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+- Physics +- Plasma physics +- Plasma waves +layout: "concept" +--- + +In the [magnetohydrodynamic](/know/concept/magnetohydrodynamics/) description of a plasma, +we split the velocity $\vb{u}$, electric current $\vb{J}$, +[magnetic field](/know/concept/magnetic-field/) $\vb{B}$ +and [electric field](/know/concept/electric-field/) $\vb{E}$ like so, +into a constant uniform equilibrium (subscript $0$) +and a small unknown perturbation (subscript $1$): + +$$\begin{aligned} + \vb{u} + = \vb{u}_0 + \vb{u}_1 + \qquad + \vb{J} + = \vb{J}_0 + \vb{J}_1 + \qquad + \vb{B} + = \vb{B}_0 + \vb{B}_1 + \qquad + \vb{E} + = \vb{E}_0 + \vb{E}_1 +\end{aligned}$$ + +Inserting this decomposition into the ideal form of the generalized Ohm's law +and keeping only terms that are first-order in the perturbation, we get: + +$$\begin{aligned} + 0 + &= (\vb{E}_0 + \vb{E}_1) + (\vb{u}_0 + \vb{u}_1) \cross (\vb{B}_0 + \vb{B}_1) + \\ + &= \vb{E}_1 + \vb{u}_1 \cross \vb{B}_0 +\end{aligned}$$ + +We do this for the momentum equation too, +assuming that $\vb{J}_0 \!=\! 0$ (to be justified later). +Note that the temperature is set to zero, such that the pressure vanishes: + +$$\begin{aligned} + \rho \pdv{\vb{u}_1}{t} + = \vb{J}_1 \cross \vb{B}_0 +\end{aligned}$$ + +Where $\rho$ is the uniform equilibrium density. +We would like an equation for $\vb{J}_1$, +which is provided by the magnetohydrodynamic form of Ampère's law: + +$$\begin{aligned} + \nabla \cross \vb{B}_1 + = \mu_0 \vb{J}_1 + \qquad \implies \quad + \vb{J}_1 + = \frac{1}{\mu_0} \nabla \cross \vb{B}_1 +\end{aligned}$$ + +Substituting this into the momentum equation, +and differentiating with respect to $t$: + +$$\begin{aligned} + \rho \pdvn{2}{\vb{u}_1}{t} + = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \pdv{}{\vb{B}1}{t} \Big) \cross \vb{B}_0 \bigg) +\end{aligned}$$ + +For which we can use Faraday's law to rewrite $\ipdv{\vb{B}_1}{t}$, +incorporating Ohm's law too: + +$$\begin{aligned} + \pdv{\vb{B}_1}{t} + = - \nabla \cross \vb{E}_1 + = \nabla \cross (\vb{u}_1 \cross \vb{B}_0) +\end{aligned}$$ + +Inserting this into the momentum equation for $\vb{u}_1$ +thus yields its final form: + +$$\begin{aligned} + \rho \pdvn{2}{\vb{u}_1}{t} + = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \big) \Big) \cross \vb{B}_0 \bigg) +\end{aligned}$$ + +Suppose the magnetic field is pointing in $z$-direction, +i.e. $\vb{B}_0 = B_0 \vu{e}_z$. +Then Faraday's law justifies our earlier assumption that $\vb{J}_0 = 0$, +and the equation can be written as: + +$$\begin{aligned} + \pdvn{2}{\vb{u}_1}{t} + = v_A^2 \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg) +\end{aligned}$$ + +Where we have defined the so-called **Alfvén velocity** $v_A$ to be given by: + +$$\begin{aligned} + \boxed{ + v_A + \equiv \sqrt{\frac{B_0^2}{\mu_0 \rho}} + } +\end{aligned}$$ + +Now, consider the following plane-wave ansatz for $\vb{u}_1$, +with wavevector $\vb{k}$ and frequency $\omega$: + +$$\begin{aligned} + \vb{u}_1(\vb{r}, t) + &= \vb{u}_1 \exp(i \vb{k} \cdot \vb{r} - i \omega t) +\end{aligned}$$ + +Inserting this into the above differential equation for $\vb{u}_1$ leads to: + +$$\begin{aligned} + \omega^2 \vb{u}_1 + = v_A^2 \bigg( \Big( \vb{k} \cross \big( \vb{k} \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg) +\end{aligned}$$ + +To evaluate this, we rotate our coordinate system around the $z$-axis +such that $\vb{k} = (0, k_\perp, k_\parallel)$, +i.e. the wavevector's $x$-component is zero. +Calculating the cross products: + +$$\begin{aligned} + \omega^2 \vb{u}_1 + &= v_A^2 \bigg( \Big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} + \cross \big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} + \cross ( \begin{bmatrix} u_{1x} \\ u_{1y} \\ u_{1z} \end{bmatrix} + \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} ) \big) \Big) + \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg) + \\ + &= v_A^2 \bigg( \Big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} + \cross \big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} + \cross \begin{bmatrix} u_{1y} \\ -u_{1x} \\ 0 \end{bmatrix} \big) \Big) + \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg) + \\ + &= v_A^2 \bigg( \Big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix} + \cross \begin{bmatrix} k_\parallel u_{1x} \\ k_\parallel u_{1y} \\ -k_\perp u_{1y} \end{bmatrix} \Big) + \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg) + \\ + &= v_A^2 \bigg( \begin{bmatrix} -(k_\perp^2 \!+ k_\parallel^2) u_{1y} \\ k_\parallel^2 u_{1x} \\ -k_\perp k_\parallel u_{1x} \end{bmatrix} + \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg) + \\ + &= v_A^2 \begin{bmatrix} k_\parallel^2 u_{1x} \\ (k_\perp^2 \!+ k_\parallel^2) u_{1y} \\ 0 \end{bmatrix} +\end{aligned}$$ + +We rewrite this equation in matrix form, +using that $k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$: + +$$\begin{aligned} + \begin{bmatrix} + \omega^2 - v_A^2 k_\parallel^2 & 0 & 0 \\ + 0 & \omega^2 - v_A^2 k^2 & 0 \\ + 0 & 0 & \omega^2 + \end{bmatrix} + \vb{u}_1 + = 0 +\end{aligned}$$ + +This has the form of an eigenvalue problem for $\omega^2$, +meaning we must find non-trivial solutions, +where we cannot simply choose the components of $\vb{u}_1$ to satisfy the equation. +To achieve this, we demand that the matrix' determinant is zero: + +$$\begin{aligned} + \big(\omega^2 - v_A^2 k_\parallel^2\big) \: \big(\omega^2 - v_A^2 k^2\big) \: \omega^2 + = 0 +\end{aligned}$$ + +This equation has three solutions for $\omega^2$, +one for each of its three factors being zero. +The simplest case $\omega^2 = 0$ is of no interest to us, +because we are looking for waves. + +The first interesting case is $\omega^2 = v_A^2 k_\parallel^2$, +yielding the following dispersion relation: + +$$\begin{aligned} + \boxed{ + \omega + = \pm v_A k_\parallel + } +\end{aligned}$$ + +The resulting waves are called **shear Alfvén waves**. +From the eigenvalue problem, we see that in this case +$\vb{u}_1 = (u_{1x}, 0, 0)$, meaning $\vb{u}_1 \cdot \vb{k} = 0$: +these waves are **transverse**. +The phase velocity $v_p$ and group velocity $v_g$ are as follows, +where $\theta$ is the angle between $\vb{k}$ and $\vb{B}_0$: + +$$\begin{aligned} + v_p + = \frac{|\omega|}{k} + = v_A \frac{k_\parallel}{k} + = v_A \cos(\theta) + \qquad \qquad + v_g + = \pdv{|\omega|}{k} + = v_A +\end{aligned}$$ + +The other interesting case is $\omega^2 = v_A^2 k^2$, +which leads to so-called **compressional Alfvén waves**, +with the simple dispersion relation: + +$$\begin{aligned} + \boxed{ + \omega + = \pm v_A k + } +\end{aligned}$$ + +Looking at the eigenvalue problem reveals that $\vb{u}_1 = (0, u_{1y}, 0)$, +meaning $\vb{u}_1 \cdot \vb{k} = u_{1y} k_\perp$, +so these waves are not necessarily transverse, nor longitudinal (since $k_\parallel$ is free). +The phase velocity $v_p$ and group velocity $v_g$ are given by: + +$$\begin{aligned} + v_p + = \frac{|\omega|}{k} + = v_A + \qquad \qquad + v_g + = \pdv{|\omega|}{k} + = v_A +\end{aligned}$$ + +The mechanism behind both of these oscillations is magnetic tension: +the waves are "ripples" in the field lines, +which get straightened out by Faraday's law, +but the ions' inertia causes them to overshoot and form ripples again. + + + +## References +1. M. Salewski, A.H. Nielsen, + *Plasma physics: lecture notes*, + 2021, unpublished. + diff --git a/source/know/concept/archimedes-principle/index.md b/source/know/concept/archimedes-principle/index.md new file mode 100644 index 0000000..4bab87b --- /dev/null +++ b/source/know/concept/archimedes-principle/index.md @@ -0,0 +1,89 @@ +--- +title: "Archimedes' principle" +date: 2021-04-10 +categories: +- Fluid statics +- Fluid mechanics +- Physics +layout: "concept" +--- + +Many objects float when placed on a liquid, +but some float higher than others, +and some do not float at all, sinking instead. +**Archimedes' principle** balances the forces, +and predicts how much of a body is submerged, +and how much is non-submerged. + +In truth, there is no real distinction between +the submerged and non-submerged parts, +since the latter is surrounded by another fluid (air), +which has a pressure and thus affects it. +The right thing to do is treat the entire body as being +submerged in a fluid with varying properties. + +Let us consider a volume $V$ completely submerged in such a fluid. +This volume will experience a downward force due to gravity, given by: + +$$\begin{aligned} + \va{F}_g + = \int_V \va{g} \rho_\mathrm{b} \dd{V} +\end{aligned}$$ + +Where $\va{g}$ is the gravitational field, +and $\rho_\mathrm{b}$ is the density of the body. +Meanwhile, the pressure $p$ of the surrounding fluid exerts a force +on the entire surface $S$ of $V$: + +$$\begin{aligned} + \va{F}_p + = - \oint_S p \dd{\va{S}} + = - \int_V \nabla p \dd{V} +\end{aligned}$$ + +Where we have used the divergence theorem. +Assuming [hydrostatic equilibrium](/know/concept/hydrostatic-pressure/), +we replace $\nabla p$, +leading to the definition of the **buoyant force**: + +$$\begin{aligned} + \boxed{ + \va{F}_p + = - \int_V \va{g} \rho_\mathrm{f} \dd{V} + } +\end{aligned}$$ + +For the body to be at rest, we require $\va{F}_g + \va{F}_p = 0$. +Concretely, the equilibrium condition is: + +$$\begin{aligned} + \boxed{ + \int_V \va{g} (\rho_\mathrm{b} - \rho_\mathrm{f}) \dd{V} + = 0 + } +\end{aligned}$$ + +It is commonly assumed that $\va{g}$ is constant everywhere, with magnitude $\mathrm{g}$. +If we also assume that $\rho_\mathrm{f}$ is constant on the "submerged" side, +and zero on the "non-submerged" side, we find: + +$$\begin{aligned} + 0 + = \mathrm{g} (m_\mathrm{b} - m_\mathrm{f}) +\end{aligned}$$ + +In other words, the mass $m_\mathrm{b}$ of the entire body +is equal to the mass $m_\mathrm{f}$ of the fluid it displaces. +This is the best-known version of Archimedes' principle. + +Note that if $\rho_\mathrm{b} > \rho_\mathrm{f}$, +then the displaced mass $m_\mathrm{f} < m_\mathrm{b}$ +even if the entire body is submerged, +and the object will therefore continue to sink. + + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/source/know/concept/bb84-protocol/index.md b/source/know/concept/bb84-protocol/index.md new file mode 100644 index 0000000..95ba720 --- /dev/null +++ b/source/know/concept/bb84-protocol/index.md @@ -0,0 +1,233 @@ +--- +title: "BB84 protocol" +date: 2021-03-06 +categories: +- Quantum information +- Cryptography +layout: "concept" +--- + +The **BB84** or **Bennett-Brassard 1984** protocol is +a *quantum key distribution* (QKD) protocol, +whose purpose is to securely transmit a string of random bits +over a quantum channel for later use as a one-time pad. +It is provably information-secure, thanks to the fact that +quantum channels cannot be eavesdropped without interfering with the signal. + +Alice wants to send a secret key to Bob. +Between them, they have one quantum channel and one classical channel. +Both channels may have eavesdroppers without compromising the security of the BB84 protocol, +as long as Alice and Bob authenticate all data sent over the classical channel. + +First, Alice securely generates a sequence of random (classical) bits. +Note that the BB84 protocol is only suitable for random (high-entropy) data, +because the later stages of the protocol involve revealing parts of the data +over the (insecure) classical channel. + +For each bit, Alice randomly chooses a qubit basis, +either $\{ \Ket{0}, \Ket{1} \}$ (eigenstates of the $z$-spin $\hat{\sigma}_z$) +or $\{ \Ket{-}, \Ket{+} \}$ (eigenstates of the $x$-spin $\hat{\sigma}_x$). +Using the basis she chose, she then transmits the bits to Bob over the quantum channel, +encoding them as follows: + +$$\begin{aligned} + 0 \:\:\rightarrow\:\: \Ket{0} \:\mathrm{or}\: \Ket{+} + \qquad \quad + 1 \:\:\rightarrow\:\: \Ket{1} \:\mathrm{or}\: \Ket{-} +\end{aligned}$$ + +Crucially, Bob has no idea which basis Alice used for any of the bits. +For every bit, he chooses $\hat{\sigma}_z$ or $\hat{\sigma}_x$ at random, +and makes a measurement of the qubit, yielding 0 or 1. +If he guessed the basis correctly, he gets the bit value intended by Alice, +but if he guessed incorrectly, he randomly gets 0 or 1 with a 50-50 probability: + +$$\begin{aligned} + | \Inprod{0}{+} |^2 = | \Inprod{0}{-} |^2 = | \Inprod{1}{+} |^2 = | \Inprod{1}{-} |^2 = \frac{1}{2} +\end{aligned}$$ + +After Alice has sent all her qubits, +the next step is **basis reconciliation**: +over the classical channel, Bob announces, for each bit, +which basis he chose, and Alice tells him if he was right or wrong. +Bob discards all bits where he guessed wrongly. +If their quantum channel did not have any noise or eavesdroppers, +Alice and Bob now have a perfectly correlated secret string of bits. + + +## Eavesdropper detection + +But what if there is actually an eavesdropper? +Consider a third party, Eve, who wants to intercept Alice' secret string. +The main advantage of using QKD compared to classical protocols +is that such an eavesdropper can be detected. + +Suppose that Eve is performing an *intercept-resend attack* +(not very effective, but simple), +where she listens on the quantum channel. +For each qubit received from Alice, Eve chooses +$\hat{\sigma}_z$ or $\hat{\sigma}_x$ at random and measures it. +She records her results and resends the qubits to Bob +using the basis she chose, which may or may not be what Alice intended. + +If Eve guesses Alice' basis correctly, her presence is not revealed, +and the protocol proceeds as normal. +However, if she guesses wrongly (which has a probability of 50%), +her bit value might be incorrect, and she will resend whatever she measured +to Bob encoded in the wrong basis. + +If we assume that Eve chose wrongly, +Bob might measure using Eve's basis, +which will yield a 50% error rate due to Eve's mistake. +Otherwise, Bob might measure using Alice' basis, +which will also yield a 50% error rate due to the disagreement with Eve. + +In the end, the probability is 0.5 that Eve chose correctly, +which, multiplied by the 0.5 error rate from Bob's choice, +will result in a 0.25 error rate due to Eve's presence. +To detect this, after basis reconciliation, +Bob reveals a part of his secret string as a sacrifice, +which allows Alice to estimate the error rate. +If the rate is too high, Eve is detected, and the protocol can be aborted, +although that is not mandatory. + +This was an intercept-resend attack. +There exist other attacks, but similar logic holds. +It has been proven by Shor and Preskill in 2000 +that as long as the error rate is below 11%, +the BB84 protocol is fully secure, i.e. there cannot be any eavesdroppers. + + +## Error correction + +In practice, even without Eve, quantum channels are imperfect, +and will introduce some errors in the qubits received by Bob. +Suppose that after basis reconciliation, +Alice and Bob have the strings +$\{a_1, ..., a_N\}$ and $\{b_1, ..., b_N\}$, respectively. +We define $p$ as the probability that Alice and Bob agree on the $n$th bit, +which we assume to be greater than 50%: + +$$\begin{aligned} + p = P(a_n = b_n) > \frac{1}{2} +\end{aligned}$$ + +Ideally, $p = 1$. To improve $p$, the following simple scheme can be used: +starting at $n = 1$, Alice and Bob reveal $A$ and $B$ over the classical channel, +where $\oplus$ is an XOR: + +$$\begin{aligned} + A = a_n \oplus a_{n+1} + \qquad \quad + B = b_n \oplus b_{n+1} +\end{aligned}$$ + +If $A = B$, then $a_{n+1}$ and $b_{n+1}$ are discarded to prevent +a listener on the classical channel from learning anything about the string. +If $A \neq B$, all of $a_n$, $b_n$, $a_{n+1}$ and $b_{n+1}$ are discarded, +and then Alice and Bob move on to $n = 3$, etc. + +Given that $A = B$, the probability that $a_n = b_n$, +which is what we want, is given by: + +$$\begin{aligned} + P(a_{n} = b_{n} | A = B) + &= \frac{P(a_{n} = b_{n} \land A = B)}{P(A = B)} + \\ + &= \frac{P(a_{n} = b_{n} \land a_{n+1} = b_{n+1})}{P(a_{n} = b_{n} \land a_{n+1} = b_{n+1}) + P(a_{n} \neq b_{n} \land a_{n+1} \neq b_{n+1})} + \\ + &= \frac{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1})}{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1}) + P(a_{n} \neq b_{n}) \: P(a_{n+1} \neq b_{n+1})} +\end{aligned}$$ + +We use the definition of $p$ to get the following inequality, +which can be verified by plotting: + +$$\begin{aligned} + P(a_{n} = b_{n} | A = B) + = \frac{p^2}{p^2 + (1 - p)^2} + > p +\end{aligned}$$ + +Alice and Bob can repeat this error correction scheme multiple times, +until their estimate of $p$ is satisfactory. +This involves discarding many bits, +so the length $N_\mathrm{new}$ of the string they end up with +after one iteration is given by: + +$$\begin{aligned} + N_\mathrm{new} + = \frac{1}{2} N_\mathrm{old} P(A = B) + = \frac{1}{2} N_\mathrm{old} \big( p^2 + (1 - p)^2 \big) +\end{aligned}$$ + +More efficient schemes exist, which do not consume so many bits. + + +## Privacy amplification + +Suppose that after the error correction step, $p = 1$, +so Alice and Bob fully agree on the random string. +However, in the meantime, Eve has been listening, +and has been doing a good job +building up her own string $\{e_1, ..., e_N\}$, +such that she knows more that 50% of the bits: + +$$\begin{aligned} + q = P(e_n = a_n) > \frac{1}{2} +\end{aligned}$$ + +**Privacy amplification** is an optional final step of the BB84 protocol +which aims to reduce Eve's $q$. +Alice and Bob use their existing strings to generate a new one +$\{a_1', ..., a_M'\}$: + +$$\begin{aligned} + a_1' + = a_1 \oplus a_2 = b_1 \oplus b_2 + \qquad + a_2' + = a_3 \oplus a_4 = b_3 \oplus b_4 + \qquad + \cdots +\end{aligned}$$ + +Note that this halves the string's length; +more efficient schemes exist, which consume less. + +To see why this improves Alice and Bob's privacy, +suppose that Eve is following along, +and creates a new string $\{e_1', ..., e_M'\}$ +where $e_m' = e_{2m - 1} \oplus e_{2m}$. +The probability that Eve's result agrees with +Alice and Bob's string is given by: + +$$\begin{aligned} + P(e_m' = a_m') + &= P(e_1 = a_1 \land e_2 = a_2) + P(e_1 \neq a_1 \land e_2 \neq a_2) + \\ + &= P(e_1 = a_1) \: P(e_2 = a_2) + P(e_1 \neq a_1) \: P(e_2 \neq a_2) +\end{aligned}$$ + +Recognizing $q$ 's definition, +we find the following inequality, +which can be verified by plotting: + +$$\begin{aligned} + P(e_m' = a_m') + = q^2 + (1 - q)^2 + < q +\end{aligned}$$ + +After repeating this step several times, $q$ will be close to 1/2, +which is the ideal value: for $q =$ 0.5, +Eve would only know 50% of the bits, +which is equivalent to her guessing at random. + + +## References +1. N. Brunner, + *Quantum information theory: lecture notes*, + 2019, unpublished. +2. J.B. Brask, + *Quantum information: lecture notes*, + 2021, unpublished. diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md new file mode 100644 index 0000000..15b916d --- /dev/null +++ b/source/know/concept/bell-state/index.md @@ -0,0 +1,93 @@ +--- +title: "Bell state" +date: 2021-03-09 +categories: +- Quantum mechanics +- Quantum information +layout: "concept" +--- + +In quantum information, the **Bell states** are a set of four two-qubit states +which are simple and useful examples of [quantum entanglement](/know/concept/quantum-entanglement/). +They are given by: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \ket{\Phi^{\pm}} + &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{0}_B \pm \Ket{1}_A \Ket{1}_B \Big) + \\ + \ket{\Psi^{\pm}} + &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{1}_B \pm \Ket{1}_A \Ket{0}_B \Big) + \end{aligned} + } +\end{aligned}$$ + +Where e.g. $\Ket{0}_A \Ket{1}_B = \Ket{0}_A \otimes \Ket{1}_B$ +is the tensor product of qubit $A$ in state $\Ket{0}$ and $B$ in $\Ket{1}$. +These states form an orthonormal basis for the two-qubit +[Hilbert space](/know/concept/hilbert-space/). + +More importantly, however, +is that the Bell states are maximally entangled, +which we prove here for $\ket{\Phi^{+}}$. +Consider the following pure [density operator](/know/concept/density-operator/): + +$$\begin{aligned} + \hat{\rho} + = \ket{\Phi^{+}} \bra{\Phi^{+}} + &= \frac{1}{2} \Big( \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big) \Big( \Bra{0}_A \Bra{0}_B + \Bra{1}_A \Bra{1}_B \Big) +\end{aligned}$$ + +The reduced density operator $\hat{\rho}_A$ of qubit $A$ is then calculated as follows: + +$$\begin{aligned} + \hat{\rho}_A + &= \Tr_B(\hat{\rho}) + = \sum_{b = 0, 1} \Bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \Ket{b}_B + \\ + &= \sum_{b = 0, 1} \Big( \Ket{0}_A \Inprod{b}{0}_B + \Ket{1}_A \Inprod{b}{1}_B \Big) + \Big( \Bra{0}_A \Inprod{0}{b}_B + \Bra{1}_A \Inprod{1}{b}_B \Big) + \\ + &= \frac{1}{2} \Big( \Ket{0}_A \Bra{0}_A + \Ket{1}_A \Bra{1}_A \Big) + = \frac{1}{2} \hat{I} +\end{aligned}$$ + +This result is maximally mixed, therefore $\ket{\Phi^{+}}$ is maximally entangled. +The same holds for the other three Bell states, +and is equally true for qubit $B$. + +This means that a measurement of qubit $A$ +has a 50-50 chance to yield $\Ket{0}$ or $\Ket{1}$. +However, due to the entanglement, +measuring $A$ also has consequences for qubit $B$: + +$$\begin{aligned} + \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{0}{0}_B + \Inprod{0}{1}_A \Inprod{0}{1}_B \Big)^2 + = \frac{1}{2} + \\ + \big| \Bra{0}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{1}{0}_B + \Inprod{0}{1}_A \Inprod{1}{1}_B \Big)^2 + = 0 + \\ + \big| \Bra{1}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{0}{0}_B + \Inprod{1}{1}_A \Inprod{0}{1}_B \Big)^2 + = 0 + \\ + \big| \Bra{1}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{1}{0}_B + \Inprod{1}{1}_A \Inprod{1}{1}_B \Big)^2 + = \frac{1}{2} +\end{aligned}$$ + +As an example, if $A$ collapses into $\Ket{0}$ due to a measurement, +then $B$ instantly also collapses into $\Ket{0}$, never $\Ket{1}$, +even if it was not measured. +This was a specific example for $\ket{\Phi^{+}}$, +but analogous results can be found for the other Bell states. + + +## References +1. J.B. Brask, + *Quantum information: lecture notes*, + 2021, unpublished. diff --git a/source/know/concept/bells-theorem/index.md b/source/know/concept/bells-theorem/index.md new file mode 100644 index 0000000..1752854 --- /dev/null +++ b/source/know/concept/bells-theorem/index.md @@ -0,0 +1,372 @@ +--- +title: "Bell's theorem" +date: 2021-03-28 +categories: +- Physics +- Quantum mechanics +- Quantum information +layout: "concept" +--- + +**Bell's theorem** states that the laws of quantum mechanics +cannot be explained by theories built on +so-called **local hidden variables** (LHVs). + +Suppose that we have two spin-1/2 particles, called $A$ and $B$, +in an entangled [Bell state](/know/concept/bell-state/): + +$$\begin{aligned} + \Ket{\Psi^{-}} + = \frac{1}{\sqrt{2}} \Big( \Ket{\uparrow \downarrow} - \Ket{\downarrow \uparrow} \Big) +\end{aligned}$$ + +Since they are entangled, +if we measure the $z$-spin of particle $A$, and find e.g. $\Ket{\uparrow}$, +then particle $B$ immediately takes the opposite state $\Ket{\downarrow}$. +The point is that this collapse is instant, +regardless of the distance between $A$ and $B$. + +Einstein called this effect "action-at-a-distance", +and used it as evidence that quantum mechanics is an incomplete theory. +He said that there must be some **hidden variable** $\lambda$ +that determines the outcome of measurements of $A$ and $B$ +from the moment the entangled pair is created. +However, according to Bell's theorem, he was wrong. + +To prove this, let us assume that Einstein was right, and some $\lambda$, +which we cannot understand, let alone calculate or measure, controls the results. +We want to know the spins of the entangled pair +along arbitrary directions $\vec{a}$ and $\vec{b}$, +so the outcomes for particles $A$ and $B$ are: + +$$\begin{aligned} + A(\vec{a}, \lambda) = \pm 1 + \qquad \quad + B(\vec{b}, \lambda) = \pm 1 +\end{aligned}$$ + +Where $\pm 1$ are the eigenvalues of the Pauli matrices +in the chosen directions $\vec{a}$ and $\vec{b}$: + +$$\begin{aligned} + \hat{\sigma}_a + &= \vec{a} \cdot \vec{\sigma} + = a_x \hat{\sigma}_x + a_y \hat{\sigma}_y + a_z \hat{\sigma}_z + \\ + \hat{\sigma}_b + &= \vec{b} \cdot \vec{\sigma} + = b_x \hat{\sigma}_x + b_y \hat{\sigma}_y + b_z \hat{\sigma}_z +\end{aligned}$$ + +Whether $\lambda$ is a scalar or a vector does not matter; +we simply demand that it follows an unknown probability distribution $\rho(\lambda)$: + +$$\begin{aligned} + \int \rho(\lambda) \dd{\lambda} = 1 + \qquad \quad + \rho(\lambda) \ge 0 +\end{aligned}$$ + +The product of the outcomes of $A$ and $B$ then has the following expectation value. +Note that we only multiply $A$ and $B$ for shared $\lambda$-values: +this is what makes it a **local** hidden variable: + +$$\begin{aligned} + \Expval{A_a B_b} + = \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda} +\end{aligned}$$ + +From this, two inequalities can be derived, +which both prove Bell's theorem. + + +## Bell inequality + +If $\vec{a} = \vec{b}$, then we know that $A$ and $B$ always have opposite spins: + +$$\begin{aligned} + A(\vec{a}, \lambda) + = A(\vec{b}, \lambda) + = - B(\vec{b}, \lambda) +\end{aligned}$$ + +The expectation value of the product can therefore be rewritten as follows: + +$$\begin{aligned} + \Expval{A_a B_b} + = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} +\end{aligned}$$ + +Next, we introduce an arbitrary third direction $\vec{c}$, +and use the fact that $( A(\vec{b}, \lambda) )^2 = 1$: + +$$\begin{aligned} + \Expval{A_a B_b} - \Expval{A_a B_c} + &= - \int \rho(\lambda) \Big( A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) - A(\vec{a}, \lambda) \: A(\vec{c}, \lambda) \Big) \dd{\lambda} + \\ + &= - \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} +\end{aligned}$$ + +Inside the integral, the only factors that can be negative +are the last two, and their product is $\pm 1$. +Taking the absolute value of the whole left, +and of the integrand on the right, we thus get: + +$$\begin{aligned} + \Big| \Expval{A_a B_b} - \Expval{A_a B_c} \Big| + &\le \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) + \: \Big| A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \Big| \dd{\lambda} + \\ + &\le \int \rho(\lambda) \dd{\lambda} - \int \rho(\lambda) A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \dd{\lambda} +\end{aligned}$$ + +Since $\rho(\lambda)$ is a normalized probability density function, +we arrive at the **Bell inequality**: + +$$\begin{aligned} + \boxed{ + \Big| \Expval{A_a B_b} - \Expval{A_a B_c} \Big| + \le 1 + \Expval{A_b B_c} + } +\end{aligned}$$ + +Any theory involving an LHV $\lambda$ must obey this inequality. +The problem, however, is that quantum mechanics dictates the expectation values +for the state $\Ket{\Psi^{-}}$: + +$$\begin{aligned} + \Expval{A_a B_b} = - \vec{a} \cdot \vec{b} +\end{aligned}$$ + +Finding directions which violate the Bell inequality is easy: +for example, if $\vec{a}$ and $\vec{b}$ are orthogonal, +and $\vec{c}$ is at a $\pi/4$ angle to both of them, +then the left becomes $0.707$ and the right $0.293$, +which clearly disagrees with the inequality, +meaning that LHVs are impossible. + + +## CHSH inequality + +The **Clauser-Horne-Shimony-Holt** or simply **CHSH inequality** +takes a slightly different approach, and is more useful in practice. + +Consider four spin directions, two for $A$ called $\vec{a}_1$ and $\vec{a}_2$, +and two for $B$ called $\vec{b}_1$ and $\vec{b}_2$. +Let us introduce the following abbreviations: + +$$\begin{aligned} + A_1 &= A(\vec{a}_1, \lambda) + \qquad \quad + A_2 = A(\vec{a}_2, \lambda) + \\ + B_1 &= B(\vec{b}_1, \lambda) + \qquad \quad + B_2 = B(\vec{b}_2, \lambda) +\end{aligned}$$ + +From the definition of the expectation value, +we know that the difference is given by: + +$$\begin{aligned} + \Expval{A_1 B_1} - \Expval{A_1 B_2} + = \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \Big) \dd{\lambda} +\end{aligned}$$ + +We introduce some new terms and rearrange the resulting expression: + +$$\begin{aligned} + \Expval{A_1 B_1} - \Expval{A_1 B_2} + &= \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \pm A_1 B_1 A_2 B_2 \mp A_1 B_1 A_2 B_2 \Big) \dd{\lambda} + \\ + &= \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} + - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} +\end{aligned}$$ + +Taking the absolute value of both sides +and invoking the triangle inequality then yields: + +$$\begin{aligned} + \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big| + &= \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} + - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg| + \\ + &\le \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} \!\bigg| + + \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg| +\end{aligned}$$ + +Using the fact that the product of $A$ and $B$ is always either $-1$ or $+1$, +we can reduce this to: + +$$\begin{aligned} + \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big| + &\le \int \rho(\lambda) \Big| A_1 B_1 \Big| \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} + + \!\int \rho(\lambda) \Big| A_1 B_2 \Big| \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} + \\ + &\le \int \rho(\lambda) \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} + + \!\int \rho(\lambda) \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} +\end{aligned}$$ + +Evaluating these integrals gives us the following inequality, +which holds for both choices of $\pm$: + +$$\begin{aligned} + \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big| + &\le 2 \pm \Expval{A_2 B_2} \pm \Expval{A_2 B_1} +\end{aligned}$$ + +We should choose the signs such that the right-hand side is as small as possible, that is: + +$$\begin{aligned} + \Bi