To expand upon my question, allow me to introduce a problem that I'm attempting:

Suppose $\lvert+\rangle = \frac{1}{\sqrt{2}}(\lvert0\rangle + \lvert1\rangle)$ and $\lvert-\rangle = \frac{1}{\sqrt{2}}(\lvert0\rangle - \lvert1\rangle)$. Find the action of the CZ gate on the basis states $\lvert+\rangle\lvert0\rangle$, $\lvert-\rangle\lvert0\rangle$, $\lvert+\rangle\lvert1\rangle$, $\lvert-\rangle\lvert1\rangle$ and $\lvert0\rangle\lvert+\rangle$, $\lvert0\rangle\lvert-\rangle$, $\lvert1\rangle\lvert+\rangle$, $\lvert1\rangle\lvert-\rangle$.

For clarification:

$$ \lvert0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \lvert1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

In this question, I am asked to calculate the action of the CZ matrix on the different basis states. This is a straightforward task, but what I am confused about is the mathematical form of the CZ matrix/operator.

The CZ gate is defined to be a gate that applies the Z Pauli matrix on the target qubit when the control qubit is $\lvert1\rangle$. For the basis states $\lvert00\rangle$, $\lvert01\rangle$, $\lvert10\rangle$ and $\lvert11\rangle$, the CZ gate has the following matrix form:

$$CZ = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix} \tag{1} $$

As you can see from the question that I'm working on, the basis states are different. How am I supposed to tackle this question? I have two approaches in mind:

Approach 1: The definition of any logic gate is defined with respect to the basis states $\lvert0\rangle$ and $\lvert1\rangle$. Therefore, the matrix in equation 1 is still valid for my basis states and I should compute $CZ\lvert+\rangle\lvert0\rangle$, $CZ\lvert-\rangle\lvert0\rangle$, $CZ\lvert+\rangle\lvert1\rangle$, etc.

Approach 2: The definition of a logic gate changes according to the basis states. The matrix form in equation 1 is no longer valid. Instead, I should apply the Z Pauli matrix on the target qubits if the control qubit is $\lvert1\rangle$ i.e. I apply the Z Pauli matrix on the target qubits in $\lvert1\rangle\lvert+\rangle$ and $\lvert1\rangle\lvert-\rangle$ as they are the only basis states which have the control qubit $\lvert1\rangle$.

I am new to this topic, so if I have missed out anything important or not explained my point clearly, please comment below so I can update the question.

  • 1
    $\begingroup$ to answer the question in the title: yes, of course it does. For example, any gate can be represented as a diagonal matrix with phases in its diagonal, in some basis (the basis of its eigenvectors). $\endgroup$
    – glS
    Jan 18, 2021 at 13:13

1 Answer 1


Firstly calculate vector description of state $|+0\rangle$ in computational basis $$ |+0\rangle = |+\rangle \otimes |0\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1 \\0 \end{pmatrix} $$ and now apply your CZ gate, i.e. $CZ|+0\rangle$. Similarly for other states.

Note that you can avoid using tensor product: $$ |+0\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)|0\rangle= \frac{1}{\sqrt{2}}(|00\rangle+|10\rangle) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ 1 \\0 \end{pmatrix} $$

If you use one of these approaches you get representation of the state in computational basis and since matrix representation of quantum gates is ussually also in computational basis, your inputs to calculation are consistent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.