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diff --git a/content/know/concept/bell-state/index.pdc b/content/know/concept/bell-state/index.pdc new file mode 100644 index 0000000..5e147e2 --- /dev/null +++ b/content/know/concept/bell-state/index.pdc @@ -0,0 +1,93 @@ +--- +title: "Bell state" +firstLetter: "B" +publishDate: 2021-03-09 +categories: +- Quantum mechanics +- Quantum information + +date: 2021-03-09T17:31:29+01:00 +draft: false +markup: pandoc +--- + +# Bell state + +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: + +$$\begin{aligned} + \hat{\rho}_A + &= \Tr_B(\hat{\rho}) + = \sum_{b = 0, 1} \bra{b}_B \Big( \ket*{\Phi^{+}} \bra*{\Phi^{+}} \Big) \ket{b}_B + \\ + &= \sum_{b = 0, 1} \Big( \ket{0}_A \braket{b}{0}_B + \ket{1}_A \braket{b}{1}_B \Big) + \Big( \bra{0}_A \braket{0}{b}_B + \bra{1}_A \braket{1}{b}_B \Big) + \\ + &= \frac{1}{2} \Big( \ket{0}_A \bra{0}_A + \ket{1}_A \bra{1}_A \Big) + = \frac{1}{2} \hat{I} +\end{aligned}$$ + +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 $B$: + +$$\begin{aligned} + \big| \bra{0}_A \bra{0}_B \: \ket*{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \braket{0}{0}_A \braket{0}{0}_B + \braket{0}{1}_A \braket{0}{1}_B \Big)^2 + = \frac{1}{2} + \\ + \big| \bra{0}_A \bra{1}_B \: \ket*{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \braket{0}{0}_A \braket{1}{0}_B + \braket{0}{1}_A \braket{1}{1}_B \Big)^2 + = 0 + \\ + \big| \bra{1}_A \bra{0}_B \: \ket*{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \braket{1}{0}_A \braket{0}{0}_B + \braket{1}{1}_A \braket{0}{1}_B \Big)^2 + = 0 + \\ + \big| \bra{1}_A \bra{1}_B \: \ket*{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \braket{1}{0}_A \braket{1}{0}_B + \braket{1}{1}_A \braket{1}{1}_B \Big)^2 + = \frac{1}{2} +\end{aligned}$$ + +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. diff --git a/content/know/concept/binomial-distribution/index.pdc b/content/know/concept/binomial-distribution/index.pdc index 68dec8c..70cc897 100644 --- a/content/know/concept/binomial-distribution/index.pdc +++ b/content/know/concept/binomial-distribution/index.pdc @@ -191,3 +191,9 @@ $$\begin{aligned} \lim_{N \to \infty} P_N(n) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\!\Big(\!-\!\frac{(n - \mu)^2}{2 \sigma^2} \Big) } \end{aligned}$$ + + +## References +1. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. diff --git a/content/know/concept/bloch-sphere/bloch-small.jpg b/content/know/concept/bloch-sphere/bloch-small.jpg Binary files differnew file mode 100644 index 0000000..e99c0e1 --- /dev/null +++ b/content/know/concept/bloch-sphere/bloch-small.jpg diff --git a/content/know/concept/bloch-sphere/bloch.jpg b/content/know/concept/bloch-sphere/bloch.jpg Binary files differnew file mode 100644 index 0000000..9515d84 --- /dev/null +++ b/content/know/concept/bloch-sphere/bloch.jpg diff --git a/content/know/concept/bloch-sphere/index.pdc b/content/know/concept/bloch-sphere/index.pdc new file mode 100644 index 0000000..843de1f --- /dev/null +++ b/content/know/concept/bloch-sphere/index.pdc @@ -0,0 +1,129 @@ +--- +title: "Bloch sphere" +firstLetter: "B" +publishDate: 2021-03-09 +categories: +- Quantum mechanics +- Quantum information + +date: 2021-03-09T15:35:33+01:00 +draft: false +markup: pandoc +--- + +# Bloch sphere + +In quantum mechanics, particularly quantum information, +the **Bloch sphere** is an invaluable tool to visualize qubits. +All pure qubit states are represented by a point on the sphere's surface: + +<a href="bloch.jpg"> +<img src="bloch-small.jpg" style="width:50%;display:block;margin:auto;"> +</a> + +The $x$, $y$ and $z$-axes represent the components of a spin-1/2-alike system, +and their extremes are the eigenstates of the Pauli matrices: + +$$\begin{aligned} + \hat{\sigma}_z + \to \{\ket{0}, \ket{1}\} + \qquad + \hat{\sigma}_x + \to \{\ket{+}, \ket{-}\} + \qquad + \hat{\sigma}_y + \to \{\ket{+i}, \ket{-i}\} +\end{aligned}$$ + +Where the latter two states are expressed as follows in the conventional $z$-basis: + +$$\begin{aligned} + \ket{\pm} + = \frac{\ket{0} \pm \ket{1}}{\sqrt{2}} + \qquad \quad + \ket{\pm i} + = \frac{\ket{0} \pm i \ket{1}}{\sqrt{2}} +\end{aligned}$$ + +More generally, every point on the surface of the sphere +describes a pure qubit state in terms of the angles $\theta$ and $\varphi$, +respectively the elevation and azimuth: + +$$\begin{aligned} + \ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \ket{1} +\end{aligned}$$ + +We can generalize this further by describing points using the **Bloch vector** $\vec{r}$, +with radius $r \le 1$: + +$$\begin{aligned} + \boxed{ + \vec{r} + = \begin{bmatrix} r_x \\ r_y \\ r_z \end{bmatrix} + = \begin{bmatrix} r \sin\theta \cos\varphi \\ r \sin\theta \sin\varphi \\ r \cos\theta \end{bmatrix} + } +\end{aligned}$$ + +Note that $\vec{r}$ is not actually a qubit state, +but rather an implicit description of one, +meaning that it does not need to be normalized. +The main point of the Bloch vector is that it allows us +to describe the qubit using a [density operator](/know/concept/density-operator/): + +$$\begin{aligned} + \boxed{ + \hat{\rho} + = \frac{1}{2} \Big( \hat{I} + \vec{r} \cdot \vec{\sigma} \Big) + } +\end{aligned}$$ + +Where $\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$ is the Pauli "vector". +Now, we know that $\hat{\rho}$ represents a pure ensemble +if and only if it is idempotent, i.e. $\hat{\rho}^2 = \hat{\rho}$: + +$$\begin{aligned} + \hat{\rho}^2 + &= \frac{1}{4} \Big( \hat{I}^2 + 2 \hat{I} (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big) + = \frac{1}{4} \Big( \hat{I} + 2 (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big) +\end{aligned}$$ + +You can easily convince yourself that if $(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$, +then we get $\hat{\rho}$ again, and the state is pure: + +$$\begin{aligned} + (\vec{r} \cdot \vec{\sigma})^2 + &= (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z)^2 + \\ + &= r_x^2 \hat{\sigma}_x^2 + r_x r_y \hat{\sigma}_x \hat{\sigma}_y + r_x r_z \hat{\sigma}_x \hat{\sigma}_z + + r_x r_y \hat{\sigma}_y \hat{\sigma}_x + r_y^2 \hat{\sigma}_y^2 + \\ + &\quad + r_y r_z \hat{\sigma}_y \hat{\sigma}_z + r_x r_z \hat{\sigma}_z \hat{\sigma}_x + + r_y r_z \hat{\sigma}_z \hat{\sigma}_y + r_z^2 \hat{\sigma}_z^2 + \\ + &= r_x^2 \hat{I} + r_y^2 \hat{I} + r_z^2 \hat{I} + + r_x r_y \{ \hat{\sigma}_x, \hat{\sigma}_y \} + + r_y r_z \{ \hat{\sigma}_y, \hat{\sigma}_z \} + + r_x r_z \{ \hat{\sigma}_x, \hat{\sigma}_z \} + \\ + &= (r_x^2 + r_y^2 + r_z^2) \hat{I} + = r^2 \hat{I} +\end{aligned}$$ + +Therefore, if the radius $r = 1$, the ensemble is pure, +else if $r < 1$ it is mixed. + +Another useful property of the Bloch vector +is that the expectation value of the Pauli matrices +are given by the corresponding component of $\vec{r}$, +for example for $\hat{\sigma}_z$: + +$$\begin{aligned} + \expval{\hat{\sigma}_z} + &= \Tr(\hat{\rho} \hat{\sigma}_z) + = \frac{1}{2} \Tr\big(\hat{\sigma}_z + (\vec{r} \cdot \vec{\sigma}) \hat{\sigma}_z \big) + = \frac{1}{2} \Tr\big( (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z) \hat{\sigma}_z \big) + \\ + &= \frac{1}{2} \Tr\big( r_x \hat{\sigma}_x \hat{\sigma}_z + r_y \hat{\sigma}_y \hat{\sigma}_z + r_z \hat{\sigma}_z^2 \big) + = \frac{1}{2} \Tr\big( r_z \hat{I} \big) + = r_z +\end{aligned}$$ diff --git a/content/know/concept/central-limit-theorem/index.pdc b/content/know/concept/central-limit-theorem/index.pdc new file mode 100644 index 0000000..270bb0b --- /dev/null +++ b/content/know/concept/central-limit-theorem/index.pdc @@ -0,0 +1,208 @@ +--- +title: "Central limit theorem" +firstLetter: "C" +publishDate: 2021-03-09 +categories: +- Statistics + +date: 2021-03-09T20:39:38+01:00 +draft: false +markup: pandoc +--- + +# Central limit theorem + +In statistics, the **central limit theorem** states that +the sum of many independent variables tends towards a normal distribution, +even if the individual variables $x_n$ follow different distributions. + +For example, by taking $M$ samples of size $N$ from a population, +and calculating $M$ averages $\mu_m$ (which involves summing over $N$), +the resulting means $\mu_m$ are normally distributed +across the $M$ samples if $N$ is sufficiently large. + +More formally, for $N$ independent variables $x_n$ with probability distributions $p(x_n)$, +the central limit theorem states the following, +where we define the sum $S$: + +$$\begin{aligned} + S = \sum_{n = 1}^N x_n + \qquad + \mu_S = \sum_{n = 1}^N \mu_n + \qquad + \sigma_S^2 = \sum_{n = 1}^N \sigma_n^2 +\end{aligned}$$ + +And crucially, it states that the probability distribution $p_N(S)$ of $S$ for $N$ variables +will become a normal distribution when $N$ goes to infinity: + +$$\begin{aligned} + \boxed{ + \lim_{N \to \infty} \!\big(p_N(S)\big) + = \frac{1}{\sigma_S \sqrt{2 \pi}} \: \exp\!\Big( -\frac{(\mu_S - S)^2}{2 \sigma_S^2} \Big) + } +\end{aligned}$$ + +We prove this below, +but first we need to introduce some tools. +Given a probability density $p(x)$, its [Fourier transform](/know/concept/fourier-transform/) +is called the **characteristic function** $\phi(k)$: + +$$\begin{aligned} + \phi(k) = \int_{-\infty}^\infty p(x) \exp(i k x) \dd{x} +\end{aligned}$$ + +Note that $\phi(k)$ can be interpreted as the average of $\exp(i k x)$. +We take its Taylor expansion in two separate ways, +where an overline denotes the mean: + +$$\begin{aligned} + \phi(k) + = \sum_{n = 0}^\infty \frac{k^n}{n!} \: \phi^{(n)}(0) + \qquad + \phi(k) + = \overline{\exp(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n} +\end{aligned}$$ + +By comparing the coefficients of these two power series, +we get a useful relation: + +$$\begin{aligned} + \phi^{(n)}(0) = i^n \: \overline{x^n} +\end{aligned}$$ + +Next, the **cumulants** $C^{(n)}$ are defined from the Taylor expansion of $\ln\!\big(\phi(k)\big)$: + +$$\begin{aligned} + \ln\!\big( \phi(k) \big) + = \sum_{n = 1}^\infty \frac{(ik)^n}{n!} C^{(n)} + \quad \mathrm{where} \quad + C^{(n)} = \frac{1}{i^n} \: \dv[n]{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0} +\end{aligned}$$ + +The first two cumulants $C^{(1)}$ and $C^{(2)}$ are of particular interest, +since they turn out to be the mean and the variance respectively, +using our earlier relation: + +$$\begin{aligned} + C^{(1)} + &= - i \dv{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0} + = - i \frac{\phi'(0)}{\exp(0)} + = \overline{x} + \\ + C^{(2)} + &= - \dv[2]{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0} + = \frac{\big(\phi'(0)\big)^2}{\exp(0)^2} - \frac{\phi''(0)}{\exp(0)} + = - \overline{x}^2 + \overline{x^2} = \sigma^2 +\end{aligned}$$ + +Let us now define $S$ as the sum of $N$ independent variables $x_n$, in other words: + +$$\begin{aligned} + S = \sum_{n = 1}^N x_n = x_1 + x_2 + ... + x_N +\end{aligned}$$ + +The probability density of $S$ is then as follows, where $p(x_n)$ are +the densities of all the individual variables and $\delta$ is +the [Dirac delta function](/know/concept/dirac-delta-function/): + +$$\begin{aligned} + p(S) + &= \idotsint_{-\infty}^\infty \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( S - \sum_{n = 1}^N x_n \Big) \dd{x_1} \cdots \dd{x_N} + \\ + &= \Big( p_1 * \big( p_2 * ( ... * (p_N * \delta))\big)\Big)(S) +\end{aligned}$$ + +In other words, the integrals pick out all combinations of $x_n$ which +add up to the desired $S$-value, and multiply the probabilities +$p(x_1) p(x_2) \cdots p(x_N)$ of each such case. This is a convolution, +so the [convolution theorem](/know/concept/convolution-theorem/) +states that it is a product in the Fourier domain: + +$$\begin{aligned} + \phi_S(k) = \prod_{n = 1}^N \phi_n(k) +\end{aligned}$$ + +By taking the logarithm of both sides, the product becomes a sum, +which we further expand: + +$$\begin{aligned} + \ln\!\big(\phi_S(k)\big) + = \sum_{n = 1}^N \ln\!\big(\phi_n(k)\big) + = \sum_{n = 1}^N \sum_{m = 1}^{\infty} \frac{(ik)^m}{m!} C_n^{(m)} +\end{aligned}$$ + +Consequently, the cumulants $C^{(m)}$ stack additively for the sum $S$ +of independent variables $x_m$, and therefore +the means $C^{(1)}$ and variances $C^{(2)}$ do too: + +$$\begin{aligned} + C_S^{(m)} = \sum_{n = 1}^N C_n^{(m)} = C_1^{(m)} + C_2^{(m)} + ... + C_N^{(m)} +\end{aligned}$$ + +We now introduce the scaled sum $z$ as the new combined variable: + +$$\begin{aligned} + z = \frac{S}{\sqrt{N}} = \frac{1}{\sqrt{N}} (x_1 + x_2 + ... + x_N) +\end{aligned}$$ + +Its characteristic function $\phi_z(k)$ is then as follows, +with $\sqrt{N}$ appearing in the arguments of $\phi_n$: + +$$\begin{aligned} + \phi_z(k) + &= \idotsint + \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp(i k z) + \dd{x_1} \cdots \dd{x_N} + \\ + &= \idotsint + \Big( \prod_{n = 1}^N p(x_n) \Big) \exp\!\Big( i \frac{k}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) + \dd{x_1} \cdots \dd{x_N} + \\ + &= \prod_{n = 1}^N \phi_n\Big(\frac{k}{\sqrt{N}}\Big) +\end{aligned}$$ + +By expanding $\ln\!\big(\phi_z(k)\big)$ in terms of its cumulants $C^{(m)}$ +and introducing $\kappa = k / \sqrt{N}$, we see that the higher-order terms +become smaller for larger $N$: + +$$\begin{gathered} + \ln\!\big( \phi_z(k) \big) + = \sum_{m = 1}^\infty \frac{(ik)^m}{m!} C^{(m)} + \\ + C^{(m)} + = \frac{1}{i^m} \dv[m]{k} \sum_{n = 1}^N \ln\!\bigg( \phi_n\Big(\frac{k}{\sqrt{N}}\Big) \bigg) + = \frac{1}{i^m N^{m/2}} \dv[m]{\kappa} \sum_{n = 1}^N \ln\!\big( \phi_n(\kappa) \big) +\end{gathered}$$ + +For sufficiently large $N$, we can therefore approximate it using just the first two terms: + +$$\begin{aligned} + \ln\!\big( \phi_z(k) \big) + &\approx i k C^{(1)} - \frac{k^2}{2} C^{(2)} + = i k \overline{z} - \frac{k^2}{2} \sigma_z^2 + \\ + \phi_z(k) + &\approx \exp(i k \overline{z}) \exp\!(- k^2 \sigma_z^2 / 2) +\end{aligned}$$ + +We take its inverse Fourier transform to get the density $p(z)$, +which turns out to be a Gaussian normal distribution, +which is even already normalized: + +$$\begin{aligned} + p(z) + = \hat{\mathcal{F}}^{-1} \{\phi_z(k)\} + &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp(- k^2 \sigma_z^2 / 2) \dd{k} + \\ + &= \frac{1}{\sqrt{2 \pi \sigma_z^2}} \exp\!\Big(\!-\! \frac{(z - \overline{z})^2}{2 \sigma_z^2} \Big) +\end{aligned}$$ + +Therefore, the sum of many independent variables tends to a normal distribution, +regardless of the densities of the individual variables. + + +## References +1. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. diff --git a/content/know/concept/density-operator/index.pdc b/content/know/concept/density-operator/index.pdc index 39c2e85..5126f31 100644 --- a/content/know/concept/density-operator/index.pdc +++ b/content/know/concept/density-operator/index.pdc @@ -81,15 +81,17 @@ $$\begin{aligned} This can be used to find out whether a given $\hat{\rho}$ represents a pure or mixed ensemble. -Next, we define the ensemble average $\expval*{\expval*{\hat{L}}}$ -as the mean of the expectation values for states in the ensemble, -which can be calculated like so: +Next, we define the ensemble average $\expval*{\hat{O}}$ +as the mean of the expectation values of $\hat{O}$ for states in the ensemble. +We use the same notation as for the pure expectation value, +since this is only a small extension of the concept to mixed ensembles. +It is calculated like so: $$\begin{aligned} \boxed{ - \expval*{\expval*{\hat{L}}} - = \sum_{n} p_n \matrixel{\Psi_n}{\hat{L}}{\Psi_n} - = \mathrm{Tr}(\hat{L} \hat{\rho}) + \expval*{\hat{O}} + = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O}}{\Psi_n} + = \mathrm{Tr}(\hat{\rho} \hat{O}) } \end{aligned}$$ @@ -97,13 +99,13 @@ To prove the latter, we write out the trace $\mathrm{Tr}$ as the sum of the diagonal elements, so: $$\begin{aligned} - \mathrm{Tr}(\hat{L} \hat{\rho}) - &= \sum_{j} \matrixel{j}{\hat{L} \hat{\rho}}{j} - = \sum_{j} \sum_{n} p_n \matrixel{j}{\hat{L}}{\Psi_n} \braket{\Psi_n}{j} + \mathrm{Tr}(\hat{\rho} \hat{O}) + &= \sum_{j} \matrixel{j}{\hat{\rho} \hat{O}}{j} + = \sum_{j} \sum_{n} p_n \braket{j}{\Psi_n} \matrixel{\Psi_n}{\hat{O}}{j} \\ - &= \sum_{n} \sum_{j} p_n \braket{\Psi_n}{j} \matrixel{j}{\hat{L}}{\Psi_n} - = \sum_{n} p_n \matrixel{\Psi_n}{\hat{I} \hat{L}}{\Psi_n} - = \expval*{\expval*{\hat{L}}} + &= \sum_{n} \sum_{j} p_n\matrixel{\Psi_n}{\hat{O}}{j} \braket{j}{\Psi_n} + = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O} \hat{I}}{\Psi_n} + = \expval*{\hat{O}} \end{aligned}$$ In both the pure and mixed cases, diff --git a/content/know/concept/dirac-delta-function/index.pdc b/content/know/concept/dirac-delta-function/index.pdc index 3982afc..6c9f2b0 100644 --- a/content/know/concept/dirac-delta-function/index.pdc +++ b/content/know/concept/dirac-delta-function/index.pdc @@ -13,8 +13,10 @@ markup: pandoc # Dirac delta function -The **Dirac delta function** $\delta(x)$, often just called the **delta function**, -is an infinitely narrow discontinuous "spike" at $x = 0$ whose area is +The **Dirac delta function** $\delta(x)$, often just the **delta function**, +is a function (or, more accurately, a [Schwartz distribution](/know/concept/schwartz-distribution/)) +that is commonly used in physics. +It is an infinitely narrow discontinuous "spike" at $x = 0$ whose area is defined to be 1: $$\begin{aligned} @@ -29,7 +31,7 @@ $$\begin{aligned} } \end{aligned}$$ -It is sometimes also called the **sampling function**, due to its most +It is sometimes also called the **sampling function**, thanks to its most important property: the so-called **sampling property**: $$\begin{aligned} diff --git a/content/know/concept/impulse-response/index.pdc b/content/know/concept/impulse-response/index.pdc new file mode 100644 index 0000000..012a2c3 --- /dev/null +++ b/content/know/concept/impulse-response/index.pdc @@ -0,0 +1,64 @@ +--- +title: "Impulse response" +firstLetter: "I" +publishDate: 2021-03-09 +categories: +- Mathematics +- Physics + +date: 2021-03-09T20:34:38+01:00 +draft: false +markup: pandoc +--- + +# Impulse response + +The **impulse response** $u_p(t)$ of a system whose behaviour is described +by a linear operator $\hat{L}$, is defined as the reponse of the system +when forced by the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(t)$: + +$$\begin{aligned} + \boxed{ + \hat{L} \{ u_p(t) \} = \delta(t) + } +\end{aligned}$$ + +This can be used to find the response $u(t)$ of $\hat{L}$ to +*any* forcing function $f(t)$, i.e. not only $\delta(t)$, +by simply taking the convolution with $u_p(t)$: + +$$\begin{aligned} + \boxed{ + \hat{L} \{ u(t) \} = f(t) + \quad \implies \quad + u(t) = (f * u_p)(t) + } +\end{aligned}$$ + +*__Proof.__ Starting from the definition of $u_p(t)$, +we shift the argument by some constant $\tau$, +and multiply both sides by the constant $f(\tau)$:* + +$$\begin{aligned} + \hat{L} \{ u_p(t - \tau) \} &= \delta(t - \tau) + \\ + \hat{L} \{ f(\tau) \: u_p(t - \tau) \} &= f(\tau) \: \delta(t - \tau) +\end{aligned}$$ + +*Where $f(\tau)$ can be moved inside using the +linearity of $\hat{L}$. Integrating over $\tau$ then gives us:* + +$$\begin{aligned} + \int_0^\infty \hat{L} \{ f(\tau) \: u_p(t - \tau) \} \dd{\tau} + &= \int_0^\infty f(\tau) \: \delta(t - \tau) \dd{\tau} + = f(t) +\end{aligned}$$ + +*The integral and $\hat{L}$ are operators of different variables, so we reorder them:* + +$$\begin{aligned} + \hat{L} \int_0^\infty f(\tau) \: u_p(t - \tau) \dd{\tau} + &= (f * u_p)(t) = \hat{L}\{ u(t) \} = f(t) +\end{aligned}$$ + +*__Q.E.D.__* diff --git a/content/know/concept/quantum-teleportation/index.pdc b/content/know/concept/quantum-teleportation/index.pdc index 3287544..b4394c9 100644 --- a/content/know/concept/quantum-teleportation/index.pdc +++ b/content/know/concept/quantum-teleportation/index.pdc @@ -97,7 +97,7 @@ $$\begin{aligned} &= \alpha \ket{1} + \beta \ket{0} \qquad \quad \hat{\sigma}_x \hat{\sigma}_z \ket{q} - = \alpha \ket{0} - \beta \ket{1} + = \alpha \ket{1} - \beta \ket{0} \end{aligned}$$ Consequently, Alice and Bob are sharing (or, to be precise, seeing different sides of) @@ -143,3 +143,7 @@ the entangled $\ket*{\Phi^{+}}_{AB}$ state must be distributed in advance, and Alice' declaration of her result is sent classically. Before receiving that, Bob only sees his side of the maximally entangled Bell state $\ket*{\Phi^{+}}_{AB}$, which contains nothing of $\ket{q}$. + + +## References +1. N. Brunner, *Quantum information theory: lecture notes*, 2019. |
