# Why does measuring one qubit after the other in this entangled system alter the result?

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

• 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. – Bob Person Dec 7 '19 at 19:16 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.