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| -rw-r--r-- | source/know/concept/dielectric-function/index.md | 65 | ||||
| -rw-r--r-- | source/know/concept/einstein-coefficients/index.md | 5 | ||||
| -rw-r--r-- | source/know/concept/fermis-golden-rule/index.md | 2 | ||||
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| -rw-r--r-- | source/know/concept/hellmann-feynman-theorem/index.md | 31 | ||||
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| -rw-r--r-- | source/know/concept/time-dependent-perturbation-theory/index.md | 73 | ||||
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diff --git a/source/know/concept/amplitude-rate-equations/index.md b/source/know/concept/amplitude-rate-equations/index.md new file mode 100644 index 0000000..0ca3248 --- /dev/null +++ b/source/know/concept/amplitude-rate-equations/index.md @@ -0,0 +1,94 @@ +--- +title: "Amplitude rate equations" +sort_title: "Amplitude rate equations" +date: 2023-01-03 +categories: +- Physics +- Quantum mechanics +layout: "concept" +--- + +In quantum mechanics, the **amplitude rate equations** give +the evolution of a quantum state's superposition coefficients through time. +They are known as the precursors for +[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/), +but by themselves they are exact and widely applicable. + +Let $$\hat{H}_0$$ be a "simple" time-independent part +of the full Hamiltonian, +and $$\hat{H}_1$$ a time-varying other part, +whose contribution need not be small: + +$$\begin{aligned} + \hat{H}(t) = \hat{H}_0 + \hat{H}_1(t) +\end{aligned}$$ + +We assume that the time-independent problem +$$\hat{H}_0 \Ket{n} = E_n \Ket{n}$$ has already been solved, +such that its general solution is a superposition as follows: + +$$\begin{aligned} + \Ket{\Psi_0(t)} = \sum_{n} c_n \Ket{n} e^{- i E_n t / \hbar} +\end{aligned}$$ + +Since these $$\Ket{n}$$ form a complete basis, +the full solution for $$\hat{H}_0 + \hat{H}_1$$ can be written in the same form, +but now with time-dependent coefficients $$c_n(t)$$: + +$$\begin{aligned} + \Ket{\Psi(t)} = \sum_{n} c_n(t) \Ket{n} e^{- i E_n t / \hbar} +\end{aligned}$$ + +We put this ansatz into the full Schrödinger equation, +and use the known solution for $$\hat{H}_0$$: + +$$\begin{aligned} + 0 + &= \hat{H}_0 \Ket{\Psi(t)} + \hat{H}_1 \Ket{\Psi(t)} - i \hbar \dv{}{t}\Ket{\Psi(t)} + \\ + &= \sum_{n} + \Big( c_n \hat{H}_0 \Ket{n} + c_n \hat{H}_1 \Ket{n} - c_n E_n \Ket{n} - i \hbar \dv{c_n}{t} \Ket{n} \Big) e^{- i E_n t / \hbar} + \\ + &= \sum_{n} \Big( c_n \hat{H}_1 \Ket{n} - i \hbar \dv{c_n}{t} \Ket{n} \Big) e^{- i E_n t / \hbar} +\end{aligned}$$ + +We then take the inner product with an arbitrary stationary basis state $$\Ket{m}$$: + +$$\begin{aligned} + 0 + &= \sum_{n} \Big( c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \dv{c_n}{t} \inprod{m}{n} \Big) e^{- i E_n t / \hbar} +\end{aligned}$$ + +Thanks to orthonormality, this moves the latter term outside the summation: + +$$\begin{aligned} + i \hbar \dv{c_m}{t} e^{- i E_m t / \hbar} + &= \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} e^{- i E_n t / \hbar} +\end{aligned}$$ + +We divide by the left-hand exponential and define +$$\omega_{mn} \equiv (E_m - E_n) / \hbar$$ to arrive at +the desired set of amplitude rate equations, +one for each basis state $$\ket{m}$$: + +$$\begin{aligned} + \boxed{ + i \hbar \dv{c_m}{t} + = \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} e^{i \omega_{mn} t} + } +\end{aligned}$$ + +We have not made any approximations, +so it is possible to exactly solve for $$c_n(t)$$ in some simple systems. +This is worth pointing out, because these equations' most famous uses +are for deriving time-dependent-perturbation theory +(by making a truncated power series approximation) +and [Rabi oscillation](/know/concept/rabi-oscillation/) +(by making the [rotating wave approximation](/know/concept/rotating-wave-approximation/)). + + + +## References +1. D.J. Griffiths, D.F. Schroeter, + *Introduction to quantum mechanics*, 3rd edition, + Cambridge. diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md index f454264..fa289de 100644 --- a/source/know/concept/bell-state/index.md +++ b/source/know/concept/bell-state/index.md @@ -16,16 +16,16 @@ $$\begin{aligned} \boxed{ \begin{aligned} \ket{\Phi^{\pm}} - &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{0}_B \pm \Ket{1}_A \Ket{1}_B \Big) + &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{0}_B \pm \ket{1}_A \ket{1}_B \Big) \\ \ket{\Psi^{\pm}} - &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{1}_B \pm \Ket{1}_A \Ket{0}_B \Big) + &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{1}_B \pm \ket{1}_A \ket{0}_B \Big) \end{aligned} } \end{aligned}$$ -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 $$\Ket{1}$$. +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/). @@ -37,7 +37,7 @@ Consider the following pure [density operator](/know/concept/density-operator/): $$\begin{aligned} \hat{\rho} = \ket{\Phi^{+}} \bra{\Phi^{+}} - &= \frac{1}{2} \Big( \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big) \Big( \Bra{0}_A \Bra{0}_B + \Bra{1}_A \Bra{1}_B \Big) + &= \frac{1}{2} \Big( \ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big) \Big( \bra{0}_A \bra{0}_B + \bra{1}_A \bra{1}_B \Big) \end{aligned}$$ The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated as follows: @@ -45,12 +45,12 @@ The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated $$\begin{aligned} \hat{\rho}_A &= \Tr_B(\hat{\rho}) - = \sum_{b = 0, 1} \Bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \Ket{b}_B + = \sum_{b = 0, 1} \bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \ket{b}_B \\ - &= \sum_{b = 0, 1} \Big( \Ket{0}_A \Inprod{b}{0}_B + \Ket{1}_A \Inprod{b}{1}_B \Big) - \Big( \Bra{0}_A \Inprod{0}{b}_B + \Bra{1}_A \Inprod{1}{b}_B \Big) + &= \sum_{b = 0, 1} \Big( \ket{0}_A \inprod{b}{0}_B + \ket{1}_A \inprod{b}{1}_B \Big) + \Big( \bra{0}_A \inprod{0}{b}_B + \bra{1}_A \inprod{1}{b}_B \Big) \\ - &= \frac{1}{2} \Big( \Ket{0}_A \Bra{0}_A + \Ket{1}_A \Bra{1}_A \Big) + &= \frac{1}{2} \Big( \ket{0}_A \bra{0}_A + \ket{1}_A \bra{1}_A \Big) = \frac{1}{2} \hat{I} \end{aligned}$$ @@ -59,35 +59,36 @@ 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 $$\Ket{1}$$. +has a 50-50 chance to yield $$\ket{0}$$ or $$\ket{1}$$. However, due to the entanglement, measuring $$A$$ also has consequences for qubit $$B$$: $$\begin{aligned} - \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 - &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{0}{0}_B + \Inprod{0}{1}_A \Inprod{0}{1}_B \Big)^2 + \big| \bra{0}_A \! \bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \inprod{0}{0}_A \inprod{0}{0}_B + \inprod{0}{1}_A \inprod{0}{1}_B \Big)^2 = \frac{1}{2} \\ - \big| \Bra{0}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 - &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{1}{0}_B + \Inprod{0}{1}_A \Inprod{1}{1}_B \Big)^2 + \big| \bra{0}_A \! \bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \inprod{0}{0}_A \inprod{1}{0}_B + \inprod{0}{1}_A \inprod{1}{1}_B \Big)^2 = 0 \\ - \big| \Bra{1}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 - &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{0}{0}_B + \Inprod{1}{1}_A \Inprod{0}{1}_B \Big)^2 + \big| \bra{1}_A \! \bra{0}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \inprod{1}{0}_A \inprod{0}{0}_B + \inprod{1}{1}_A \inprod{0}{1}_B \Big)^2 = 0 \\ - \big| \Bra{1}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 - &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{1}{0}_B + \Inprod{1}{1}_A \Inprod{1}{1}_B \Big)^2 + \big| \bra{1}_A \! \bra{1}_B \cdot \ket{\Phi^{+}} \big|^2 + &= \frac{1}{2} \Big( \inprod{1}{0}_A \inprod{1}{0}_B + \inprod{1}{1}_A \inprod{1}{1}_B \Big)^2 = \frac{1}{2} \end{aligned}$$ -As an example, if $$A$$ collapses into $$\Ket{0}$$ due to a measurement, -then $$B$$ instantly also collapses into $$\Ket{0}$$, never $$\Ket{1}$$, +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*, diff --git a/source/know/concept/dielectric-function/index.md b/source/know/concept/dielectric-function/index.md index 529ce2a..d55cc91 100644 --- a/source/know/concept/dielectric-function/index.md +++ b/source/know/concept/dielectric-function/index.md @@ -13,7 +13,8 @@ The **dielectric function** or **relative permittivity** $$\varepsilon_r$$ is a measure of how strongly a given medium counteracts [electric fields](/know/concept/electric-field/) compared to a vacuum. Let $$\vb{D}$$ be the applied external field, -and $$\vb{E}$$ the effective field inside the material: +and $$\vb{E}$$ the effective field inside the material, +then $$\varepsilon_r$$ is defined such that: $$\begin{aligned} \boxed{ @@ -23,7 +24,7 @@ $$\begin{aligned} If $$\varepsilon_r$$ is large, then $$\vb{D}$$ is strongly suppressed, because the material's electrons and nuclei move to create an opposing field. -In order for $$\varepsilon_r$$ to be well defined, we only consider linear media, +In order for $$\varepsilon_r$$ to be well-defined, we only consider *linear* media, where the induced polarization $$\vb{P}$$ is proportional to $$\vb{E}$$. We would like to find an alternative definition of $$\varepsilon_r$$. @@ -54,13 +55,10 @@ $$\begin{aligned} } \end{aligned}$$ - -## From induced charge density - -A common way to calculate $$\varepsilon_r$$ is from +In practice, a common way to calculate $$\varepsilon_r$$ is from the induced charge density $$\rho_\mathrm{ind}$$, i.e. the offset caused by the material's particles responding to the field. -We start from [Gauss' law](/know/concept/maxwells-equations/) for $$\vb{P}$$: +Starting from [Gauss' law](/know/concept/maxwells-equations/) for $$\vb{P}$$: $$\begin{aligned} \nabla \cdot \vb{P} @@ -68,27 +66,27 @@ $$\begin{aligned} = - \rho_\mathrm{ind}(\vb{r}) \end{aligned}$$ -This is Poisson's equation, which has the following well-known -[Fourier transform](/know/concept/fourier-transform/): +This is Poisson's equation, which has a well-known solution +via [Fourier transformation](/know/concept/fourier-transform/): $$\begin{aligned} \Phi_\mathrm{ind}(\vb{q}) = \frac{\rho_\mathrm{ind}(\vb{q})}{\varepsilon_0 |\vb{q}|^2} - = V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q}) + \equiv V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q}) \end{aligned}$$ Where $$V(\vb{q})$$ represents Coulomb interactions, -and $$V(0) = 0$$ to ensure overall neutrality: +and $$V(0) \equiv 0$$ to ensure overall neutrality: $$\begin{aligned} V(\vb{q}) - = \frac{1}{\varepsilon_0 |\vb{q}|^2} + \equiv \frac{1}{\varepsilon_0 |\vb{q}|^2} \qquad \implies \qquad V(\vb{r} - \vb{r}') = \frac{1}{4 \pi \varepsilon_0 |\vb{r} - \vb{r}'|} \end{aligned}$$ -The [convolution theorem](/know/concept/convolution-theorem/) +Note that the [convolution theorem](/know/concept/convolution-theorem/) then gives us the solution $$\Phi_\mathrm{ind}$$ in the $$\vb{r}$$-domain: $$\begin{aligned} @@ -97,37 +95,56 @@ $$\begin{aligned} = \int_{-\infty}^\infty V(\vb{r} - \vb{r}') \: \rho_\mathrm{ind}(\vb{r}') \dd{\vb{r}'} \end{aligned}$$ -To proceed, we need to find an expression for $$\rho_\mathrm{ind}$$ -that is proportional to $$\Phi_\mathrm{tot}$$ or $$\Phi_\mathrm{ext}$$, +To proceed to calculate $$\varepsilon_r$$ from $$\rho_\mathrm{ind}$$, +one needs an expression for $$\rho_\mathrm{ind}$$ +that is proportional to $$\Phi_\mathrm{tot}$$ or $$\Phi_\mathrm{ext}$$ or some linear combination thereof. -Such an expression must exist for a linear material. +Such an expression must exist for a linear medium, +but the details depend on the physics being considered +and are thus beyond our current scope; +we will just show the general form of $$\varepsilon_r$$ +once such an expression has been found. -Suppose we can show that $$\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$$, -for some $$C_\mathrm{ext}$$, which may depend on $$\vb{q}$$. Then: +Suppose we know that $$\rho_\mathrm{ind} = c_\mathrm{ext} \Phi_\mathrm{ext}$$ +for some factor $$c_\mathrm{ext}$$, which may depend on $$\vb{q}$$. +Then, since $$\Phi_\mathrm{tot} = \Phi_\mathrm{ext} \!+\! \Phi_\mathrm{ind}$$, +we find in the $$\vb{q}$$-domain: $$\begin{aligned} \Phi_\mathrm{tot} - = (1 + C_\mathrm{ext} V) \Phi_\mathrm{ext} + = (1 + c_\mathrm{ext} V) \Phi_\mathrm{ext} \quad \implies \quad \boxed{ \varepsilon_r(\vb{q}) - = \frac{1}{1 + C_\mathrm{ext}(\vb{q}) V(\vb{q})} + = \frac{1}{1 + c_\mathrm{ext}(\vb{q}) V(\vb{q})} } \end{aligned}$$ -Similarly, suppose we can show that $$\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$$, -for some quantity $$C_\mathrm{tot}$$, then: +Likewise, suppose we can instead show that +$$\rho_\mathrm{ind} = c_\mathrm{tot} \Phi_\mathrm{tot}$$ +for some quantity $$c_\mathrm{tot}$$, then: $$\begin{aligned} \Phi_\mathrm{ext} - = (1 - C_\mathrm{tot} V) \Phi_\mathrm{tot} + = (1 - c_\mathrm{tot} V) \Phi_\mathrm{tot} \quad \implies \quad \boxed{ \varepsilon_r(\vb{q}) - = 1 - C_\mathrm{tot}(\vb{q}) V(\vb{q}) + = 1 - c_\mathrm{tot}(\vb{q}) V(\vb{q}) } \end{aligned}$$ +And in the unlikely event that an expression of the form +$$\rho_\mathrm{ind} = c_\mathrm{ext} \Phi_\mathrm{ext} \!+\! c_\mathrm{tot} \Phi_\mathrm{tot}$$ is found: + +$$\begin{aligned} + (1 - c_\mathrm{tot} V) \Phi_\mathrm{tot} + = (1 + c_\mathrm{ext} V) \Phi_\mathrm{ext} + \quad \implies \quad + \varepsilon_r(\vb{q}) + = \frac{1 - c_\mathrm{tot}(\vb{q}) V(\vb{q})}{1 + c_\mathrm{ext}(\vb{q}) V(\vb{q})} +\end{aligned}$$ + ## References diff --git a/source/know/concept/einstein-coefficients/index.md b/source/know/concept/einstein-coefficients/index.md index 179d866..ad75ce5 100644 --- a/source/know/concept/einstein-coefficients/index.md +++ b/source/know/concept/einstein-coefficients/index.md @@ -159,7 +159,8 @@ $$\begin{aligned} This form of $$\hat{H}_1$$ is a well-known case for [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/), -which tells us that the transition probability from $$\Ket{a}$$ to $$\Ket{b}$$ is: +which tells us that the transition probability +from $$\ket{a}$$ to $$\ket{b}$$ is (to first order): $$\begin{aligned} P_{ab} @@ -167,7 +168,7 @@ $$\begin{aligned} \end{aligned}$$ If the nucleus is at $$z = 0$$, -then generally $$\Ket{1}$$ and $$\Ket{2}$$ will be even or odd functions of $$z$$, +then generally $$\ket{1}$$ and $$\ket{2}$$ will be even or odd functions of $$z$$, meaning that $$\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$$ (see also [Laporte's selection rule](/know/concept/selection-rules/)), leading to: diff --git a/source/know/concept/fermis-golden-rule/index.md b/source/know/concept/fermis-golden-rule/index.md index 021c8e4..ea58ee6 100644 --- a/source/know/concept/fermis-golden-rule/index.md +++ b/source/know/concept/fermis-golden-rule/index.md @@ -20,7 +20,7 @@ time. From [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/), we know that the transition probability for a particle in state $$\Ket{a}$$ to go to $$\Ket{b}$$ -is as follows for a periodic perturbation at frequency $$\omega$$: +is as follows for a sinusoidal perturbation at frequency $$\omega$$: $$\begin{aligned} P_{ab} diff --git a/source/know/concept/ghz-paradox/index.md b/source/know/concept/ghz-paradox/index.md index a59ccfe..9951883 100644 --- a/source/know/concept/ghz-paradox/index.md +++ b/source/know/concept/ghz-paradox/index.md @@ -12,43 +12,46 @@ layout: "concept" The **Greenberger-Horne-Zeilinger** or **GHZ paradox** is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/) that does not use inequalities, -but the three-particle entangled **GHZ state** $$\Ket{\mathrm{GHZ}}$$ instead, +but the three-particle entangled **GHZ state** $$\ket{\mathrm{GHZ}}$$ instead, $$\begin{aligned} \boxed{ - \Ket{\mathrm{GHZ}} - = \frac{1}{\sqrt{2}} \Big( \Ket{000} + \Ket{111} \Big) + \ket{\mathrm{GHZ}} + = \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big) } \end{aligned}$$ -Where $$\Ket{0}$$ and $$\Ket{1}$$ are qubit states, -for example, the eigenvalues of the Pauli matrix $$\hat{\sigma}_z$$. +Where $$\ket{0}$$ and $$\ket{1}$$ are qubit states, +specifically the eigenvalues of the Pauli matrix $$\hat{\sigma}_z$$. If we now apply certain products of the Pauli matrices $$\hat{\sigma}_x$$ and $$\hat{\sigma}_y$$ -to the three particles, we find: +as [quantum gates](/know/concept/quantum-gate/) +to this three-particle state, we find: $$\begin{aligned} - \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \Ket{\mathrm{GHZ}} - &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0} - + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \Big) + \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \ket{\mathrm{GHZ}} + &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0} + + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \Big) \\ - &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes \Ket{1} \otimes \Ket{1} + \Ket{0} \otimes \Ket{0} \otimes \Ket{0} \Big) - = \Ket{\mathrm{GHZ}} + &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes \ket{1} \otimes \ket{1} + \ket{0} \otimes \ket{0} \otimes \ket{0} \Big) \\ - \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \Ket{\mathrm{GHZ}} - &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_y \Ket{0} \otimes \hat{\sigma}_y \Ket{0} - + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \Big) + &= \ket{\mathrm{GHZ}} \\ - &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes i \Ket{1} \otimes i \Ket{1} + \Ket{0} \otimes i \Ket{0} \otimes i \Ket{0} \Big) - = - \Ket{\mathrm{GHZ}} + \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \ket{\mathrm{GHZ}} + &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_y \ket{0} \otimes \hat{\sigma}_y \ket{0} + + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_y \ket{1} \otimes \hat{\sigma}_y \ket{1} \Big) + \\ + &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes i \ket{1} \otimes i \ket{1} + \ket{0} \otimes i \ket{0} \otimes i \ket{0} \Big) + \\ + &= - \ket{\mathrm{GHZ}} \end{aligned}$$ In other words, the GHZ state is a simultaneous eigenstate of these composite operators, with eigenvalues $$+1$$ and $$-1$$, respectively. -Let us introduce two other product operators, -such that we have a set of four observables, -for which $$\Ket{\mathrm{GHZ}}$$ gives these eigenvalues: +Let us introduce two more operators in the same way, +so that we have a set of four observables, +for which $$\ket{\mathrm{GHZ}}$$ gives these eigenvalues: $$\begin{aligned} \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x diff --git a/source/know/concept/hellmann-feynman-theorem/index.md b/source/know/concept/hellmann-feynman-theorem/index.md index e18acc2..c6bf720 100644 --- a/source/know/concept/hellmann-feynman-theorem/index.md +++ b/source/know/concept/hellmann-feynman-theorem/index.md @@ -9,20 +9,20 @@ layout: "concept" --- Consider the time-independent Schrödinger equation, -where the Hamiltonian $$\hat{H}$$ depends on a general parameter $$\lambda$$, -whose meaning or type we will not specify: +where the Hamiltonian $$\hat{H}$$ depends on some parameter $$\lambda$$ +whose meaning we will not specify: $$\begin{aligned} - \hat{H}(\lambda) \Ket{\psi_n(\lambda)} - = E_n(\lambda) \Ket{\psi_n(\lambda)} + \hat{H}(\lambda) \ket{\psi_n(\lambda)} + = E_n(\lambda) \ket{\psi_n(\lambda)} \end{aligned}$$ -Assuming all eigenstates $$\Ket{\psi_n}$$ are normalized, +Assuming all eigenstates $$\ket{\psi_n}$$ are normalized, this gives us the following basic relation: $$\begin{aligned} \matrixel{\psi_m}{\hat{H}}{\psi_n} - = E_n \Inprod{\psi_m}{\psi_n} + = E_n \inprod{\psi_m}{\psi_n} = \delta_{mn} E_n \end{aligned}$$ @@ -38,33 +38,32 @@ $$\begin{aligned} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n} + \matrixel{\psi_m}{\hat{H}}{\nabla_\lambda \psi_n} \\ - &= E_m \Inprod{\psi_m}{\nabla_\lambda \psi_n} + E_n \Inprod{\nabla_\lambda \psi_m}{\psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n} + &= E_m \inprod{\psi_m}{\nabla_\lambda \psi_n} + E_n \inprod{\nabla_\lambda \psi_m}{\psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n} \end{aligned}$$ In order to simplify this, we differentiate the orthogonality relation -$$\Inprod{\psi_m}{\psi_n} = \delta_{mn}$$, -which ends up telling us that -$$\Inprod{\nabla_\lambda \psi_m}{\psi_n} = - \Inprod{\psi_m}{\nabla_\lambda \psi_n}$$: +$$\inprod{\psi_m}{\psi_n} = \delta_{mn}$$: $$\begin{aligned} 0 = \nabla_\lambda \delta_{mn} - = \nabla_\lambda \Inprod{\psi_m}{\psi_n} - = \Inprod{\nabla_\lambda \psi_m}{\psi_n} + \Inprod{\psi_m}{\nabla_\lambda \psi_n} + = \nabla_\lambda \inprod{\psi_m}{\psi_n} + = \inprod{\nabla_\lambda \psi_m}{\psi_n} + \inprod{\psi_m}{\nabla_\lambda \psi_n} \end{aligned}$$ -Using this result to replace $$\Inprod{\nabla_\lambda \psi_m}{\psi_n}$$ +Meaning that $$\inprod{\nabla_\lambda \psi_m}{\psi_n} = - \inprod{\psi_m}{\nabla_\lambda \psi_n}$$. +Using this result to replace $$\inprod{\nabla_\lambda \psi_m}{\psi_n}$$ in the previous equation leads to: $$\begin{aligned} \delta_{mn} \nabla_\lambda E_n - &= (E_m - E_n) \Inprod{\psi_m}{\nabla_\lambda \psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n} + &= (E_m - E_n) \inprod{\psi_m}{\nabla_\lambda \psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n} \end{aligned}$$ For $$m = n$$, we therefore arrive at the **Hellmann-Feynman theorem**, which is useful when doing numerical calculations -to minimize energies with respect to $$\lambda$$: +that often involve minimizing energies with respect to $$\lambda$$: $$\begin{aligned} \boxed{ @@ -79,7 +78,7 @@ the [Berry phase](/know/concept/berry-phase/): $$\begin{aligned} \boxed{ - (E_n - E_m) \Inprod{\psi_m}{\nabla_\lambda \psi_n} + (E_n - E_m) \inprod{\psi_m}{\nabla_\lambda \psi_n} = \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n} } \end{aligned}$$ diff --git a/source/know/concept/maxwell-bloch-equations/index.md b/source/know/concept/maxwell-bloch-equations/index.md index ba8a677..0252b5c 100644 --- a/source/know/concept/maxwell-bloch-equations/index.md +++ b/source/know/concept/maxwell-bloch-equations/index.md @@ -12,13 +12,13 @@ layout: "concept" --- For an electron in a two-level system with time-independent states -$$\Ket{g}$$ (ground) and $$\Ket{e}$$ (excited), +$$\ket{g}$$ (ground) and $$\ket{e}$$ (excited), consider the following general solution -to the full Schrödinger equation: +to the time-dependent Schrödinger equation: $$\begin{aligned} - \Ket{\Psi} - &= c_g \: \Ket{g} \exp(-i E_g t / \hbar) + c_e \: \Ket{e} \exp(-i E_e t / \hbar) + \ket{\Psi} + &= c_g \ket{g} \exp(-i E_g t / \hbar) + c_e \ket{e} \exp(-i E_e t / \hbar) \end{aligned}$$ Perturbing this system with @@ -87,15 +87,16 @@ $$\begin{aligned} \end{aligned}$$ + ## Optical Bloch equations -For $$\Ket{\Psi}$$ as defined above, +For $$\ket{\Psi}$$ as defined above, the corresponding pure [density operator](/know/concept/density-operator/) $$\hat{\rho}$$ is as follows: $$\begin{aligned} \hat{\rho} - = \Ket{\Psi} \Bra{\Psi} + = \ket{\Psi} \bra{\Psi} = \begin{bmatrix} c_e c_e^* & c_e c_g^* \exp(-i \omega_0 t) \\ @@ -159,11 +160,10 @@ $$\begin{aligned} These equations are correct if nothing else is affecting $$\hat{\rho}$$. But in practice, these quantities decay due to various processes, -e.g. spontaneous emission (see [Einstein coefficients](/know/concept/einstein-coefficients/)). +e.g. [spontaneous emission](/know/concept/einstein-coefficients/). -Let $$\rho_{ee}$$ decays with rate $$\gamma_e$$. -Since the total probability $$\rho_{ee} + \rho_{gg} = 1$$, -we thus have: +Suppose $$\rho_{ee}$$ decays with rate $$\gamma_e$$. +Because the total probability $$\rho_{ee} + \rho_{gg} = 1$$, we have: $$\begin{aligned} \Big( \dv{\rho_{ee}}{t} \Big)_{e} @@ -220,10 +220,11 @@ $$\begin{aligned} } \end{aligned}$$ -Many authors simplify these equations a bit by choosing +Some authors simplify these equations a bit by choosing $$\gamma_g = 0$$ and $$\gamma_\perp = \gamma_e / 2$$. + ## Including Maxwell's equations This two-level system has a dipole moment $$\vb{p}$$ as follows, @@ -286,7 +287,7 @@ $$\begin{aligned} We can rewrite the first two terms in the following intuitive form, which describes a decay with rate $$\gamma_\parallel \equiv \gamma_g + \gamma_e$$ -towards an equilbrium $$d_0$$: +towards an equilibrium $$d_0$$: $$\begin{aligned} 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee} diff --git a/source/know/concept/no-cloning-theorem/index.md b/source/know/concept/no-cloning-theorem/index.md index a91ae6f..840a598 100644 --- a/source/know/concept/no-cloning-theorem/index.md +++ b/source/know/concept/no-cloning-theorem/index.md @@ -62,6 +62,7 @@ This is clearly not the same as before: we have a contradiction, which implies that such a general cloning machine cannot ever exist. + ## References 1. N. Brunner, *Quantum information theory: lecture notes*, diff --git a/source/know/concept/quantum-gate/index.md b/source/know/concept/quantum-gate/index.md index 9704e53..dd198f2 100644 --- a/source/know/concept/quantum-gate/index.md +++ b/source/know/concept/quantum-gate/index.md @@ -17,15 +17,15 @@ so we only consider the most important examples here. ## One-qubit gates -As an example, consider the following must general single-qubit state $$\Ket{\psi}$$: +As an example, consider the following most general single-qubit state $$\ket{\psi}$$: $$\begin{aligned} - \Ket{\psi} - = \alpha \Ket{0} + \beta \Ket{1} + \ket{\psi} + = \alpha \ket{0} + \beta \ket{1} = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \end{aligned}$$ -Arguably the most famous and/or most fundamental quantum gates are the **Pauli matrices**: +Arguably the most famous and most fundamental quantum gates are the **Pauli matrices**: $$\begin{aligned} \boxed{ @@ -53,19 +53,19 @@ $$\begin{aligned} } \end{aligned}$$ -They have the following effect on $$\Ket{\psi}$$. +They have the following effect on $$\ket{\psi}$$. Note that $$X$$ is equivalent to the classical $$\mathrm{NOT}$$ gate (and is often given that name), and $$Z$$ is sometimes called the **phase-flip gate**: $$\begin{aligned} - X \Ket{\psi} + X \ket{\psi} = \begin{bmatrix} \beta \\ \alpha \end{bmatrix} \qquad - Y \Ket{\psi} + Y \ket{\psi} = \begin{bmatrix} -i \beta \\ i \alpha \end{bmatrix} \qquad - Z \Ket{\psi} + Z \ket{\psi} = \begin{bmatrix} \alpha \\ -\beta \end{bmatrix} \end{aligned}$$ @@ -87,7 +87,7 @@ For $$\phi = \pi$$, we recover the Pauli-$$Z$$ gate. In general, the action of $$R_\phi$$ is as follows: $$\begin{aligned} - R_\phi \Ket{\psi} + R_\phi \ket{\psi} = \begin{bmatrix} \alpha \\ e^{i \phi} \beta \end{bmatrix} \end{aligned}$$ @@ -128,10 +128,11 @@ $$\begin{aligned} \end{aligned}$$ Its action consists of rotating the qubit -by $$\pi$$ around the axis $$(X + Z) / \sqrt{2}$$ of the Bloch sphere: +by $$\pi$$ around the axis $$(X + Z) / \sqrt{2}$$ of +the [Bloch sphere](/know/concept/bloch-sphere/): $$\begin{aligned} - H \Ket{\psi} + H \ket{\psi} = \frac{1}{\sqrt{2}} \begin{bmatrix} \alpha + \beta \\ \alpha - \beta \end{bmatrix} \end{aligned}$$ @@ -139,20 +140,20 @@ Notably, it maps the eigenstates of $$X$$ and $$Z$$ to each other, and is its own inverse (i.e. unitary): $$\begin{aligned} - H \Ket{0} = \Ket{+} + H \ket{0} = \ket{+} \qquad - H \Ket{1} = \Ket{-} + H \ket{1} = \ket{-} \qquad - H \Ket{+} = \Ket{0} + H \ket{+} = \ket{0} \qquad - H \Ket{-} = \Ket{1} + H \ket{-} = \ket{1} \end{aligned}$$ The **Clifford gates** are a set including $$X$$, $$Y$$, $$Z$$, $$H$$ and $$S$$, or more generally any gates that rotate by multiples of $$\pi/2$$ around the Bloch sphere. -This set is **not universal**, meaning that if we start from $$\Ket{0}$$, -we can only reach $$\Ket{0}$$, $$\Ket{1}$$, $$\Ket{+}$$, $$\Ket{-}$$, $$\Ket{+i}$$ $$\Ket{-i}$$ using these gates. +This set is **not universal**, meaning that if we start from $$\ket{0}$$, +we can only reach $$\ket{0}$$, $$\ket{1}$$, $$\ket{+}$$, $$\ket{-}$$, $$\ket{+i}$$ $$\ket{-i}$$ using these gates. If we add *any* non-Clifford gate, for example $$T$$, then we can reach any point on the Bloch sphere, @@ -170,15 +171,15 @@ any state can be approximated. ## Two-qubit gates As an example, let us consider -the following two pure one-qubit states $$\Ket{\psi_1}$$ and $$\Ket{\psi_2}$$: +the following two pure one-qubit states $$\ket{\psi_1}$$ and $$\ket{\psi_2}$$: $$\begin{aligned} - \Ket{\psi_1} - = \alpha_1 \Ket{0} + \beta_1 \Ket{1} + \ket{\psi_1} + = \alpha_1 \ket{0} + \beta_1 \ket{1} = \begin{bmatrix} \alpha_1 \\ \beta_1 \end{bmatrix} \qquad \quad - \Ket{\psi_2} - = \alpha_2 \Ket{0} + \beta_2 \Ket{1} + \ket{\psi_2} + = \alpha_2 \ket{0} + \beta_2 \ket{1} = \begin{bmatrix} \alpha_2 \\ \beta_2 \end{bmatrix} \end{aligned}$$ @@ -186,23 +187,22 @@ The composite state of both qubits, assuming they are pure, is then their tensor product $$\otimes$$: $$\begin{aligned} - \Ket{\psi_1 \psi_2} - = \Ket{\psi_1} \otimes \Ket{\psi_2} - &= \alpha_1 \alpha_2 \Ket{00} + \alpha_1 \beta_2 \Ket{01} + \beta_1 \alpha_2 \Ket{10} + \beta_1 \beta_2 \Ket{11} + \ket{\psi_1 \psi_2} + = \ket{\psi_1} \otimes \ket{\psi_2} + &= \alpha_1 \alpha_2 \ket{00} + \alpha_1 \beta_2 \ket{01} + \beta_1 \alpha_2 \ket{10} + \beta_1 \beta_2 \ket{11} \\ - &= c_{00} \Ket{00} + c_{01} \Ket{01} + c_{10} \Ket{10} + c_{11} \Ket{11} + &= c_{00} \ket{00} + c_{01} \ket{01} + c_{10} \ket{10} + c_{11} \ket{11} \end{aligned}$$ Note that a two-qubit system may be [entangled](/know/concept/quantum-entanglement/), in which case the coefficients $$c_{00}$$ etc. cannot be written as products, -i.e. $$\Ket{\psi_2}$$ cannot be expressed separately from $$\Ket{\psi_1}$$, and vice versa. +i.e. $$\ket{\psi_2}$$ cannot be expressed separately from $$\ket{\psi_1}$$, and vice versa. +In other words, the action of a two-qubit gate +can be expressed in the basis of $$\ket{00}$$, $$\ket{01}$$, $$\ket{10}$$ and $$\ket{11}$$, +but not always in the basis of $$\ket{0}_1$$, $$\ket{1}_1$$, $$\ket{0}_2$$ and $$\ket{1}_2$$. -In other words, the general action of a two-qubit quantum gate -can be expressed in the basis of $$\Ket{00}$$, $$\Ket{01}$$, $$\Ket{10}$$ and $$\Ket{11}$$, -but not always in the basis of $$\Ket{0}_1$$, $$\Ket{1}_1$$, $$\Ket{0}_2$$ and $$\Ket{1}_2$$. - -With that said, the first two-qubit gate is $$\mathrm{SWAP}$$, -which simply swaps $$\Ket{\psi_1}$$ and $$\Ket{\psi_2}$$: +With this noted, the first two-qubit gate is $$\mathrm{SWAP}$$, +which simply swaps $$\ket{\psi_1}$$ and $$\ket{\psi_2}$$: {% include image.html file="swap.png" width="22%" alt="SWAP gate diagram" %} @@ -219,18 +219,18 @@ $$\begin{aligned} } \end{aligned}$$ -This matrix is given in the basis of $$\Ket{00}$$, $$\Ket{01}$$, $$\Ket{10}$$ and $$\Ket{11}$$. +This matrix is given in the basis of $$\ket{00}$$, $$\ket{01}$$, $$\ket{10}$$ and $$\ket{11}$$. Note that $$\mathrm{SWAP}$$ cannot generate entanglement, so if its input is separable, its output is too. In any case, its effect is clear: $$\begin{aligned} - \mathrm{SWAP} \Ket{\psi_1 \psi_2} - &= c_{00} \Ket{00} + c_{10} \Ket{01} + c_{01} \Ket{10} + c_{11} \Ket{11} + \mathrm{SWAP} \ket{\psi_1 \psi_2} + &= c_{00} \ket{00} + c_{10} \ket{01} + c_{01} \ket{10} + c_{11} \ket{11} \end{aligned}$$ Next, there is the **controlled NOT gate** $$\mathrm{CNOT}$$, -which "flips" (applies $$X$$ to) $$\Ket{\psi_2}$$ if $$\Ket{\ |