I am looking at a certain measure for macroscopic quantum states, namely the one in http://dx.doi.org/10.1103/PhysRevA.89.012116. I use the notation from https://arxiv.org/abs/1706.06173 (p. 15-16, chapter "11"). there, macroscopicity of states in cat form $|\mathcal{A}\rangle+|\mathcal{D}\rangle$ is defined by how well one can tell $|\mathcal{A}\rangle$ and $|\mathcal{D}\rangle$ apart when using a coarse grained classical measurement.

The measurement is formalized in the following way (see arXiv source for more details):

  1. Take an observable $A=\sum_a a\cdot P_a$ where $p_{\mathcal{A}}^0=\langle\mathcal{A}|P_a|\mathcal{A}\rangle$ is the probability of finding measurement result $a$ when measuring state $|\mathcal{A}\rangle$ (same for $\mathcal{D}$). This measurement so far obviously isn't coarse grained, but gives a discrete probability distribution.
  2. Now, to achieve coarse-grainedness, one (graphically speaking) puts gaussian density functions on top of the discrete "peaks", turning the probability distribution continuous. Formally, one calculates $p_{\mathcal{A}}^{\sigma}(x)=\sum_a g_{a,\sigma}(x)p_{\mathcal{A}}^{0}(a)$, where $g$ is the gaussian density function with mean $a$ and width $\sigma$. The $\sigma$ gives the coarse-grainedness or resolution of the measurement.
  3. One then carries on in defining $$P^\sigma=[|\mathcal{A}\rangle,\mathcal{D}\rangle]=\frac{1}{2}+\frac{1}{4}\int_{-\infty}^{\infty}|p_{\mathcal{A}}^{\sigma}(x)-p_{\mathcal{D}}^{\sigma}(x)|\text{d}x$$ as the probability of being able to tell the states apart by said measurement.


I am trying to show that for $|\mathcal{A}\rangle=|0\rangle^{\otimes n}$ and $|\mathcal{D}\rangle=|1\rangle^{\otimes n}$, the resolution necessariy to tell both apart (with high probability) by a single measurement is comparatively small - making the GHZ state macroscopic according to above definition.

I struggle with this and face the following problems:

According to the arXiv source (p. 18 lower right) one is said to use the observable $$S_z=\sum_{s=1}^n\left(\bigotimes_{s_1=1}^{s-1} I\otimes \sigma_z\otimes\bigotimes_{s_2=s+1}^n I\right).$$ And in the more general case of replacing $|1\rangle$ with the more general $|\epsilon\rangle=\cos(\epsilon)|0\rangle+\sin(\epsilon)|1\rangle$, one is supposed to use the observable $\cos(\epsilon)S_x+\sin(\epsilon)S_z$ ($\sigma_{x,y,z}$ being the pauli matrices).

How can I show that the GHZ state only needs a small resolution compared to others? Even in the case if I use a self picked observable asking only for the orthogonal states $|0\rangle^{\otimes n}$ and $|1\rangle^{\otimes n}$, mathematically this isn't as straight forward.

I even tried it intuitively using this simpler observable, to at least motivate it. But is this somewhat acccurate (at least as a motivation?). Some feedback on this would also be greatly appreaciated:

The simpler observable $|0\rangle^{\otimes n}\langle 0|^{\otimes n}-|1\rangle^{\otimes n}\langle 1|^{\otimes n}$ e.g. describes observing a cat being in position 1 or position 2 (both being somewhat apart) and therefore the measurement "looking at the situation". The cat is modelled by qubits in that case. In the situation of $\epsilon=\pi/2$, one deals with the ghz-state. It's two parts mean (in the cat context) that the cat is in position 1 or position 2. Whether the cat is in pos 1 or pos 2 can easily be observed, even if the observer has blurry sight (low measurement resolution). In a different case, e.g. distinguising the states $|0101...01\rangle$ and $|1010...10\rangle$ the exact same measurement wouldn't be able to tell them apart as easily with a low resolution. The reason is that those two states mean, that "half of the atoms making up the cat are in position 1" and "half of the atoms making up the cat are in position 2" (and vice versa for the other state). So an observer would see somewhat of 2 cats (with holes as in a cheese) - but different holes whether $|0101...01\rangle$ or $|1010...10\rangle$ is measured. To tell these two situations apart, the observer needs a (very) good eyesight and therefore a high resolution of the measurement.


1 Answer 1


To get more intuition about what is happening, let's start with $n=1$.


The observable $S_z$ you defined will be just $S_z=\sigma_z$ and we have

$$ S_z = \left | 0 \right \rangle \left \langle 0 \right | - \left | 1 \right \rangle \left \langle 1 \right | $$

and $\left | \mathcal A \right \rangle = \left | 0 \right \rangle$, $\left | \mathcal B \right \rangle = \left | 1 \right \rangle$. It is not hard to check that $p_\mathcal A (a) \neq 0$ if and only if $a=1$, $p_\mathcal B(a) \neq 0$ if and only if $a=-1$ and $$ p^\sigma_\mathcal A(x) = g_{1,\sigma}(x), \quad p^\sigma_\mathcal B(x) = g_{-1,\sigma}(x). $$ This implies that we can write $$ P^\sigma = \frac{1}{2} + \frac{1}{4}\int_{-\infty}^{\infty} |g_{1,\sigma}(x) - g_{-1,\sigma}(x)|d x. $$ The last integral can be simplified and we obtain

$$ P^\sigma = \frac{1}{2} \left[1+ \text{Erf}\left(\frac{1}{\sqrt{2}\sigma}\right)\right] $$ where $\text{Erf(x)}$ is the error function.


The observable $S_z$ will be $\mathbb I\otimes \sigma_z + \sigma_z \otimes \mathbb I$ and is not hard to see that the eigenvectors of $S_z$ will be vectors of the form $$ \left |00\right \rangle, \left |01\right \rangle, \left |10\right \rangle, \left |11\right \rangle $$ i.e., they are just the tensor computational basis. However, the eigenvalues will be $2,0,-2$ and we can write $$ S_z = 2\left |00\right \rangle \left \langle 00 \right | - 2\left |11\right \rangle \left \langle 11 \right | $$

Now, $\left | \mathcal A \right \rangle = \left | 00 \right \rangle$, $\left | \mathcal B \right \rangle = \left | 11 \right \rangle$ and $p_\mathcal A (a) \neq 0$ if and only if $a=2$, $p_\mathcal B(a) \neq 0$ if and only if $a=-2$. This implies $$ p^\sigma_\mathcal A(x) = g_{2,\sigma}(x), \quad p^\sigma_\mathcal B(x) = g_{-2,\sigma}(x). $$ This implies that we can write $$ P^\sigma = \frac{1}{2} + \frac{1}{4}\int_{-\infty}^{\infty} |g_{2,\sigma}(x) - g_{-2,\sigma}(x)|d x. $$ Again, we simplify the last integral to obtain $$ P^\sigma = \frac{1}{2} \left[1+ \text{Erf}\left(\frac{2}{\sqrt{2}\sigma}\right)\right] $$

(General n case)

It is not hard to see that the eigenvectors of $S_z$ will always be like $\left |z \right \rangle, z\in \{0,1\}^n$. The eigenvalues will be $$ n,n-2,n-4,\dots,-n. $$

Now, the states $\left | \mathcal A \right \rangle = \left | 0\right \rangle ^{\otimes n}$ and $\left | \mathcal B \right \rangle =\left | 0\right \rangle ^{\otimes n}$ are actually eigenvectors of such observable, which means $p_\mathcal A(a) \neq 0$ if and only if $a=n$, $p_\mathcal B(a) \neq 0$ if and only if $a=-n$. This implies $$ p^\sigma_\mathcal A(x) = g_{n,\sigma}(x), \quad p^\sigma_\mathcal B(x) = g_{-n,\sigma}(x), $$ and finally $$ P^\sigma = \frac{1}{2} \left[1+ \text{Erf}\left(\frac{n}{\sqrt{2}\sigma}\right)\right]. $$

In the limit $n>>\sigma$, $\text{Erf}\left(\frac{n}{\sqrt{2}\sigma}\right) \rightarrow 1$, so $P^\sigma \rightarrow 1$.


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