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Diffstat (limited to 'content/know/concept')
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diff --git a/content/know/concept/calculus-of-variations/index.pdc b/content/know/concept/calculus-of-variations/index.pdc index c5280e5..576863c 100644 --- a/content/know/concept/calculus-of-variations/index.pdc +++ b/content/know/concept/calculus-of-variations/index.pdc @@ -234,3 +234,107 @@ $$\begin{aligned} 0 = \pdv{L}{f} - \sum_{n} \dv{x_n} \Big( \pdv{L}{f_{x_n}} \Big) } \end{aligned}$$ + + +## Constraints + +So far, for multiple functions $f_1, ... f_N$, +we have been assuming that all $f_n$ are independent, and by extension all $\eta_n$. +Suppose that we now have $M < N$ constraints $\phi_m$ +that all $f_n$ need to obey, introducing implicit dependencies between them. + +Let us consider constraints $\phi_m$ of the two forms below. +It is important that they are **holonomic**, +meaning they do not depend on any derivatives of any $f_n(x)$: + +$$\begin{aligned} + \phi_m(f_1, ..., f_N, x) = 0 + \qquad + \int_{x_0}^{x_1} \phi_m(f_1, ..., f_N, x) \dd{x} = C_m +\end{aligned}$$ + +Where $C_m$ is a constant. +Note that the first form can also be used for $\phi_m = C_m \neq 0$, +by simply redefining the constraint as $\phi_m^0 = \phi_m - C_m = 0$. + +To solve this constrained optimization problem for $f_n(x)$, +we introduce [Lagrange multipliers](/know/concept/lagrange-multiplier/) $\lambda_m$. +In the former case $\lambda_m(x)$ is a function of all $x$, while in the +latter case $\lambda_m$ is constant: + +$$\begin{aligned} + \int \lambda_m(x_i) \: \phi_m(\{f_n\}, x) \dd{x} = 0 + \qquad + \lambda_m \int \phi_m(\{f_n\}, x) \dd{x} = \lambda_m C_m +\end{aligned}$$ + +The reason for this distinction in $\lambda_m$ +is that we need to find the stationary points with respect to $\varepsilon$ +of both constraint types. Written in the variational form, this is: + +$$\begin{aligned} + \delta \int \lambda_m \: \phi_m \dd{x} = 0 +\end{aligned}$$ + +From this, we define a new Lagrangian $\Lambda$ for the functional $J$, +with the contraints built in: + +$$\begin{aligned} + J[f_n] + &= \int \Lambda(f_1, ..., f_N; f_1', ..., f_N'; \lambda_1, ..., \lambda_M; x) \dd{x} + \\ + &= \int L + \sum_{m} \lambda_m \phi_m \dd{x} +\end{aligned}$$ + +Then we derive the Euler-Lagrange equation as usual for $\Lambda$ instead of $L$: + +$$\begin{aligned} + 0 + &= \delta \int \Lambda \dd{x} + = \int \pdv{\Lambda}{\varepsilon} \dd{x} + = \int \sum_n \Big( \pdv{\Lambda}{f_n} \pdv{f_n}{\varepsilon} + \pdv{\Lambda}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x} + \\ + &= \int \sum_n \Big( \pdv{\Lambda}{f_n} \eta_n + \pdv{\Lambda}{f_n'} \eta_n' \Big) \dd{x} + \\ + &= \Big[ \sum_n \pdv{\Lambda}{f_n'} \eta_n \Big]_{x_0}^{x_1} + + \int \sum_n \eta_n \bigg( \pdv{\Lambda}{f_n} - \dv{x} \Big( \pdv{\Lambda}{f_n'} \Big) \bigg) \dd{x} +\end{aligned}$$ + +Using the same logic as before, we end up with a set of Euler-Lagrange equations with $\Lambda$: + +$$\begin{aligned} + 0 + = \pdv{\Lambda}{f_n} - \dv{x} \Big( \pdv{\Lambda}{f_n'} \Big) +\end{aligned}$$ + +By inserting the definition of $\Lambda$, we then get the following. +Recall that $\phi_m$ is holonomic, and thus independent of all derivatives $f_n'$: + +$$\begin{aligned} + \boxed{ + 0 + = \pdv{L}{f_n} - \dv{x} \Big( \pdv{L}{f_n'} \Big) + \sum_{m} \lambda_m \pdv{\phi_m}{f_n} + } +\end{aligned}$$ + +These are **Lagrange's equations of the first kind**, +with their second-kind counterparts being the earlier Euler-Lagrange equations. +Note that there are $N$ separate equations, one for each $f_n$. + +Due to the constraints $\phi_m$, the functions $f_n$ are not independent. +This is solved by choosing $\lambda_m$ such that $M$ of the $N$ equations hold, +i.e. solving a system of $M$ equations for $\lambda_m$: + +$$\begin{aligned} + \dv{x} \Big( \pdv{L}{f_n'} \Big) - \pdv{L}{f_n} + = \sum_{m} \lambda_m \pdv{\phi_m}{f_n} +\end{aligned}$$ + +And then the remaining $N - M$ equations can be solved in the normal unconstrained way. + + + +## References +1. G.B. Arfken, H.J. Weber, + *Mathematical methods for physicists*, 6th edition, 2005, + Elsevier. diff --git a/content/know/concept/density-operator/index.pdc b/content/know/concept/density-operator/index.pdc index 84c2d74..39c2e85 100644 --- a/content/know/concept/density-operator/index.pdc +++ b/content/know/concept/density-operator/index.pdc @@ -45,11 +45,15 @@ $$\begin{aligned} \end{aligned}$$ However, from the special case where $\ket{\Psi_n}$ are indeed basis vectors, -we can conclude that $\hat{\rho}$ is Hermitian, +we can conclude that $\hat{\rho}$ is positive semidefinite and Hermitian, and that its trace (i.e. the total probability) is 100%: $$\begin{gathered} \boxed{ + \hat{\rho} \ge 0 + } + \qquad \qquad + \boxed{ \hat{\rho}^\dagger = \hat{\rho} } \qquad \qquad diff --git a/content/know/concept/holomorphic-function/index.pdc b/content/know/concept/holomorphic-function/index.pdc index 3e7a91e..1077060 100644 --- a/content/know/concept/holomorphic-function/index.pdc +++ b/content/know/concept/holomorphic-function/index.pdc @@ -196,7 +196,7 @@ $$\begin{aligned} \end{aligned}$$ **Cauchy's residue theorem** generalizes Cauchy's integral theorem -to meromorphic functions, and states that the integral of a contour $C$, +to meromorphic functions, and states that the integral of a contour $C$ depends on the simple poles $p$ it encloses: $$\begin{aligned} @@ -206,7 +206,7 @@ $$\begin{aligned} \end{aligned}$$ *__Proof__*. *From the definition of a meromorphic function, -we know that we can decompose $f(z)$ as follows, +we know that we can decompose $f(z)$ like so, where $h(z)$ is holomorphic and $p$ are all its poles:* $$\begin{aligned} @@ -228,5 +228,5 @@ This theorem might not seem very useful, but in fact, thanks to some clever mathematical magic, it allows us to evaluate many integrals along the real axis, most notably [Fourier transforms](/know/concept/fourier-transform/). -It can also be used to derive the Kramers-Kronig relations. +It can also be used to derive the [Kramers-Kronig relations](/know/concept/kramers-kronig-relations). diff --git a/content/know/concept/quantum-entanglement/index.pdc b/content/know/concept/quantum-entanglement/index.pdc new file mode 100644 index 0000000..f8d0b76 --- /dev/null +++ b/content/know/concept/quantum-entanglement/index.pdc @@ -0,0 +1,152 @@ +--- +title: "Quantum entanglement" +firstLetter: "Q" +publishDate: 2021-03-07 +categories: +- Physics +- Quantum mechanics +- Quantum information + +date: 2021-03-07T15:36:15+01:00 +draft: false +markup: pandoc +--- + +# Quantum entanglement + +Consider a composite quantum system which consists of two subsystems $A$ and $B$, +respectively with basis states $\ket{a_n}$ and $\ket{b_n}$. +All accessible states of the sytem $\ket{\Psi}$ lie in +the tensor product of the subsystems' +[Hilbert spaces](/know/concept/hilbert-space/) $\mathbb{H}_A$ and $\mathbb{H}_B$: + +$$\begin{aligned} + \ket{\Psi} \in \mathbb{H}_A \otimes \mathbb{H}_B +\end{aligned}$$ + +A subset of these states can be written as the tensor product (i.e. Kronecker product in a basis) +of a state $\ket{\alpha}$ in $A$ and a state $\ket{\beta}$ in $B$, +often abbreviated as $\ket{\alpha} \ket{\beta}$: + +$$\begin{aligned} + \ket{\Psi} + = \ket{\alpha} \ket{\beta} + = \ket{\alpha} \otimes \ket{\beta} +\end{aligned}$$ + +The states that can be written in this way are called **separable**, +and states that cannot are called **entangled**. +Therefore, we are dealing with **quantum entanglement** +if the state of subsystem $A$ cannot be fully described +independently of the state of subsystem $B$, and vice versa. + +To detect and quantify entanglement, +we can use the [density operator](/know/concept/density-operator/) $\hat{\rho}$. +For a pure ensemble in a given (possibly entangled) state $\ket{\Psi}$, +$\hat{\rho}$ is given by: + +$$\begin{aligned} + \hat{\rho} = \ket{\Psi} \bra{\Psi} +\end{aligned}$$ + +From this, we would like to extract the corresponding state of subsystem $A$. +For that purpose, we define the **reduced density operator** $\hat{\rho}_A$ of subsystem $A$ as follows: + +$$\begin{aligned} + \boxed{ + \hat{\rho}_A + = \Tr_B(\hat{\rho}) + = \sum_m \bra{b_m} \Big( \hat{\rho} \Big) \ket{b_m} + } +\end{aligned}$$ + +Where $\Tr_B(\hat{\rho})$ is called the **partial trace** of $\hat{\rho}$, +which basically eliminates subsystem $B$ from $\hat{\rho}$. +For a pure composite state $\ket{\Psi}$, +the resulting $\hat{\rho}_A$ describes a pure state in $A$ if $\ket{\Psi}$ is separable, +else, if $\ket{\Psi}$ is entangled, it describes a mixed state in $A$. +In the former case we simply find: + +$$\begin{aligned} + \boxed{ + \ket{\Psi} = \ket{\alpha} \otimes \ket{\beta} + \quad \implies \quad + \hat{\rho}_A = \ket{\alpha} \bra{\alpha} + } +\end{aligned}$$ + +We call $\ket{\Psi}$ **maximally entangled** +if its reduced density operators are **maximally mixed**, +where $N$ is the dimension of $\mathbb{H}_A$ and $\hat{I}$ is the identity matrix: + +$$\begin{aligned} + \hat{\rho}_A + = \frac{1}{N} \hat{I} +\end{aligned}$$ + +Suppose that we are given an entangled pure state +$\ket{\Psi} \neq \ket{\alpha} \otimes \ket{\beta}$. +Then the partial traces $\hat{\rho}_A$ and $\hat{\rho}_B$ +of $\hat{\rho} = \ket{\Psi} \bra{\Psi}$ are mixed states with the same probabilities $p_n$ +(assuming $\mathbb{H}_A$ and $\mathbb{H}_B$ have the same dimensions, +which is usually the case): + +$$\begin{aligned} + \hat{\rho}_A + = \Tr_B(\hat{\rho}) + = \sum_n p_n \ket{a_n} \bra{a_n} + \qquad \quad + \hat{\rho}_B + = \Tr_A(\hat{\rho}) + = \sum_n p_n \ket{b_n} \bra{b_n} +\end{aligned}$$ + +There exists an orthonormal choice +of the subsystem basis states $\ket{a_n}$ and $\ket{b_n}$, +such that $\ket{\Psi}$ can be written as follows, +where $p_n$ are the probabilities in the reduced density operators: + +$$\begin{aligned} + \ket{\Psi} + = \sum_n \sqrt{p_n} \Big( \ket{a_n} \otimes \ket{b_n} \Big) +\end{aligned}$$ + +This is the **Schmidt decomposition**, +and the **Schmidt number** is the number of nonzero terms in the summation, +which can be used to determine if the state $\ket{\Psi}$ +is entangled (greater than one) or separable (equal to one). + +By looking at the Schmidt decomposition, we can notice that, +if $\hat{O}_A$ and $\hat{O}_B$ are the subsystem observables +with basis eigenstates $\ket{a_n}$ and $\ket{b_n}$, +then measurement results of these operators +will be perfectly correlated across $A$ and $B$. +This is a general property of entangled systems, +but beware: correlation does not imply entanglement! + +But what if the composite system is in a mixed state $\hat{\rho}$? +The state is separable if and only if: + +$$\begin{aligned} + \boxed{ + \hat{\rho} + = \sum_m p_m \Big( \hat{\rho}_A \otimes \hat{\rho}_B \Big) + } +\end{aligned}$$ + +Where $p_m$ are probabilities, +and $\hat{\rho}_A$ and $\hat{\rho}_B$ can be any subsystem states. +In reality, it is very hard to determine, using this criterium, +whether an arbitrary given $\hat{\rho}$ is separable or not. + +As a final side note, the expectation value +of an obervable $\hat{O}_A$ acting only on $A$ is given by: + +$$\begin{aligned} + \expval*{\hat{O}_A} + = \Tr\big(\hat{\rho} \hat{O}_A\big) + = \Tr_A\big(\Tr_B(\hat{\rho} \hat{O}_A)\big) + = \Tr_A\big(\Tr_B(\hat{\rho}) \hat{O}_A)\big) + = \Tr_A\big(\hat{\rho}_A \hat{O}_A\big) +\end{aligned}$$ + diff --git a/content/know/concept/quantum-teleportation/index.pdc b/content/know/concept/quantum-teleportation/index.pdc new file mode 100644 index 0000000..3287544 --- /dev/null +++ b/content/know/concept/quantum-teleportation/index.pdc @@ -0,0 +1,145 @@ +--- +title: "Quantum teleportation" +firstLetter: "Q" +publishDate: 2021-03-07 +categories: +- Quantum information + +date: 2021-03-07T20:30:30+01:00 +draft: false +markup: pandoc +--- + +# Quantum teleportation + +**Quantum teleportation** is a method to transfer quantum information +between systems without the use of a quantum channel. +It is based on [quantum entanglement](/know/concept/quantum-entanglement/). + +Suppose that Alice has a qubit $\ket{q}_{A'}$ that she wants to send to Bob. +Since she has not measured it yet, she does not know $\alpha$ or $\beta$; +she just wants Bob to get the same qubit: + +$$\begin{aligned} + \ket{q} + = \alpha \ket{0}_{A'} + \beta \ket{1}_{A'} +\end{aligned}$$ + +She can only directly communicate with Bob over a classical channel. +This is not enough: even if Alice did know $\alpha$ and $\beta$ exactly +(which would need her having infinitely many copies to measure), +sending an arbitrary real number requires an infinite amount of classical data. + +However, between them, she and Bob also have an entangled Bell state, +e.g. $\ket*{\Phi^+}_{AB}$ (it does not matter which Bell state it is) +The state of the composite system is then as follows, +with $A'$ being Alice' qubit, $A$ her side of the Bell state, and $B$ Bob's side: + +$$\begin{aligned} + \ket{q}_{A'} \otimes \ket*{\Phi^+}_{AB} + &= \frac{1}{\sqrt{2}} \Big( \alpha \ket{0} + \beta \ket{1} \Big)_{A'} \Big( \ket{00} + \ket{11} \Big)_{AB} + \\ + &= \frac{1}{\sqrt{2}} \Big( \alpha \ket{000} + \beta \ket{100} + + \alpha \ket{011} + \beta \ket{111} \Big)_{A'AB} +\end{aligned}$$ + +Now, observe that we can write any combination of $\ket{0}$ and $\ket{1}$ +in the Bell basis like so: + +$$\begin{aligned} + \ket{00} + &= \frac{\ket{\Phi^{+}} + \ket{\Phi^{-}}}{\sqrt{2}} + \qquad \quad + \ket{11} + = \frac{\ket{\Phi^{+}} - \ket{\Phi^{-}}}{\sqrt{2}} + \\ + \ket{01} + &= \frac{\ket{\Psi^{+}} + \ket{\Psi^{-}}}{\sqrt{2}} + \qquad \quad + \ket{10} + = \frac{\ket{\Psi^{+}} - \ket{\Psi^{-}}}{\sqrt{2}} +\end{aligned}$$ + +Using this, we can rewrite our previous result in terms of the Bell states as follows: + +$$\begin{aligned} + \ket{q}_{A'} \ket*{\Phi^+}_{AB} + &= \frac{\alpha}{2} \Big( \ket*{\Phi^{+}} + \ket*{\Phi^{-}} \Big)_{A'A} \ket{0}_B + + \frac{\beta}{2} \Big( \ket*{\Psi^{+}} - \ket*{\Psi^{-}} \Big)_{A'A} \ket{0}_B + \\ + &+ \frac{\alpha}{2} \Big( \ket*{\Psi^{+}} + \ket*{\Psi^{-}} \Big)_{A'A} \ket{1}_B + + \frac{\beta}{2} \Big( \ket*{\Phi^{+}} - \ket*{\Phi^{-}} \Big)_{A'A} \ket{1}_B +\end{aligned}$$ + +If we group all terms according to the Bell states, +we end up with an interesting expression: + +$$\begin{aligned} + \ket{q}_{A'} \ket*{\Phi^+}_{AB} + = \frac{1}{2} \bigg( &\ket*{\Phi^{+}}_{A'A} \Big( \alpha \ket{0} + \beta \ket{1} \Big)_{B} + + \ket*{\Phi^{-}}_{A'A} \Big( \alpha \ket{0} - \beta \ket{1} \Big)_{B} + \\ + + &\ket*{\Psi^{+}}_{A'A} \Big( \alpha \ket{1} + \beta \ket{0} \Big)_{B} + + \ket*{\Psi^{-}}_{A'A} \Big( \alpha \ket{1} - \beta \ket{0} \Big)_{B} \bigg) +\end{aligned}$$ + +Thus, purely due to entanglement, +Bob's qubit $B$ is in a superposition of the following states: + +$$\begin{aligned} + \ket{q} + &= \alpha \ket{0} + \beta \ket{1} + \qquad \quad + \quad \hat{\sigma}_z \ket{q} + = \alpha \ket{0} - \beta \ket{1} + \\ + \hat{\sigma}_x \ket{q} + &= \alpha \ket{1} + \beta \ket{0} + \qquad \quad + \hat{\sigma}_x \hat{\sigma}_z \ket{q} + = \alpha \ket{0} - \beta \ket{1} +\end{aligned}$$ + +Consequently, Alice and Bob are sharing (or, to be precise, seeing different sides of) +the following entangled three-qubit state: + +$$\begin{aligned} + \ket{q}_{A'} \ket*{\Phi^+}_{AB} + = \frac{1}{2} \bigg( &\ket*{\Phi^{+}}_{A'A} \Big( \ket{q} \Big)_B \quad\, + \ket*{\Phi^{-}}_{A'A} \Big( \hat{\sigma}_z \ket{q} \Big)_B + \\ + + &\ket*{\Psi^{+}}_{A'A} \Big( \hat{\sigma}_x \ket{q} \Big)_B + \ket*{\Psi^{-}}_{A'A} \Big( \hat{\sigma}_x \hat{\sigma}_z \ket{q} \Big)_B \bigg) +\end{aligned}$$ + +The point is that, thanks to the initial entanglement between Alice and Bob, +adding $\ket{q}_{A'}$ into the mix somehow "teleports" that information to Bob, +although it is not in a usable form yet. + +To finish the process, Alice measures her side $A'A$ in the Bell basis. +Consequently, $A'A$ collapses into one of +$\ket*{\Phi^{+}}$, $\ket*{\Phi^{-}}$, $\ket*{\Psi^{+}}$, $\ket*{\Psi^{-}}$ +with equal probability, and she knows which. +This collapse leaves Bob's side $B$ in $\ket{q}$, $\hat{\sigma}_z \ket{q}$, +$\hat{\sigma}_x \ket{q}$, or $\hat{\sigma}_x \hat{\sigma}_z \ket{q}$, respectively. +The entanglement between $A$ and $B$ is thus broken, +and instead Alice has local entanglement between $A'$ and $A$. + +She then uses the classical channel to tell Bob her result, +who then either does nothing (for $\ket{q}$), +applies $\hat{\sigma}_z$ (for $\hat{\sigma}_z \ket{q}$), +applies $\hat{\sigma}_x$ (for $\hat{\sigma}_x \ket{q}$), +or applies $\hat{\sigma}_z \hat{\sigma}_x$ (for $\hat{\sigma}_x \hat{\sigma}_z \ket{q}$). +Then, due to the fact that $\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$, +he recovers $\ket{q}$ in his local qubit $B$. + +This is not violating the [no-cloning theorem](/know/concept/no-cloning-theorem) +because Alice does not require any knowledge of $\ket{q}$, +and after the measurement, her qubit $A'$ will no longer be in that state. +In other words, quantum teleportation *moves* states, +rather than copying them. + +Nor does this conflict with Einstein's relativity, +since the information travels no faster than light: +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}$. diff --git a/content/know/concept/time-dependent-perturbation-theory/index.pdc b/content/know/concept/time-dependent-perturbation-theory/index.pdc new file mode 100644 index 0000000..fbb71b2 --- /dev/null +++ b/content/know/concept/time-dependent-perturbation-theory/index.pdc @@ -0,0 +1,122 @@ +--- +title: "Time-dependent perturbation theory" +firstLetter: "T" +publishDate: 2021-03-07 +categories: +- Physics +- Quantum mechanics +- Perturbation + +date: 2021-03-07T11:08:14+01:00 +draft: false +markup: pandoc +--- + +# Time-dependent perturbation theory + +In quantum mechanics, **time-dependent perturbation theory** exists to deal +with time-varying perturbations to the Schrödinger equation. +This is in contrast to [time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/), +where the perturbation is is stationary. + +Let $\hat{H}_0$ be the base time-independent +Hamiltonian, and $\hat{H}_1$ be a time-varying perturbation, with +"bookkeeping" parameter $\lambda$: + +$$\begin{aligned} + \hat{H}(t) = \hat{H}_0 + \lambda \hat{H}_1(t) +\end{aligned}$$ + +We assume that the unperturbed time-independent problem +$\hat{H}_0 \ket{n} = E_n \ket{n}$ has already been solved, such that the +full solution is: + +$$\begin{aligned} + \ket{\Psi_0(t)} = \sum_{n} c_n \ket{n} \exp(- i E_n t / \hbar) +\end{aligned}$$ + +Since these $\ket{n}$ form a complete basis, the perturbed wave function +can be written in the same form, but with time-dependent coefficients $c_n(t)$: + +$$\begin{aligned} + \ket{\Psi(t)} = \sum_{n} c_n(t) \ket{n} \exp(- i E_n t / \hbar) +\end{aligned}$$ + +We insert this ansatz in the time-dependent Schrödinger equation, and +reduce it using the known unperturbed time-independent problem: + +$$\begin{aligned} + 0 + &= \hat{H}_0 \ket{\Psi(t)} + \lambda \hat{H}_1 \ket{\Psi(t)} - i \hbar \dv{t} \ket{\Psi(t)} + \\ + &= \sum_{n} + \Big( c_n \hat{H}_0 \ket{n} + \lambda c_n \hat{H}_1 \ket{n} - c_n E_n \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- i E_n t / \hbar) + \\ + &= \sum_{n} \Big( \lambda c_n \hat{H}_1 \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- 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( \lambda c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \frac{d c_n}{dt} \braket{m}{n} \Big) \exp(- i E_n t / \hbar) +\end{aligned}$$ + +Thanks to orthonormality, this removes the latter term from the summation: + +$$\begin{aligned} + i \hbar \frac{d c_m}{dt} \exp(- i E_m t / \hbar) + &= \lambda \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} \exp(- i E_n t / \hbar) +\end{aligned}$$ + +We divide by the left-hand exponential and define +$\omega_{mn} = (E_m - E_n) / \hbar$ to get: + +$$\begin{aligned} + \boxed{ + i \hbar \frac{d c_m}{dt} + = \lambda \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} \exp(i \omega_{mn} t) + } +\end{aligned}$$ + +So far, we have not invoked any approximation, +so we can analytically find $c_n(t)$ for some simple systems. +Furthermore, it is useful to write this equation in integral form instead: + +$$\begin{aligned} + c_m(t) + = c_m(0) - \lambda \frac{i}{\hbar} \sum_{n} \int_0^t c_n(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau} +\end{aligned}$$ + +If this cannot be solved exactly, we must approximate it. We expand +$c_m(t)$ in the usual way, with the initial condition $c_m^{(j)}(0) = 0$ +for $j > 0$: + +$$\begin{aligned} + c_m(t) = c_m^{(0)} + \lambda c_m^{(1)}(t) + \lambda^2 c_m^{(2)}(t) + ... +\end{aligned}$$ + +We then insert this into the integral and collect the non-zero orders of $\lambda$: + +$$\begin{aligned} + c_m^{(1)}(t) + &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(0)} \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau} + \\ + c_m^{(2)}(t) + &= - \frac{i}{\hbar} \sum_{n} + \int_0^t c_n^{(1)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau} + \\ + c_m^{(3)}(t) + &= - \frac{i}{\hbar} \sum_{n} + \int_0^t c_n^{(2)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau} +\end{aligned}$$ + +And so forth. The pattern here is clear: we can calculate the $(j\!+\!1)$th +correction using only our previous result for the $j$th correction. +We cannot go any further than this without considering a specific perturbation $\hat{H}_1(t)$. + + +## References +1. D.J. Griffiths, D.F. Schroeter, + *Introduction to quantum mechanics*, 3rd edition, + Cambridge. diff --git a/content/know/concept/time-independent-perturbation-theory/index.pdc b/content/know/concept/time-independent-perturbation-theory/index.pdc index 2035fc2..3be3cd5 100644 --- a/content/know/concept/time-independent-perturbation-theory/index.pdc +++ b/content/know/concept/time-independent-perturbation-theory/index.pdc @@ -159,22 +159,22 @@ $$\begin{aligned} Here it is clear why this is only valid in the non-degenerate case: otherwise we would divide by zero in the denominator. -Next, to find the second-order correction to the energy $E_n^{(2)}$, we -take the corresponding equation and put $\bra{n}$ in front of it: +Next, to find the second-order energy correction $E_n^{(2)}$, +we take the corresponding equation and put $\bra{n}$ in front of it: $$\begin{aligned} - \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} - &= E_n^{(2)} \braket{n}{n} + E_n^{(1)} \braket{n}{\psi_n^{(1)}} + \varepsilon_n \braket{n}{\psi_n^{(2)}} + \matrixel*{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel*{n}{\hat{H}_0}{\psi_n^{(2)}} + &= E_n^{(2)} \braket{n}{n} + E_n^{(1)} \braket*{n}{\psi_n^{(1)}} + \varepsilon_n \braket*{n}{\psi_n^{(2)}} \end{aligned}$$ Because $\hat{H}_0$ is Hermitian, we know that -$\matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \braket{n}{\psi_n^{(2)}}$, +$\matrixel*{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \braket*{n}{\psi_n^{(2)}}$, i.e. we apply it to the bra, which lets us eliminate two terms. Also, since $\ket{n}$ is normalized, we find: $$\begin{aligned} E_n^{(2)} - = \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket{n}{\psi_n^{(1)}} + = \matrixel*{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket*{n}{\psi_n^{(1)}} \end{aligned}$$ We explicitly removed the $\ket{n}$-dependence of $\ket*{\psi_n^{(1)}}$, @@ -221,8 +221,8 @@ take the equation at order $\lambda^1$ and prepend an arbitrary eigenspace basis vector $\bra{n, \delta}$: $$\begin{aligned} - \matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel{n, \delta}{\hat{H}_0}{\psi_n^{(1)}} - &= E_n^{(1)} \braket{n, \delta}{n} + \varepsilon_n \braket{n, \delta}{\psi_n^{(1)}} + \matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel*{n, \delta}{\hat{H}_0}{\psi_n^{(1)}} + &= E_n^{(1)} \braket{n, \delta}{n} + \varepsilon_n \braket*{n, \delta}{\psi_n^{(1)}} \end{aligned}$$ Since $\hat{H}_0$ is Hermitian, we use the same trick as before to @@ -320,11 +320,18 @@ $\ket{n, d_1}$ and $\ket{n, d_2}$ have distinct eigenvalues $\ell_1 \neq \ell_2$ for $d_1 \neq d_2$: $$\begin{aligned} - \hat{L} \ket{n, b_1} = \ell_1 \ket{n, b_1} + \hat{L} \ket{n, d_1} = \ell_1 \ket{n, d_1} \qquad - \hat{L} \ket{n, b_2} = \ell_2 \ket{n, b_2} + \hat{L} \ket{n, d_2} = \ell_2 \ket{n, d_2} \end{aligned}$$ -When this holds for any orthogonal choice of $\ket{n, d_1}$ and +When this condition holds for any orthogonal choice of $\ket{n, d_1}$ and $\ket{n, d_2}$, then these specific eigenvectors of $\hat{L}$ are the "good states", for any valid choice of $\hat{L}$. + + + +## References +1. D.J. Griffiths, D.F. Schroeter, + *Introduction to quantum mechanics*, 3rd edition, + Cambridge. diff --git a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc b/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc index 482650e..cf44fc8 100644 --- a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc +++ b/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc @@ -14,7 +14,7 @@ markup: pandoc # Wentzel-Kramers-Brillouin approximation In quantum mechanics, the **Wentzel-Kramers-Brillouin** or simply the **WKB -approximation** is a method to approximate the wave function $\psi(x)$ of +approximation** is a technique to approximate the wave function $\psi(x)$ of the one-dimensional time-independent Schrödinger equation. It is an example of a **semiclassical approximation**, because it tries to find a balance between classical and quantum physics. @@ -196,3 +196,12 @@ In the classical region ($E > V$), the wave function oscillates, and in the quantum-mechanical region ($E < V$) it is exponential. Note that for $E \approx V$ the approximation breaks down, due to the appearance of $p(x)$ in the denominator. + + +## References +1. D.J. Griffiths, D.F. Schroeter, + *Introduction to quantum mechanics*, 3rd edition, + Cambridge. +2. R. Shankar, + *Principles of quantum mechanics*, 2nd edition, + Springer. |