Multi-Qubit Programs Solutions

Solutions for the exercises in ​Try Your Hand​.

1. 1. The amplitude for is the second element of the vector:

      

   2. The amplitude for is the first element of the vector:

      

2. No. The columns of a matrix for a quantum gate correspond to the idealized states. The number of idealized states is a power of 2. Since 3 is not a power of 2, a matrix can’t be a quantum gate matrix.

3. The correct expression is c. This is a single qubit with two pentagon qubelets and a single triangle qubelet rotated a quarter turn anticlockwise. Thus, the correct way to express it is:

   

4. 1. Probability of collapsing to each of the four idealized states is calculated in the following table:

      Idealized State	Amplitude	Conjugate of Amplitude	Probability of Collapsing

   2. For the classical register to record a 1 when you collapse the second qubit means that the quantum state had to collapse to either a or a state. Thus, from the probabilities calculated in the previous part, the probability of logging a 1 is:

      

   3. Yes, the quantum state of the system will change.

      Since the second qubit collapses to , the new state of the system can only have the and states. These must be normalized as follows to get the new state, :

      

      In other words, , and the state before and after the measurement of the second qubit are different.

      This example shows that the act of measuring a qubit changes the state of the system. This effect, in fact, is a key defining feature of quantum mechanics and underpins Heisenberg’s uncertainty principle.^[119]

      Thus, if you arbitrarily place Measure gates in your code to help you see whether it’s behaving as expected, you’ll end up actually destroying the effect you’re trying to see.

5. This matrix describes a quantum operation that is like a gate as shown by the following circuit:

   

   When the control qubit is , any pentagon qubelets in the target qubit’s state are left alone but the triangle qubelets are given a quarter turn clockwise.

6. 1. Since this gate works on three qubits, its gate matrix’s dimensions will be .

   2. The quantum states on the bottom two qubits are swapped only when is . That is,

      

      Specifically, only the following states are affected by this gate:

      

      Even though the control qubit is in and , swapping the second and third states doesn’t change the overall quantum state. For all other cases, the control qubit is , and hence, the gate doesn’t modify any of those states. Thus, the gate matrix is:

      

7. 1. Since this circuit has two qubits, it’ll have a gate matrix. To obtain the gate matrix, recall that the H gate splits the and qubits as follows:

      

      In these equations, and are the actions of the H gate on the and qubits, respectively.

      Now, work out what this circuit does to each of the four idealized states, , , , and :

      Idealized State:

      Both the top and bottom H gates split the qubit.

      Thus, the mega-qubit formed by this idealized state is:

      

      This state corresponds to the following vector:

      

      Idealized State:

      The top H gate splits the qubit, and the bottom H gate splits the qubit.

      Thus, the mega-qubit formed by this idealized state is:

      

      This state corresponds to the following vector:

      

      Idealized State:

      The top H gate splits the qubit, and the bottom H gate splits the qubit.

      Thus, the mega-qubit formed by this idealized state is:

      

      This state corresponds to the following vector:

      

      Idealized State:

      Both the top and bottom H gates split the qubit.

      Thus, the mega-qubit formed by this idealized state is:

      

      This state corresponds to the following vector:

      

      The previous four vectors correspond to the columns of the matrix representing this circuit:

      

   2. While the S gate rotates triangle qubelets a quarter turn anticlockwise, the gate rotates the triangle qubelets clockwise. Both gates leave pentagon qubelets alone.

      Thus, to obtain the gate matrix for the given circuit, replace with :

      

8. 1. To calculate the matrix for this circuit, start by breaking it up as shown in the following figure:

      

      Then the matrix for the entire circuit is calculated as follows:

      

      is the matrix for the CNOT gate. and are the matrices for the part of the circuit where each qubit is operated on by an H gate, respectively.

      The matrix for the CNOT gate is:

      

      The and matrices were obtained in the previous part. That is,

      

      Thus, the matrix for the entire circuit is:

      

      This circuit, then, modifies the idealized states as follows:

      

      It leaves the and states alone but affects the and states. Specifically, when the second qubit is , it switches the first qubit.

      This circuit acts like an upside down CNOT gate where the first qubit is the target and the second qubit is the control, as shown here:

      

   2. To calculate the matrix for this circuit, start by breaking it up as shown in the following figure:

      

      Thus, the matrix for the entire circuit can be calculated as follows:

      

      is the matrix for the upside down CNOT gate calculated in the previous part:

      

      The and are the matrices for the pass-through H gate. This matrix was calculated in ​Working with Blended States: Mega-Qubit as a Tensor​:

      

      Thus, the matrix for the entire circuit is:

      

      This is the matrix for the Control Z gate shown in the following figure:

      

      This gate leaves the , , and alone but inverts the triangle qubelets in the target qubit if the control qubit is . That is:

      

9. 1. The triangle qubelet in the bottom cell of the qubelet combination on the left is rotated 90° anticlockwise but non-inverted in the bottom cell of that on the right. That is, the bottom triangle qubelet on the left is rotated 90° clockwise on the right. So for the qubelet combination on the right to have the same quantum state as that on the left, the top triangle qubelet in the right combination must be rotated 90° anticlockwise, as shown in the following figure:

      

   2. The triangle qubelet in the bottom cell of the qubelet combination on the left is inverted but non-inverted in the bottom cell of that on the right. So that the right combination has the same quantum state as that on the left, the triangle qubelet in the top cell of the left combination is given a 180° rotation, as shown in the following figure:

      

10. 1. Expand the tensor product to get the quantum state as follows:

       

       This corresponds to a quantum state vector having elements. All the elements are except its twenty-third element which is .

       The associated mega-qubit is:

       

       The mega-qubit contains just a single qubelet combination. Thus, even an idealized state can be expressed as a tensor product.

    2. Expand the tensor product to get the quantum state, as follows:

       

       This corresponds to the following vector:

       

       The associated mega-qubit is shown in the figure.

       

       The triangle qubelet in the second qubelet combination is rotated 90° due to the amplitude of being .

    3. Expand the tensor product to get the quantum state as follows:

       

       This corresponds to the following vector:

       

       The associated mega-qubit is the following:

       

       The anticlockwise quarter-turn triangle qubelet in the second qubelet combination contributes to its amplitude coefficent. That is, . Likewise, the clockwise quarter-turn triangle qubelet in the third qubelet combination contributes to its amplitude coefficient. That is, .

       In the last qubelet combination, top triangle qubelet contributes and the bottom triangle qubelet contributes . That is, the overall amplitude coefficient is the product of these terms: . In other words, the fourth qubelet combination is equivalent to one where both triangle qubelets are not rotated. You can also see this by giving both qubelets the same rotation but in opposite directions so that any sign changes are canceled out: rotate the top triangle qubelet a quarter turn anticlockwise and the bottom triangle qubelet a quarter turn clockwise.

       Thus, the mega-qubit can also be drawn as in the following figure:

       

       In this figure, the fourth qubelet combination has non-rotated triangle qubelets.

    4. The H gate splits the qubit, and the X gate switches the qubit as follows::

       

       Thus, the given tensor product is:

       

       Expand this tensor product to get the quantum state as follows:

       

       This corresponds to the following vector:

       

       This vector has a in the fifth and sixth positions, and elsewhere.

       The associated mega-qubit is:

       

11. 1. To see whether the two qubits are entangled, try factoring the quantum state as follows:

       

       This quantum state can be factored as the tensor product of two quantum states. Hence, the qubits are not entangled.

    2. The given quantum state can’t be factored as a product of tensor products. Hence, the qubits are entangled.

       You can also directly see this from the quantum state itself. If the first qubit collapses to, say, , then the quantum state has collapsed to . Thus, the second qubit is forced to collapse to . An analogous result holds if the first qubit collapses to . Furthermore, you’ll see the same behavior had you collapsed the second qubit before the first.

12. To identify the three missing qubelets, first expand the tensor product of the three qubits to obtain the quantum state of the mega-qubit, as follows:

    

    The missing qubelet is in the bottom cell of the fourth qubelet combination which corresponds to the term in the quantum state specified in the above equation. This combination is formed by taking the pentagon qubelet from the top qubit, the triangle qubelet from the middle qubit, and the inverted triangle qubit from the bottom qubit. The inverted qubelet gives the negative sign associated with this combination. Thus, the first missing qubelet is an inverted triangle qubelet.

    The missing qubelet is in the top cell of the sixth qubelet combination, which corresponds to the term in the quantum state specified in the equation for this exercise. This combination is formed by taking the 90°-rotated triangle qubelet in the first qubit, the pentagon qubelet from the middle qubit, and the inverted triangle qubelet from the bottom qubit. The inverted qubelet gives the negative sign and the 90°-rotated top qubelet gives the complex number associated with this combination. Thus, the second missing qubelet is a triangle qubelet rotated a quarter turn anticlockwise.

    The missing qubelet is in the top cell of the last qubelet combination, which corresponds to the term in the quantum state specified in this equation. This combination is formed by taking the the 90°-rotated triangle qubelet in the first qubit, the triangle qubelet from the middle qubit, and the inverted triangle qubelet from the bottom qubit. The inverted qubelet gives the negative sign and the 90°-rotated top qubelet gives the complex number associated with this combination. Thus, this qubelet combination should be drawn as in the following figure:

    

    But the qubelet combination shown in the given mega-qubit has an unrotated triangle qubelet in the middle and bottom cells. Hence, we need to bring the qubelet combination shown in the previous figure to the desired form by rotating qubelets without modifying the combination’s quantum state. Specifically, invert the bottom triangle qubelet so it’s unrotated, while simultaneously rotating the top qubelet 180° so the top triangle qubelet is now rotated a quarter turn clockwise, as shown in the following figure:

    

    The quantum state of the qubelet combination on the right is still . (The faded qubelets in the top and bottom cells indicate the original position of those qubelets, respectively.)

    The final mega-qubit is shown in the following figure:

    

13. 1. This mega-qubit can collapse in the following four ways:

       Collapsed Quantum State	Probability	State Logged in Classical register
       		00
       		01
       		10
       		11

       Notice that no rotation information is recorded in the classical register.

    2. Since the probability of each collapsed state is , the magnitude of each amplitude is the square root of . Thus, the quantum state for the mega-qubit is:

       

    3. To write the quantum state as a tensor product, factor the previous equation as follows:

       

    4. The tensor product obtained in the previous part can be drawn as shown in the following figure:

       

       The top qubit on the left, , can be obtained by splitting using an H gate.

       The bottom qubit on the left, , has a 90°-rotated triangle qubelet. Hence, after splitting with an H gate, use an S gate to give the triangle qubelet a quarter turn anticlockwise.

       The quantum circuit to create this mega-qubit is shown in the following figure:

       

14. No. Once you teleport , the and qubits collapse and are no longer entangled. Furthermore, they are physically distant from each other. Thus, this circuit can no longer teleport any more quantum states—teleporting circuits are single-use circuits. Once they’re done teleporting, the qubits are no longer useful. If you want to teleport another state, you need another pair of and qubits.

15. When the and qubits collapse, the qubit is:

    

    The qubit will have one pentagon qubelet and a triangle qubelet that is rotated a quarter turn clockwise.

    Since , you’ll need to apply a Z gate to the qubit to obtain the state that will be teleported.

    The Z gate doesn’t affect the pentagon gate but turns the triangle qubelet 180° so that it ends up rotated a quarter turn anticlockwise. Thus, the state that is teleported is:

    

16. Yes, the circuit can be used to teleport a quantum state.

    The Entangler and Loader blocks, together with the , , and the quantum state to be teleported, , are labeled as shown in the following circuit: