# Are measurements in $X$ and $Z$ bases being equal enough to prove equality of quantum states?

Suppose I have two single qubit states $$\rho$$ and $$\sigma$$. In my example, they are pure but I keep it general here. Suppose I know that

$$\text{Tr}(\vert +\rangle\langle +\vert\rho) = \text{Tr}(\vert +\rangle\langle +\vert\sigma)\,,$$ $$\text{Tr}(\vert -\rangle\langle -\vert\rho) = \text{Tr}(\vert -\rangle\langle -\vert\sigma)\,,$$ $$\text{Tr}(\vert 0\rangle\langle 0\vert\rho) = \text{Tr}(\vert 0\rangle\langle 0\vert\sigma)\,,$$ $$\text{Tr}(\vert 1\rangle\langle 1\vert\rho) = \text{Tr}(\vert 1\rangle\langle 1\vert\sigma)\,.$$

Can one now conclude that $$\rho = \sigma$$?

## 2 Answers

Consider $$\frac{1}{\sqrt{2}}\big(|0\rangle + i|1\rangle\big)$$ vs $$\frac{1}{\sqrt{2}}\big(|0\rangle - i|1\rangle\big)$$.

No, it is not possible, as you can clearly see from @CraigGidney's example. However, I would like to also provide some algebra as to why that is the case.

The most general single qubit state has three degrees of freedom, which are precisely the Bloch vector coordinates in the Bloch sphere representation. The most general valid density matrix, say $$\varphi$$, can be given as:

$$\varphi = \frac{1}{2}\begin{bmatrix}1+c &a+ib\\a-ib & 1-c \end{bmatrix}\,,$$

where $$\vec{v}\equiv(a,b,c)$$ is your Bloch vector. So, to conclude that two states are equivalent, you would need to confirm that these three degrees of freedom, $$(a,b,c)$$, are equivalent.

You have given four equations; however, in reality, they only verify that two degrees of freedom are equal between $$\rho$$ and $$\sigma$$ because two of these equations are dependent on the other two.

\begin{align} \text{Tr}(|+\rangle\langle+|\cdot \varphi)+\text{Tr}(|-\rangle\langle-|\cdot \varphi)&=1\,,\\ \text{Tr}(|0\rangle\langle0|\cdot \varphi)+\text{Tr}(|1\rangle\langle1|\cdot \varphi)&=1\,. \end{align}

The equations you provided verify that $$a$$ and $$c$$ components for both $$\rho$$ and $$\sigma$$ are equal. However, $$b$$ can be different, and thus you also need measurements along the $$Y$$ axis. One might think that for pure states, since we know that $$||\vec{v}||=1$$, and thus fixing $$a$$ and $$c$$ should also determine $$b$$, but $$b$$ is determined only upto a $$\pm$$ sign. You can see that the state $$\frac{1}{\sqrt{2}}\big(|0\rangle + i|1\rangle\big)$$ corresponds to $$(a,b,c)\equiv(0,1,0)$$ and the state $$\frac{1}{\sqrt{2}}\big(|0\rangle - i|1\rangle\big)$$ corresponds to $$(a,b,c)\equiv(0,-1,0)$$. So, to determine the sign, $$Y$$ measurements are necessary, even in the case of pure states. This is exactly why @CraigGidney's counter-example works.

There are simple algebraic equations connecting measurements to $$(a,b,c)\,.$$

\begin{align} a &=\frac{\text{Tr}(|+\rangle\langle+|\cdot \varphi)-1}{2} \,,\\ b &=\frac{\text{Tr}(|+i\rangle\langle+i|\cdot \varphi)-1}{2} \,,\\ c &=\frac{\text{Tr}(|0\rangle\langle0|\cdot \varphi)-1}{2}\,.\end{align}

In general, for $$n$$-qubit case $$(d = 2^n)$$, the unit trace condition removes one degree of freedom, and hence, you would need to check $$d^2 - 1$$ real degrees of freedom to determine if two $$n$$-qubit-systems are in the same state or not.

• Perhaps worth adding that for a pure state, the length of the Bloch vector is 1. So if you've fixed $a$ and $c$, then $b$ is determined up to a $\pm$ sign? Apr 9 at 6:46
• @DaftWullie Thanks for the suggestion! I have updated my answer. Feel free to edit if you think there is further scope to improve the answer! Apr 9 at 15:58