Question

$\mathbf{|}\mathbf{\psi }\mathbf{›}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{|}\mathbf{00}\mathbf{›}\mathbf{+}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{|}\mathbf{11}\mathbf{›}$is one of the famous “Bell states,” a highly entangled state of its two qubits. In this question we examine some of its strange properties. (a) Suppose this Bell state could be decomposed as the (tensor) product of two qubits (recall the box on page ), the first in state ${\mathit{\alpha }}_{\mathbf{0}}\mathbf{|}\mathbf{0}\mathbf{›}\mathbf{+}{\mathit{\alpha }}_{\mathbf{1}}\mathbf{|}\mathbf{1}$and the second in state${\mathit{\beta }}_{\mathbf{0}}\mathbf{|}\mathbf{0}\mathbf{›}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathbf{|}\mathbf{1}$. Write four equations that the amplitudes ${\mathit{\alpha }}_{\mathbf{0}}\mathbf{,}{\mathit{\alpha }}_{\mathbf{1}}\mathbf{,}{\mathit{\beta }}_{\mathbf{0}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{\beta }}_{\mathbf{1}}$must satisfy. Conclude that the Bell state cannot be so decomposed.

(b) What is the result of measuring the first qubit of $\mathbf{|}\mathit{\psi }\mathbf{⟩}$?

(c) What is the result of measuring the second qubit after measuring the first qubit? (d) If the two qubits in state$\mathbf{|}\mathit{\psi }\mathbf{⟩}$ are very far from each other, can you see why the answer to (c) is surprising?

Short answer

a. The required four equations are:

• $|\Psi ⟩={\alpha }_0|0⟩+{\beta }_0|1⟩$
• $|{\alpha }_0{|}^2+|{\beta }_0{|}^2=1$
• $|x_1⟩={\alpha }_0|0⟩+{\alpha }_1|1⟩$
• $|x_2⟩={\beta }_0|0⟩+{\beta }_2|1⟩$

Where,$x_1⟩and|x_2⟩$andare two quantum states.

b. The measure of first qubit state $|\Psi ⟩is|0⟩$is with probability $½.$.

c. The measure of first qubit state $|\Psi ⟩is|0⟩$is with probability$½.$.

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

Step-by-step solution

Solution (a)

Now there are following four equations that satisfied the amplitudes:

• $|\Psi ⟩={\alpha }_0|0⟩+{\beta }_0|1⟩$
• $|\Psi ⟩={\alpha }_0|0⟩+{\beta }_0|1⟩$
• $|x_1⟩={\alpha }_0|0⟩+{\alpha }_1|1⟩$
• $|x_2⟩={\beta }_0|0⟩+{\beta }_2|1⟩$

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

Let, $|\Psi ⟩=\frac{1}{\sqrt{2}}\overline{)00}⧽+\frac{1}{\sqrt{2}}\overline{)11⧽}$

Similarly, $|\Psi ⟩=x_0.y_0|00⟩+x_0.y_1|01⟩+x_1$

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.

Solution (b)

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

Like, the measure of first qubit state $|\Psi ⟩is|0⟩$which have occurrence probability of $½$and for second qubit state $\overline{)1⧽}$is also $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

By using this way, we can measure the first qubit of $|\Psi ⟩$

Solution (c)

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 $|\Psi ⟩is|1⟩$is also$½.$

Solution (d)

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