾}_2|1鉄�$

Here, $|x_1鉄�and|x_2鉄�$are two quantum states.

Let, $|唯鉄�=\frac{1}{\sqrt{2}}\overline{)00}猝�+\frac{1}{\sqrt{2}}\overline{)11猝�}$

Similarly, $|唯鉄�=x_0.y_0|00鉄�+x_0.y_1|01鉄�+x_1$

Hence,

$$\begin{gathered} x_0.y_0=\frac{1}{\sqrt{2}} \\ {x}_0.y_1=x_1.y_0=0 \end{gathered}$$

Therefore,$x_1.y_1=\frac{1}{\sqrt{2}}and{x}_1.y_0=0$

But then, localid="1658915843438" $x_0.y_0=0or{x}_1.y_1=0$. This is not according to the consumption made.

Hence, we cannot be decompose in the above manner.

The qubit can be calculated in two ways with the value $鈥�0鈥�and鈥�1鈥�$which will be most common outcome. But a qubit state can be superimposed of $鈥�0鈥�and鈥�1鈥�$

Like, the measure of first qubit state $|唯鉄�is|0鉄�$which have occurrence probability of $陆$and for second qubit state $\overline{)1猝�}$is also $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

By using this way, we can measure the first qubit of $|唯鉄�$

The second qubit can be measure in same way as that of first qubit in solution(b). And the value of second qubit will be same as that of first qubit, where, the second qubit state of $|唯鉄�is|1鉄�$is also$陆.$

No matter how far the two qubits are, yet their value will be same.