$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$
CHSH Inequality¶
First created in July 2018
Basics¶
CHSH stands for John Clauser, Michael Horne, Abner Shimony, and Richard Holt, who has derived the CHSH Inequality (1969), which can be used to prove Bell's Theorem (1964), that
No physical theory of local hidden variables can ever reproduce all of the predictions of quantum mechanics.
Technically speaking, hidden variable theory is different from the principle of locality (or local reality). The former states that quantum phenomena can be explained by some yet to be known (and therefore "hidden") mechanism. The latter states that no signal can travel faster than light and therefore cannot influence events farther away than light can reach in the period of influence.
The two combined have implied that it cannot be real that that two events far apart (beyond the reach of the same light cone as described in Einstein's Special Theory of Relativity) can be correlated beyond coincidence.
CHSH Inequality puts a constraint on the statistical "coincidence" of correlation between remote measurements. If the CHSH Inequality is violated, then quantum mechanics is correct.
A simple qubit entanglement can be achieved by $~ cX(H\otimes I)\Ket{00}=cX\Rsr2(\Ket{00}+\Ket{10})=\Rsr2(\Ket{00}+\Ket{11}) .$
Now if you measure one qubit, the other will "instantly flip" to match the result. For example, measuring the first (left) qubit on $\Ket+$ will give you 50-50 chance of a detection, but once a detection occurs on the first qubit, measuring the second on $\Ket+$ will give you a detection with certainty, no matter how far the two qubits are apart.
On one side, local realists advocating local hidden variables theories would say that the two qubits are both affected by some unknown factors (hidden variables), and the qubits' fate has been pre-determined. The measurement is just to reveal them.
On the other side, quantum theorists would say that the fate of the two qubits are not determined until the measurement time. Measuring one affects the other "instantaneously" no matter how far they are apart.
Local realists do not to dispute the accuracy of quantum mechanic calculation, but they argue that such calculation can also be reproduced by local hidden variables.
For example, a pair of photons from the same source travelling in opposite direction (and therefore cannot "communicate") will still necessarily coincide in polarisation measurement, simply because the polarisation has been pre-determined. It is the knowledge of the reality that we do not have, not reality itself.
To resolve the dispute, Bell (1964) proposed a test in which the two theories would yield different results.
In the example above, if we measure the two photons on different angles, a pre-determined "hidden" polarisation would yield a probabilistic result on each side independently, which can be calculated by simple optics, without involving quantum mechanics.
If entanglement does exist, by measuring one qubit the other would "flip" to match, we would find the probability on the "flipped" side to be different. To "catch" the difference is what the Bell test is about.
The first Bell test experiment was performed by Freedman and Clauser (1972) and proved that Bell's Theorem is correct: quantum entanglement produces a correlation beyond any local hidden variables can.
(Please note that this does not disprove any global hidden variables theories, which allows the qubits to "communicate" in a speed faster than the speed of light.)
Experiment Setup¶
Here is a schematic experiment setup of a Bell test:
The source S generates photon pairs, each is sent in opposite direction to detector $A$ and $B$ respectively. On $A$ there is a polariser of setting $a$ (angle) which will send the photon to channel $D+$ or $D-$. The same is set up for $B$ of setting $b$. Both photons after passing the detectors will be collected by the coincidence monitor $CM$.
If the photon is polarised in angle $\theta$, the probability of a $D+$ detection on $A$ is $\cos^2(\theta-a)$ and $D-$ is $\sin^2(\theta-a).$ The same will happen on $B$ with probability of a $D+$ detection being $\cos^2(\theta-b)$ and $D-$ being $\sin^2(\theta-b).$
Up to now there is not deviation of the two theories. Local hidden variables can predict the same. (Being "hidden", the variable can behave the way it is required to give the same result.)
Let us code the $D+$ detection as 1 and $D-$ as -1, and record values of $\{D_A,D_B\}:=D_A\times D_B$.
Based on the outcome from $A$ and $B, (D+,D+)=(D-,D-)=1,~\text{and}~(D+,D-)=(D-,D+)=-1.$
We repeat the experiment and count the four occurances of $N_{++}, N_{+-}, N_{-+}$ and $N_{--}$. The expectation value is therefore $\displaystyle E={N_{++}-N_{+-}-N_{-+}+N_{--}\over N_{++}+N_{+-}+N_{-+}+N_{--}} .$
If there is no correlation, all four occurances are of equal chance, so $E=0$. A correlation means $N_{++}$ and $N_{--}$ have a higher value, so $E\in(0,1]$. Likewise, an anti-correlation will see $E\in[-1,0).$
In the experiment, two possible settings are used on each side, $a$ and $a'$ on $A$, and $b$ and $b'$ on $B$. For example, we set $A$ on $a$ and $B$ on $b$, conduct the measurement to determine $E(a,b)$, and so on.
We would have four expectation value $E(a,b), E(a,b'), E(a',b)$ and $E(a',b')$, from which the value of $S=E(a,b)-E(a,b')+E(a',b)+E(a',b')$ is derived.
As the value of $E\in[-1,1],~~S$ can be negative or positive. The CHSH Inequality has established the range of $S\in[-2,2]$, or $|S|\le 2$ for any local hidden variables theories.
Let us see how the CHSH Inequality is derived.
CHSH Derivation¶
The CHSH Inequality is formulated based on local hidden variables theory.
Let the "hidden variable" be $\lambda$ (represented as the photon's polarisation in our example), the setting on $A$ side is $a$ and on $B$ is $b$.
The measurement on $A$, denoted by $A(a,\lambda)$, depends entirely on the setting $a$ and the hidden variable $\lambda$, which has a density distribution of $\rho(\lambda)$ across its domain $\Lambda$ (i.e. how likely the hidden variable has the value of $\lambda$). The same applies to $B$.
For a two-channel experiement, $A(a,\lambda), B(b,\lambda)\in\{-1,0,+1\}$, with $-1$ corresponding to a detection on $D-$ and $+1$ to $D+$. $0$ being the null cases where no photons are detected in either channel (due to apparatus imperfection).
The expectation value $E(a,b)=\int_\Lambda A(a,\lambda)B(b,\lambda)\rho(\lambda)~d\lambda$ (which is just the continuous form of $E={N_{++}-N_{+-}-N_{-+}+N_{--}\over N_{++}+N_{+-}+N_{-+}+N_{--}}$ as shown above).
$E(a,b)-E(a,b')\\ =\int_\Lambda\left(A(a,\lambda)B(b,\lambda)-A(a,\lambda)B(b',\lambda)\right)\rho(\lambda)~d\lambda\\ =\int_\Lambda\left(A(a,\lambda)B(b,\lambda)\pm A(a,\lambda)B(b,\lambda)A(a',\lambda)B(b',\lambda)\right)\rho(\lambda)~d\lambda\\ ~~-\int_\Lambda\left(A(a,\lambda)B(b',\lambda)\pm A(a,\lambda)B(b,\lambda)A(a',\lambda)B(b',\lambda)\right)\rho(\lambda)~d\lambda\\ =\int_\Lambda A(a,\lambda)B(b,\lambda)\left(1\pm A(a',\lambda)B(b',\lambda)\right)\rho(\lambda)~d\lambda\\ ~~-\int_\Lambda A(a,\lambda)B(b',\lambda)\left(1\pm A(a',\lambda)B(b,\lambda)\right)\rho(\lambda)~d\lambda .$
$\lvert E(a,b)-E(a,b')\rvert=\lvert\mathrm{RHS}\rvert\\ \le \big|\int_\Lambda A(a,\lambda)B(b,\lambda)\left(1\pm A(a',\lambda)B(b',\lambda)\right)\rho(\lambda)~d\lambda\big|\\ ~~+\big|\int_\Lambda A(a,\lambda)B(b',\lambda)\left(1\pm A(a',\lambda)B(b,\lambda)\right)\rho(\lambda)~d\lambda\big|{~\tiny\ldots\text{triangle inequality}}\\ \le\int_\Lambda\big|A(a,\lambda)B(b,\lambda)\big|~\left(1\pm A(a',\lambda)B(b',\lambda)\rho(\lambda)\right)~d\lambda\\ ~~+\int_\Lambda\big|A(a,\lambda)B(b',\lambda)\big|~\left(1\pm A(a',\lambda)B(b,\lambda)\rho(\lambda)\right)~d\lambda {~\tiny\ldots\text{given}~(1\pm A(\ldots)B(\ldots))\rho(\lambda)\ge 0}\\ \le \int_\Lambda 1\pm A(a',\lambda)B(b',\lambda)\rho(\lambda)~d\lambda +\int_\Lambda 1\pm A(a',\lambda)B(b,\lambda)\rho(\lambda)~d\lambda {~\tiny\ldots\text{given}~0\le|A(\ldots)B(\ldots)|\le 1}\\ =2\pm\left(\int_\Lambda A(a',\lambda)B(b',\lambda)\rho(\lambda)~d\lambda +\int_\Lambda A(a',\lambda)B(b,\lambda)\rho(\lambda)~d\lambda\right) {~\tiny\ldots\text{given}~\int_\Lambda\rho(\lambda)~d\lambda=1}\\ =2\pm\left(E(a',b')+E(a',b)\right) .$
$\lvert E(a,b)-E(a,b')\rvert\le 2\pm\left(E(a',b')+E(a',b)\right).$
As $E\in[-1,1]$ and the inequality is applicable to both $\pm$ cases, so it must be true for the lowest value. (i.e. Take minus when the bracket is positive, and plus when the bracket is negative.)
$\lvert E(a,b)-E(a,b')\rvert\le 2-\lvert E(a',b')+E(a',b)\rvert.$
By triangle inequality,
$2\ge\lvert E(a,b)-E(a,b')\rvert+\lvert E(a',b')+E(a',b)\rvert\ge\lvert E(a,b)-E(a,b')+E(a',b')+E(a',b)\rvert.$
$\boxed{\lvert S\rvert=\lvert E(a,b)-E(a,b')+E(a',b')+E(a',b)\rvert\le 2.}$
Prediction and Measurement¶
The above derivation is generic for any two-channel experimental settings, not necessarily photon polarisation.
This section is based on the photon polarisation experiment and calculate the predicted values of $S$ based on hidden variable and quantum mechanics respectively. Similar calculations can be done in other settings following the same pattern.
In the following calculations, a constant distribution of $\theta$ is assumed, so $\rho(\theta)={1\over 2\pi}.$
Since $a$ and $b$, $A$ and $B$ are symmetrical, you can select any of the four $E$ terms to be the negative one. Two state vectors are perfectly correlated if they are equal, and perfectly anti-correlated if they are orthogonal.
Within these two extremes, we select
$\displaystyle a=0\\ a'=\pi/4~\tiny{(\text{i.e. }2\pi/8)}\\ b=\pi/8\\ b'=3\pi/8 .$
These are called the Bell test angles, which will give the highest violation of the CHSH Inequality (miximum $S$), as we choose the least correlated angles to be negative (minimum $E$), which is $a=0$ and $b'=3\pi/8$, corresponding to the $E(a,b')$ term.
Hidden Variable¶
Let us assume that the polarisation angle $\theta$ of each photons are pre-determined. While measuring result follows the projection rules, measuring one does not affect the other. This is the view of the realists.
Here is a general derivation of $E(\alpha,\beta)$ based on the local hidden variables theory, where $\alpha$ is the settig on $A$ and $\beta$ on $B$.
$\small{ N_{++}(\alpha,\beta) =\sum N{1\over2\pi}\int_0^{2\pi}\cos^2(\theta-\alpha)\cos^2(\theta-\beta)d\theta =\sum N{1\over2\pi}{\pi\over4}(2+\cos(2(\alpha-\beta))) =\sum N{1\over8}(2+\cos(2(\alpha-\beta))) .}$
$\small{ N_{+-}(\alpha,\beta) =\sum N{1\over2\pi}\int_0^{2\pi}\cos^2(\theta-\alpha)\sin^2(\theta-\beta)d\theta =\sum N{1\over2\pi}{\pi\over4}(2-\cos(2(\alpha-\beta))) =\sum N{1\over8}(2-\cos(2(\alpha-\beta))) .}$
$\small{ N_{-+}(\alpha,\beta) =\sum N{1\over2\pi}\int_0^{2\pi}\sin^2(\theta-\alpha)\cos^2(\theta-\beta)d\theta =\sum N{1\over2\pi}{\pi\over4}(2-\cos(2(\alpha-\beta))) =\sum N{1\over8}(2-\cos(2(\alpha-\beta))) .}$
$\small{ N_{--}(\alpha,\beta) =\sum N{1\over2\pi}\int_0^{2\pi}\sin^2(\theta-\alpha)\sin^2(\theta-\beta)d\theta =\sum N{1\over2\pi}{\pi\over4}(2+\cos(2(\alpha-\beta))) =\sum N{1\over8}(2+\cos(2(\alpha-\beta))) .}$
$E(\alpha,\beta)={N_{++}-N_{+-}-N_{-+}+N_{--}\over\sum N}={1\over2}\cos(2(\alpha-\beta)) .$
$E(a,b)={1\over2}\cos(2(0-\pi/8))={1\over2\sqrt2}.$
$E(a,b')={1\over2}\cos(2(0-3\pi/8))=-{1\over2\sqrt2}.$
$E(a',b)={1\over2}\cos(2(\pi/4-\pi/8))={1\over2\sqrt2}.$
$E(a',b')={1\over2}\cos(2(\pi/4-3\pi/8))={1\over2\sqrt2}.$
$S=E(a,b)-E(a,b')+E(a',b)+E(a',b')=\sqrt2,~~\text{and}~ |S|\approx 1.414 .$
The CHSH Inquality states that $|S|\le 2$, so there is no violation.
Quantum Entanglement¶
With Quantum Entanglement, once the first detection on, say, the $A$ side happened on the polariser with angle $\alpha$, the $B$ side photon polarisation will change from $\theta$ to $\alpha$ as well.
On the other hand, if the $A$ side has a non-detection on $\alpha$, the $B$ side photon polarisation will change from $\theta$ to $\alpha+\pi/2$ (orthonormal to $\alpha$).
Note: If $A$ and $B$ are far apart (i.e. the events of detection at $A$ and $B$ are not in the same light cone), one cannot tell which detection comes first, as simultaneity breaks down based on the Theory of Relativity. So there is no need to discuss if it is $A$ changing $B$ or the other way around. If the two events are in the same light cone, it is considered local as signal has time to travel between the two events, and therefore a violation of CHSH Inequality does not violate the principle of locality (or local reality).
Here is a general derivation of $E(\alpha,\beta)$ based on the quantum entanglement phenomenon, where $\alpha$ is the settig on $A$ and $\beta$ on $B$.
It is in the same pattern as the Hidden Variable, with the $B$ side (the second $\theta$) replaced by $\alpha$ for a detection at $A$ and $\alpha+\pi/2$ for a non-detection.
$N_{++}(\alpha,\beta) =\sum N{1\over2\pi}\int_0^{2\pi}\cos^2(\theta-\alpha)\cos^2(\alpha-\beta)d\theta =\sum N{1\over2}\cos^2(\alpha-\beta) .$
$N_{+-}(\alpha,\beta) =\sum N{1\over2\pi}\int_0^{2\pi}\cos^2(\theta-\alpha)\sin^2(\alpha-\beta)d\theta =\sum N{1\over2}\sin^2(\alpha-\beta) .$
$N_{-+}(\alpha,\beta) =\sum N{1\over2\pi}\int_0^{2\pi}\sin^2(\theta-\alpha)\cos^2(\alpha+\pi/2-\beta)d\theta =\sum N{1\over2}\sin^2(\alpha-\beta) .$
$N_{--}(\alpha,\beta) =\sum N{1\over2\pi}\int_0^{2\pi}\sin^2(\theta-\alpha)\sin^2(\alpha+\pi/2-\beta)d\theta =\sum N{1\over2}\cos^2(\alpha-\beta) .$
$E(\alpha,\beta)={N_{++}-N_{+-}-N_{-+}+N_{--}\over\sum N}=\cos(2(\alpha-\beta)) ,~~$
which is twice the value predicted in the local hidden variables theory.
$E(a,b)=\cos(2(0-\pi/8))={1\over\sqrt2}.$
$E(a,b')=\cos(2(0-3\pi/8))=-{1\over\sqrt2}.$
$E(a',b)=\cos(2(\pi/4-\pi/8))={1\over\sqrt2}.$
$E(a',b')=\cos(2(\pi/4-3\pi/8))={1\over\sqrt2}.$
$S=E(a,b)-E(a,b')+E(a',b)+E(a',b')=2\sqrt2,~~\text{and}~ |S|\approx 2.828 .$
The CHSH Inquality states that $|S|\le 2$, so there is a clear violation. That means the local-hidden-variables-theory-based CHSH Inequality cannot be true. Quantum entanglement is real.
Do it with Qubits¶
Instead of photons, we can use two qubits and actually carry out the experiment by operating and measuring them in a quantum computer. Instead of polarisation, we can use basis vectors as measurement settings.
For example, we can use $X$-axis and $Z$-axis to be the $A$ side, and $W=\Rsr2(Z+X)$ and $V=\Rsr2(Z-X)$ as the $B$ side. They are $\pi/4$ apart on the Bloch Sphere (or $\pi/8$ apart in the vector space).
To maximise the value of $S$, we choose $\Braket{XV}$ to be the negative term.
So $S=\Braket{XW}-\Braket{XV}+\Braket{ZW}+\Braket{ZV}.$
As explained at the beginning, a pair of entangled qubits can be generated by
$~cX(H\otimes I)\Ket{00}=cX\Rsr2(\Ket{00}+\Ket{10})=\Rsr2(\Ket{00}+\Ket{11})=\Ket+.$
We then measure the first and the second qubit. For example, for $\Braket{XW}$, the operation is $(X\otimes W)\Ket+$. The result will give $N_{++}, N_{+-}, N_{-+}\text{ and }N_{--}$ by counting $\Ket{00}, \Ket{01}, \Ket{10}\text{ and }\Ket{11}$ respectively.
With reference to Measurement on Basis Vectors, the measurements are illustrated here:
| $\quad\quad\text{Expectation Value}$ | $\qquad\qquad\text{Measurement}$ |
|---|---|
| $\Braket{X}$ | $H$ |
| $\Braket{Z}$ | $I$ |
| $\Braket{W}$ | $HTHS$ |
| $\Braket{V}$ | $HT^\dagger HS$ |
| $\Braket{XW}$ | $H\otimes HTHS$ |
| $\Braket{XV}$ | $H\otimes HT^\dagger HS$ |
| $\Braket{ZW}$ | $I\otimes HTHS$ |
| $\Braket{ZV}$ | $I\otimes HT^\dagger HS$ |
The above implementation is from IBM Q. Just a side note, an alternative implementation of $\Braket{V}$ to the above is $HTHS^\dagger$.
# chsh.py
import numpy as np
import matplotlib.pyplot as plt
from qiskit import QuantumCircuit, QuantumRegister, ClassicalRegister, execute
from qiskit.tools.visualization import plot_histogram, plot_state, circuit_drawer
# Compute the Expectation Value from circuit qc
shots = 8192
def expval(qc, shots=shots):
qc_run = prep + qc + meas
job = execute(qc_run, backend = 'local_qasm_simulator', shots=shots)
result = job.result()
data = result.get_counts(qc_run)
for i in range(4):
dataIndex = str(bin(i+4))[-2:]
print("P%s=%.3f, " % (dataIndex, data[dataIndex]/shots), end='')
#circuit_drawer(qc_run)
ev = (data['00']-data['01']-data['10']+data['11'])/shots
print("EV=%.3f" % ev)
plot_histogram(job.result().get_counts(qc_run))
return ev
# Define the Quantum and Classical Registers
q = QuantumRegister(2)
c = ClassicalRegister(2)
# Common sub-circuits
circuits = []
prep = QuantumCircuit(q, c)
prep.h(q[0])
prep.cx(q[0], q[1])
meas = QuantumCircuit(q, c)
meas.barrier()
meas.measure(q, c)
# H(x)HTHS
xw = QuantumCircuit(q, c)
xw.h(q[0])
xw.s(q[1])
xw.h(q[1])
xw.t(q[1])
xw.h(q[1])
ev_xw = expval(xw)
print("<xw>=%.3f" % ev_xw)
# H(x)HT^*HS
xv = QuantumCircuit(q, c)
xv.h(q[0])
xv.s(q[1])
xv.h(q[1])
xv.tdg(q[1])
xv.h(q[1])
ev_xv = expval(xv)
print("<xv>=%.3f" % ev_xv)
# I(x)HTHS
zw = QuantumCircuit(q, c)
zw.s(q[1])
zw.h(q[1])
zw.t(q[1])
zw.h(q[1])
ev_zw = expval(zw)
print("<zw>=%.3f" % ev_zw)
# I(x)HT^*HS
zv = QuantumCircuit(q, c)
zv.s(q[1])
zv.h(q[1])
zv.tdg(q[1])
zv.h(q[1])
ev_zv = expval(zv)
print("<zv>=%.3f" % ev_zv)
print("CHSH S=<xw>-<xv>+<zw>+<zv>=(%.3f)-(%.3f)+(%.3f)+(%.3f)=%.3f." % (ev_xw, ev_xv, ev_zw, ev_zv, ev_xw-ev_xv+ev_zw+ev_zv))
# Alternatively, for XV
# H(x)HTHS^*
xv = QuantumCircuit(q, c)
xv.h(q[0])
xv.sdg(q[1])
xv.h(q[1])
xv.t(q[1])
xv.h(q[1])
ev_xv = expval(xv)
print("<xv>=%.3f" % ev_xv)
