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I am confused how the sum of N operators will act on an N-level system of qubits. Here, lets say N=2 so the state is $|00⟩_{CD}$. Then how will this operator $ X_{C} + Z_{D} ⊗ I_{C} + X_{D}$ act on this state?

Is it distributive like this: $ (I _{C} + X_{D}) |00⟩_{CD} = I_{C}|00⟩_{CD} + X_{D}|00⟩_{CD} = |00⟩_{CD} + |01⟩_{CD}$?

And then $ X_{C} + Z_{D}$ would act on each of the terms ($|00⟩_{CD}$ + $|01⟩_{CD}$) above, giving 4 terms?

Some intuition on how/why this happens would be appreciated. How can you construct the sum of the matrices of $I _{C}$ and $X_{D}$?

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The sum of operators in your question:

$$\tag{1} X_{C} + Z_{D} ⊗ I_{C} + X_{D}, $$

is actually this:

$$ X_C \otimes I_D + I_C \otimes Z_D + I_C \otimes X_D.\tag{2} $$

To get how this behaves on the state $|00\rangle$ you just have matrix addition and multiplication: $(A + B + C)|00\rangle$.

You get 3 terms: $ |10\rangle , |00\rangle , |01\rangle $.

I don't see how you could get the 4th term ($|11\rangle$) when there's nothing flipping both qubits at the same time.

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    $\begingroup$ +1. But should the last term be $I_C \otimes X_D$? $\endgroup$ – KAJ226 Mar 3 at 16:51
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    $\begingroup$ hahaha... :) nice! $\endgroup$ – KAJ226 Mar 3 at 17:00
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    $\begingroup$ @lethobentho $(A+B) \otimes (C+D) = A \otimes C + A \otimes D + B \otimes C + B \otimes D $. But what you wrote seems to indicate that you have $A + B +C$ where $A = X_C = X_C \otimes I$, $B = Z_D \otimes I_C$ and $C = I_C \otimes X_D$. $\endgroup$ – KAJ226 Mar 3 at 19:03
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    $\begingroup$ I agree with @KAJ226 . Thanks for taking over while I was gone :) $\endgroup$ – user1271772 Mar 3 at 19:07
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    $\begingroup$ @lethobentho Yes. $\endgroup$ – KAJ226 Mar 3 at 20:02
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TL;DR: What you have written is more or less correct and your result will indeed give 4 terms, however it is not a valid quantum operator.

Firstly, I will add brackets to your operator to make it $(X_C + Z_D) \otimes (I_C + X_D)$. Given that, let us take just focus on just the $(I_C + X_D)$ term and work it through. We will see that, as you mentioned, $(I_C + X_D)|00\rangle_{CD} = |00\rangle_{CD}+|01\rangle_{CD}$.

To begin we first have to take into account what was pointed out by @user1271772: "whenever you see an operator acting on only one qubit, please understand that there's an [implicit] identity operator acting on all other qubits". Explicitly, all this means is

$$ I_C = I_C \otimes I_D \\ X_D = I_C \otimes X_D \\ $$ Now let us construct the matrices of the above two operators

$$ \begin{align*} I_C &= I_C \otimes I_D = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \end{align*} $$ and $$ \begin{align*} X_D &= I_C \otimes X_D = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{bmatrix} \end{align*} $$ If the above computations of tensor products of matrices are unfamiliar to you, have a quick read through the Wikipedia page on Kronecker products.

Now, writing $|00\rangle_{CD}$ in matrix form $\begin{bmatrix} 1 \\ 0\\ 0 \\ 0 \end{bmatrix}$, we can perform the multiplication to get $$ \begin{align*} (I_C + X_D)|00\rangle_{CD} &= \left[(I_C \otimes I_D) + (I_C \otimes X_D)\right]|00\rangle_{CD} \\ \\ &= (I_C \otimes I_D)|00\rangle_{CD} + (I_C \otimes X_D)|00\rangle_{CD} \\ \\ &= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 0\\ 0 \\ 0 \end{bmatrix} \\ \\ &= \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} = |00\rangle_{CD}+|01\rangle_{CD} \end{align*} $$

However all that being said, the operator $(I_C + X_D)$ is actually not a valid quantum operator. This is because it is non-unitary. This can be seen by taking the its determinant which is given by

$$ \begin{align*} |I_C + X_D| &= \left| \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{bmatrix}\right| \\ &= \begin{vmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 1 & 1 \end{vmatrix} = 0 \end{align*} $$ So in fact you cannot ever perform the operation $(I_C + X_D)$ on a quantum computer.

The exact same steps can be taken with the other term in your original question $(X_C + Z_D)$ and you will find a similar situation - i.e. the math will work out but the operator will not be a unitary one.

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  • $\begingroup$ But they are valid Hamiltonian operators and those can act on the 2 qubit state too. Also why did you ad those brackets there? That's not what the original question had. $\endgroup$ – user1271772 Mar 3 at 22:08
  • $\begingroup$ (1) I added the brackets to try and clear up what was meant, because the way it was written it wasn't so clear because. Another possibility might be $(X_C \otimes I_C)+(Z_D \otimes I_C)+(I_C \otimes X_D)$? But if that's the case then you've already answered the question in your answer. (2) They are individually valid operators, but the addition of two valid operators isn't necessarily a valid operator. However in the way you have written it, it could be that there's just a normalization factor missing in which case it isn't that big a deal. $\endgroup$ – Rajiv Krishnakumar Mar 4 at 8:56
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    $\begingroup$ Thanks. My question was meant to be merely illustrative of the concept I needed clarification on (the brute math behind it) so whether it was unitary or not didnt matter in this context but thanks for the clarification regardless! $\endgroup$ – lethobentho Mar 4 at 13:00

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