Suppose I have the following circuit where q0 and q1 are measured one after the other.

Circuit diagram

The simulation results state that the state 00 occurs 75% of the time, and the state 11 occurs 25% of the time. But if you measure both at the same time, you get 00 and 11 50% of the time each.

Here are my calculations:

$\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ \end{bmatrix} \otimes \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}\\ 0 \end{bmatrix}$

$ \begin{bmatrix} 1, 0, 0, 0 \\ 0, 1, 0, 0 \\ 0, 0, 0, 1 \\ 0, 0, 1, 0 \\ \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}\\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}}\\ \end{bmatrix} $

We now have an entangled state. First H gate:

$(H \otimes I) \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix} = $ $ \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \end{bmatrix}$

2nd H gate:

$(H \otimes I) \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix}$

Thus there should be a 50-50 chance of the final state either being 00 or 11, but how come measuring the results one after the other changes it?

  • $\begingroup$ The last 2 H gates in the circuit don't really do anything; I was testing a program I was making and they happened to be there. $\endgroup$
    – Bob Person
    Commented Dec 7, 2019 at 19:16

2 Answers 2


If the simulator is saying that state 00 occurs 75% of the time then the simulator has a bug. Reordering measurements can't make certain outcomes more likely in that way. It would violate the no communication theorem.

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Just some notes:

1) A symbol for tensor product is $\otimes$, so I edited your answer so.

2) After CNOT, you do not have to calculate firstly tensor product $H \otimes I$ and then $I \otimes H$ (by the way, you calculated $H \otimes I$ instead of $I \otimes H$ in step you denoted 2nd H) and simply use $H \otimes H$

3) Regarding your comment "two Hadamards gate do nothing". They in fact change basis you measure in from z-basis to x-basis. However, if you measure your state proudced by Hadamard and CNOT (this is called Bell $\beta_{00}$ state) in both bases then you get same results. But please bear in mind that in case of another measured states, the result with and without $H$s could be different.


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