diff options
Diffstat (limited to 'source')
| -rw-r--r-- | source/know/concept/blasius-boundary-layer/index.md | 82 | ||||
| -rw-r--r-- | source/know/concept/bloch-sphere/index.md | 58 | ||||
| -rw-r--r-- | source/know/concept/dirac-notation/index.md | 76 | ||||
| -rw-r--r-- | source/know/concept/gram-schmidt-method/index.md | 59 | ||||
| -rw-r--r-- | source/know/concept/matsubara-sum/index.md | 2 |
5 files changed, 157 insertions, 120 deletions
diff --git a/source/know/concept/blasius-boundary-layer/index.md b/source/know/concept/blasius-boundary-layer/index.md index de80f96..bca933f 100644 --- a/source/know/concept/blasius-boundary-layer/index.md +++ b/source/know/concept/blasius-boundary-layer/index.md @@ -12,88 +12,76 @@ layout: "concept" In fluid dynamics, the **Blasius boundary layer** is an application of the [Prandtl equations](/know/concept/prandtl-equations/), which govern the flow of a fluid -at large Reynolds number $$\mathrm{Re} \gg 1$$ -close to a surface. +at large [Reynolds number](/know/concept/reynolds-number/) +$$\mathrm{Re} \gg 1$$ close to a surface. Specifically, the Blasius layer is the solution for a half-plane approached from the edge by a fluid. -A fluid with velocity field $$\va{v} = U \vu{e}_x$$ flows to the plane, -which starts at $$y = 0$$ and exists for $$x \ge 0$$. -To describe this, we make an ansatz -for the *slip-flow* region's $$x$$-velocity $$v_x(x, y)$$: +Let the half-plane lie in the $$(x,z)$$-plane (i.e. at $$y = 0$$) +and exist for all $$x \ge 0$$, such that its edge lies on the $$z$$-axis. +A fluid with velocity field $$\va{v} = U \vu{e}_x$$ +approaches the half-plane's edge head-on. +To describe the fluid's movements around the plane, +we make an ansatz for the so-called **slip-flow region**'s $$x$$-velocity $$v_x(x, y)$$: $$\begin{aligned} v_x = U f'(s) - \qquad \quad + \qquad \qquad s \equiv \frac{y}{\delta(x)} \end{aligned}$$ +Where $$\delta(x) \equiv \sqrt{\nu x / U}$$ is the boundary layer thickness +estimate that was used to derive the Prandtl equations. Note that $$f'(s)$$ is the derivative of an unknown $$f(s)$$, -and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$. -Furthermore, $$\delta(x)$$ is the thickness of the stationary boundary layer at the surface. -To derive the Prandtl equations, -the estimate $$\delta(x) = \sqrt{\nu x / U}$$ was used, -which we will stick with. -For later use, it is worth writing the derivatives of $$s$$: +and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$, +i.e. the fluid is stationary at the half-plane's surface $$s = 0$$, +and has velocity $$U$$ far away at $$s \to \infty$$. -$$\begin{aligned} - \pdv{s}{x} - = - y \frac{\delta'}{\delta^2} - = - s \frac{\delta'}{\delta} - \qquad \quad - \pdv{s}{y} - = \frac{1}{\delta} -\end{aligned}$$ - -Inserting the ansatz for $$v_x$$ into the incompressibility condition then yields: +Inserting the ansatz into the incompressibility condition $$\nabla \cdot \va{v} = 0$$ yields: $$\begin{aligned} \pdv{v_y}{y} = - \pdv{v_x}{x} - = U s f'' \frac{\delta'}{\delta} + = - \pdv{v_x}{s} \pdv{s}{x} + = U \frac{\delta'}{\delta} s f'' \end{aligned}$$ -Which we integrate to get an expression for the $$y$$-velocity $$v_y$$, namely: +Which we integrate by parts to get an expression for the $$y$$-velocity $$v_y$$, namely: $$\begin{aligned} v_y = U \frac{\delta'}{\delta} \int s f'' \dd{y} + = U \delta' \bigg( s f' - \int f' \dd{s} \bigg) = U \delta' \: (s f' - f) \end{aligned}$$ Now, consider the main Prandtl equation, -assuming that the attack velocity $$U$$ is constant: - -$$\begin{aligned} - v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y} - = \nu \pdvn{2}{v_x}{y} -\end{aligned}$$ - -Inserting our expressions for $$v_x$$ and $$v_y$$ into this leads us to: - -$$\begin{aligned} - - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f) - = \nu U \frac{1}{\delta^2} f''' -\end{aligned}$$ - -After multiplying it by $$\delta^2 / U$$ and cancelling out some terms, -it reduces to: +assuming the attack velocity $$U$$ is constant. +Inserting our expressions for $$v_x$$ and $$v_y$$ into it gives: $$\begin{aligned} - \nu f''' + U \delta' \delta f'' f - = 0 + \nu \pdvn{2}{v_x}{y} + &= v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y} + \\ + \nu \pdvn{2}{v_x}{s} \pdvn{2}{s}{y} + &= v_x \pdv{v_x}{s} \pdv{s}{x} + v_y \pdv{v_x}{s} \pdv{s}{y} + \\ + \nu U \frac{1}{\delta^2} f''' + &= - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f) \end{aligned}$$ -Then, substituting $$\delta(x) = \sqrt{\nu x / U}$$ and $$\delta'(x) = (1/2) \sqrt{\nu / (U x)}$$ yields: +We multiply by $$\delta^2 / U$$, cancel out some terms, +and substitute $$\delta(x) \equiv \sqrt{\nu x / U}$$, leaving: $$\begin{aligned} - \nu f''' + U \frac{\nu}{2 U} f'' f - = 0 + \nu f''' + &= - U \delta' \delta f'' f + = - U \frac{\nu}{2 U} f'' f \end{aligned}$$ -Simplifying this leads us to the **Blasius equation**, +This leads us to the **Blasius equation**, which is a nonlinear ODE for $$f(s)$$: $$\begin{aligned} diff --git a/source/know/concept/bloch-sphere/index.md b/source/know/concept/bloch-sphere/index.md index 99ac45d..b348d22 100644 --- a/source/know/concept/bloch-sphere/index.md +++ b/source/know/concept/bloch-sphere/index.md @@ -21,23 +21,23 @@ and their extremes are the eigenstates of the Pauli matrices: $$\begin{aligned} \hat{\sigma}_z - \to \{\Ket{0}, \Ket{1}\} + \to \{\ket{0}, \ket{1}\} \qquad \hat{\sigma}_x - \to \{\Ket{+}, \Ket{-}\} + \to \{\ket{+}, \ket{-}\} \qquad \hat{\sigma}_y - \to \{\Ket{+i}, \Ket{-i}\} + \to \{\ket{+i}, \ket{-i}\} \end{aligned}$$ -Where the latter two states are expressed as follows in the conventional $$z$$-basis: +Where the latter two pairs are expressed as follows in the conventional $$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}} + \ket{\pm} + = \frac{\ket{0} \pm \ket{1}}{\sqrt{2}} + \qquad \qquad + \ket{\pm i} + = \frac{\ket{0} \pm i \ket{1}}{\sqrt{2}} \end{aligned}$$ More generally, every point on the surface of the sphere @@ -45,11 +45,12 @@ describes a pure qubit state in terms of the angles $$\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} + \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$$: +Another way to describe states is the **Bloch vector** $$\vec{r}$$, +which is simply the $$(x,y,z)$$-coordinates of a point on the sphere. +Let the radius $$r \le 1$$: $$\begin{aligned} \boxed{ @@ -60,8 +61,7 @@ $$\begin{aligned} \end{aligned}$$ 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. +but rather a description of one. The main point of the Bloch vector is that it allows us to describe the qubit using a [density operator](/know/concept/density-operator/): @@ -73,7 +73,7 @@ $$\begin{aligned} \end{aligned}$$ 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 +Now, we know that a density matrix represents a pure ensemble if and only if it is idempotent, i.e. $$\hat{\rho}^2 = \hat{\rho}$$: $$\begin{aligned} @@ -82,8 +82,8 @@ $$\begin{aligned} = \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}$$ again, and the state is pure: +You can easily convince yourself that, if $$(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$$, +we get $$\hat{\rho}$$ again, so the state is pure: $$\begin{aligned} (\vec{r} \cdot \vec{\sigma})^2 @@ -109,11 +109,28 @@ 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 $$\hat{\sigma}_z$$: +are given by the corresponding component of $$\vec{r}$$: $$\begin{aligned} - \Expval{\hat{\sigma}_z} + \boxed{ + \begin{aligned} + \expval{\hat{\sigma}_{x}} + &= r_{x} + \\ + \expval{\hat{\sigma}_{y}} + &= r_{y} + \\ + \expval{\hat{\sigma}_{z}} + &= r_{z} + \end{aligned} + } +\end{aligned}$$ + +This is a consequence of the above form of the density operator $$\hat{\rho}$$. +For example for $$\hat{\sigma}_z$$: + +$$\begin{aligned} + \expval{\hat{\sigma}_z} &= \Tr(\hat{\rho} \hat{\sigma}_z) = \frac{1}{2} \Tr\!\big(\hat{\sigma}_z + (\vec{r} \cdot \vec{\sigma}) \hat{\sigma}_z \big) = \frac{1}{2} \Tr\!\big( (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z) \hat{\sigma}_z \big) @@ -124,6 +141,7 @@ $$\begin{aligned} \end{aligned}$$ + ## References 1. N. Brunner, *Quantum information theory: lecture notes*, diff --git a/source/know/concept/dirac-notation/index.md b/source/know/concept/dirac-notation/index.md index 46cc325..2830a33 100644 --- a/source/know/concept/dirac-notation/index.md +++ b/source/know/concept/dirac-notation/index.md @@ -8,47 +8,49 @@ categories: layout: "concept" --- -**Dirac notation** is a notation to do calculations in a [Hilbert space](/know/concept/hilbert-space/) -without needing to worry about the space's representation. It is -basically the *lingua franca* of quantum mechanics. - -In Dirac notation there are **kets** $$\Ket{V}$$ from the Hilbert space -$$\mathbb{H}$$ and **bras** $$\Bra{V}$$ from a dual $$\mathbb{H}'$$ of the -former. Crucially, the bras and kets are from different Hilbert spaces -and therefore cannot be added, but every bra has a corresponding ket and -vice versa. +**Dirac notation** enables us to do calculations +in a general [Hilbert space](/know/concept/hilbert-space/) +without needing to worry about the space's representation. +It is the *lingua franca* of quantum mechanics. + +In Dirac notation there are +**kets** $$\ket{V}$$ from the Hilbert space $$\mathbb{H}$$ +and **bras** $$\bra{V}$$ from its dual space $$\mathbb{H}'$$. +Crucially, the bras and kets are from different Hilbert spaces +and therefore cannot be added, +but every bra has a corresponding ket and vice versa. Bras and kets can be combined in two ways: the **inner product** -$$\Inprod{V}{W}$$, which returns a scalar, and the **outer product** -$$\Ket{V} \Bra{W}$$, which returns a mapping $$\hat{L}$$ from kets $$\Ket{V}$$ -to other kets $$\Ket{V'}$$, i.e. a linear operator. Recall that the -Hilbert inner product must satisfy: +$$\inprod{V}{W}$$, which returns a scalar, and the **outer product** +$$\ket{V} \bra{W}$$, which returns a linear operator +that maps kets $$\ket{V}$$ to other kets $$\ket{V'}$$. +Recall that by definition the Hilbert inner product must satisfy: $$\begin{aligned} - \Inprod{V}{W} = \Inprod{W}{V}^* + \inprod{V}{W} = \inprod{W}{V}^* \end{aligned}$$ -So far, nothing has been said about the actual representation of bras or -kets. If we represent kets as $$N$$-dimensional columns vectors, the -corresponding bras are given by the kets' adjoints, i.e. their transpose -conjugates: +So far, nothing has been said about the actual representation of bras or kets. +If we represent kets as $$N$$-dimensional columns vectors, +the corresponding bras are given by the kets' adjoints, +i.e. their transpose conjugates: $$\begin{aligned} - \Ket{V} = + \ket{V} = \begin{bmatrix} v_1 \\ \vdots \\ v_N \end{bmatrix} \quad \implies \quad - \Bra{V} = + \bra{V} = \begin{bmatrix} v_1^* & \cdots & v_N^* \end{bmatrix} \end{aligned}$$ -The inner product $$\Inprod{V}{W}$$ is then just the familiar dot product $$V \cdot W$$: +The inner product $$\inprod{V}{W}$$ is then just the familiar dot product $$V \cdot W$$: $$\begin{gathered} - \Inprod{V}{W} + \inprod{V}{W} = \begin{bmatrix} v_1^* & \cdots & v_N^* @@ -60,10 +62,11 @@ $$\begin{gathered} = v_1^* w_1 + ... + v_N^* w_N \end{gathered}$$ -Meanwhile, the outer product $$\Ket{V} \Bra{W}$$ creates an $$N \cross N$$ matrix: +Meanwhile, the outer product $$\ket{V} \bra{W}$$ creates an $$N \cross N$$ matrix, +which can be thought of as applying an operation to any vector it multiplies: $$\begin{gathered} - \Ket{V} \Bra{W} + \ket{V} \bra{W} = \begin{bmatrix} v_1 \\ \vdots \\ v_N @@ -80,15 +83,14 @@ $$\begin{gathered} \end{bmatrix} \end{gathered}$$ -If the kets are instead represented by functions $$f(x)$$ of -$$x \in [a, b]$$, then the bras represent *functionals* $$F[u(x)]$$ which -take an unknown function $$u(x)$$ as an argument and turn it into a scalar -using integration: +If the kets are instead represented by continuous functions $$f(x)$$ of $$x \in [a, b]$$, +then the bras are *functionals* $$F[u(x)]$$ +that take an arbitrary function $$u(x)$$ as an argument and return a scalar: $$\begin{aligned} - \Ket{f} = f(x) + \ket{f} = f(x) \quad \implies \quad - \Bra{f} + \bra{f} = F[u(x)] = \int_a^b f^*(x) \: u(x) \dd{x} \end{aligned}$$ @@ -96,23 +98,25 @@ $$\begin{aligned} Consequently, the inner product is simply the following familiar integral: $$\begin{gathered} - \Inprod{f}{g} + \inprod{f}{g} = F[g(x)] = \int_a^b f^*(x) \: g(x) \dd{x} \end{gathered}$$ -However, the outer product becomes something rather abstract: +However, the outer product is then rather abstract: +a continuous analogue of a matrix: $$\begin{gathered} - \Ket{f} \Bra{g} + \ket{f} \bra{g} = f(x) \: G[u(x)] = f(x) \int_a^b g^*(\xi) \: u(\xi) \dd{\xi} \end{gathered}$$ -This result makes more sense if we surround it by a bra and a ket: +This maybe makes more sense if we surround it +by a bra $$\bra{u}$$ and a ket $$\ket{w}$$ and rearrange: $$\begin{aligned} - \Bra{u} \!\Big(\!\Ket{f} \Bra{g}\!\Big)\! \Ket{w} + \bra{u} \!\Big(\!\ket{f} \bra{g}\!\Big)\! \ket{w} &= U\big[f(x) \: G[w(x)]\big] = U\Big[ f(x) \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big] \\ @@ -120,7 +124,7 @@ $$\begin{aligned} \\ &= \Big( \int_a^b u^*(x) \: f(x) \dd{x} \Big) \Big( \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big) \\ - &= \Inprod{u}{f} \Inprod{g}{w} + &= \inprod{u}{f} \inprod{g}{w} \end{aligned}$$ diff --git a/source/know/concept/gram-schmidt-method/index.md b/source/know/concept/gram-schmidt-method/index.md index 70ad512..a62522e 100644 --- a/source/know/concept/gram-schmidt-method/index.md +++ b/source/know/concept/gram-schmidt-method/index.md @@ -8,39 +8,66 @@ categories: layout: "concept" --- -Given a set of linearly independent non-orthonormal vectors -$$\ket{V_1}, \ket{V_2}, ...$$ from a [Hilbert space](/know/concept/hilbert-space/), -the **Gram-Schmidt method** -turns them into an orthonormal set $$\ket{n_1}, \ket{n_2}, ...$$ as follows: +Given a set of $$N$$ linearly independent vectors $$\ket{V_1}, \ket{V_2}, ...$$ +from a [Hilbert space](/know/concept/hilbert-space/), +the **Gram-Schmidt method** is an algorithm that turns them +into an orthonormal set $$\ket{n_1}, \ket{n_2}, ...$$ as follows +(in [Dirac notation](/know/concept/dirac-notation/)): 1. Take the first vector $$\ket{V_1}$$ and normalize it to get $$\ket{n_1}$$: $$\begin{aligned} - \ket{n_1} = \frac{\ket{V_1}}{\sqrt{\inprod{V_1}{V_1}}} + \ket{n_1} + = \frac{\ket{V_1}}{\sqrt{\inprod{V_1}{V_1}}} \end{aligned}$$ -2. Begin loop. Take the next non-orthonormal vector $$\ket{V_j}$$, and +2. Begin loop. Take the next input vector $$\ket{V_j}$$, and subtract from it its projection onto every already-processed vector: $$\begin{aligned} - \ket{n_j'} = \ket{V_j} - \ket{n_1} \inprod{n_1}{V_j} - \ket{n_2} \inprod{n_2}{V_j} - ... - \ket{n_{j-1}} \inprod{n_{j-1}}{V_{j-1}} + \ket{g_j} + = \ket{V_j} - \ket{n_1} \inprod{n_1}{V_j} - \ket{n_2} \inprod{n_2}{V_j} - ... - \ket{n_{j-1}} \inprod{n_{j-1}}{V_j} \end{aligned}$$ - This leaves only the part of $$\ket{V_j}$$ which is orthogonal to - $$\ket{n_1}$$, $$\ket{n_2}$$, etc. This why the input vectors must be - linearly independent; otherwise $$\Ket{n_j'}$$ may become zero at some - point. + This leaves only the part of $$\ket{V_j}$$ + that is orthogonal to all previous $$\ket{n_k}$$. + This why the input vectors must be linearly independent; + otherwise $$\ket{g_j}$$ could become zero. -3. Normalize the resulting ortho*gonal* vector $$\ket{n_j'}$$ to make it + On a computer, the resulting $$\ket{g_j}$$ will + not be perfectly orthogonal due to rounding errors. + The above description of step #2 is particularly bad. + A better approach is: + + $$\begin{aligned} + \ket{g_j^{(1)}} + &= \ket{V_j} - \ket{n_1} \inprod{n_1}{V_j} + \\ + \ket{g_j^{(2)}} + &= \ket{g_j^{(1)}} - \ket{n_2} \inprod{n_2}{g_j^{(1)}} + \\ + \vdots + \\ + \ket{g_j} + = \ket{g_j^{(j-1)}} + &= \ket{g_j^{(j-2)}} - \ket{n_{j-2}} \inprod{n_{j-2}}{g_j^{(j-2)}} + \end{aligned}$$ + + In other words, instead of projecting $$\ket{V_j}$$ directly onto all $$\ket{n_k}$$, + we instead project only the part of $$\ket{V_j}$$ that has already been made orthogonal + to all previous $$\ket{n_m}$$ with $$m < k$$. + This is known as the **modified Gram-Schmidt method**. + +3. Normalize the resulting ortho*gonal* vector $$\ket{g_j}$$ to make it ortho*normal*: $$\begin{aligned} - \ket{n_j} = \frac{\ket{n_j'}}{\sqrt{\inprod{n_j'}{n_j'}}} + \ket{n_j} + = \frac{\ket{g_j}}{\sqrt{\inprod{g_j}{g_j}}} \end{aligned}$$ -4. Loop back to step 2, taking the next vector $$\ket{V_{j+1}}$$. - -If you are unfamiliar with this notation, take a look at [Dirac notation](/know/concept/dirac-notation/). +4. Loop back to step 2, taking the next vector $$\ket{V_{j+1}}$$, + until all $$N$$ have been processed. diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md index 8b903d4..aef8379 100644 --- a/source/know/concept/matsubara-sum/index.md +++ b/source/know/concept/matsubara-sum/index.md @@ -36,7 +36,7 @@ $$\begin{aligned} = \sum_{z_p} e^{z_p \tau} \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) h(z) \big) \end{aligned}$$ -Where we have applied the residue theorem +Where we have applied the [residue theorem](/know/concept/residue-theorem/) to get a sum over all simple poles $$z_p$$ of either $$g$$ or $$h$$ (but not both) enclosed by $$C$$. Clearly, we could make this look like a Matsubara sum, |