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Diffstat (limited to 'source/know/concept/berry-phase')
| -rw-r--r-- | source/know/concept/berry-phase/index.md | 88 |
1 files changed, 44 insertions, 44 deletions
diff --git a/source/know/concept/berry-phase/index.md b/source/know/concept/berry-phase/index.md index eedc548..d237ea5 100644 --- a/source/know/concept/berry-phase/index.md +++ b/source/know/concept/berry-phase/index.md @@ -8,8 +8,8 @@ categories: layout: "concept" --- -Consider a Hamiltonian $\hat{H}$ that does not explicitly depend on time, -but does depend on a given parameter $\vb{R}$. +Consider a Hamiltonian $$\hat{H}$$ that does not explicitly depend on time, +but does depend on a given parameter $$\vb{R}$$. The Schrödinger equations then read: $$\begin{aligned} @@ -20,9 +20,9 @@ $$\begin{aligned} &= E_n(\vb{R}) \Ket{\psi_n(\vb{R})} \end{aligned}$$ -The general full solution $\Ket{\Psi_n}$ has the following form, -where we allow $\vb{R}$ to evolve in time, -and we have abbreviated the traditional phase of the "wiggle factor" as $L_n$: +The general full solution $$\Ket{\Psi_n}$$ has the following form, +where we allow $$\vb{R}$$ to evolve in time, +and we have abbreviated the traditional phase of the "wiggle factor" as $$L_n$$: $$\begin{aligned} \Ket{\Psi_n(t)} @@ -31,11 +31,11 @@ $$\begin{aligned} L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'} \end{aligned}$$ -The **geometric phase** $\gamma_n(t)$ is more interesting. -It is not included in $\Ket{\psi_n}$, -because it depends on the path $\vb{R}(t)$ -rather than only the present $\vb{R}$ and $t$. -Its dynamics can be found by inserting the above $\Ket{\Psi_n}$ +The **geometric phase** $$\gamma_n(t)$$ is more interesting. +It is not included in $$\Ket{\psi_n}$$, +because it depends on the path $$\vb{R}(t)$$ +rather than only the present $$\vb{R}$$ and $$t$$. +Its dynamics can be found by inserting the above $$\Ket{\Psi_n}$$ into the time-dependent Schrödinger equation: $$\begin{aligned} @@ -58,8 +58,8 @@ $$\begin{aligned} &= \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t} \end{aligned}$$ -Front-multiplying by $i \Bra{\Psi_n}$ gives us -the equation of motion of the geometric phase $\gamma_n$: +Front-multiplying by $$i \Bra{\Psi_n}$$ gives us +the equation of motion of the geometric phase $$\gamma_n$$: $$\begin{aligned} \boxed{ @@ -68,7 +68,7 @@ $$\begin{aligned} } \end{aligned}$$ -Where we have defined the so-called **Berry connection** $\vb{A}_n$ as follows: +Where we have defined the so-called **Berry connection** $$\vb{A}_n$$ as follows: $$\begin{aligned} \boxed{ @@ -77,9 +77,9 @@ $$\begin{aligned} } \end{aligned}$$ -Importantly, note that $\vb{A}_n$ is real, -provided that $\Ket{\psi_n}$ is always normalized for all $\vb{R}$. -To prove this, we start from the fact that $\nabla_\vb{R} 1 = 0$: +Importantly, note that $$\vb{A}_n$$ is real, +provided that $$\Ket{\psi_n}$$ is always normalized for all $$\vb{R}$$. +To prove this, we start from the fact that $$\nabla_\vb{R} 1 = 0$$: $$\begin{aligned} 0 @@ -91,14 +91,14 @@ $$\begin{aligned} = 2 \Imag\{ \vb{A}_n \} \end{aligned}$$ -Consequently, $\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is always real, -because $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary. +Consequently, $$\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is always real, +because $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary. -Suppose now that the parameter $\vb{R}(t)$ is changed adiabatically +Suppose now that the parameter $$\vb{R}(t)$$ is changed adiabatically (i.e. so slow that the system stays in the same eigenstate) -for $t \in [0, T]$, along a circuit $C$ with $\vb{R}(0) \!=\! \vb{R}(T)$. -Integrating the phase $\gamma_n(t)$ over this contour $C$ then yields -the **Berry phase** $\gamma_n(C)$: +for $$t \in [0, T]$$, along a circuit $$C$$ with $$\vb{R}(0) \!=\! \vb{R}(T)$$. +Integrating the phase $$\gamma_n(t)$$ over this contour $$C$$ then yields +the **Berry phase** $$\gamma_n(C)$$: $$\begin{aligned} \boxed{ @@ -107,9 +107,9 @@ $$\begin{aligned} } \end{aligned}$$ -But we have a problem: $\vb{A}_n$ is not unique! +But we have a problem: $$\vb{A}_n$$ is not unique! Due to the Schrödinger equation's gauge invariance, -any function $f(\vb{R}(t))$ can be added to $\gamma_n(t)$ +any function $$f(\vb{R}(t))$$ can be added to $$\gamma_n(t)$$ without making an immediate physical difference to the state. Consider the following general gauge transformation: @@ -118,9 +118,9 @@ $$\begin{aligned} \equiv \exp(i f(\vb{R})) \: \Ket{\psi_n(\vb{R})} \end{aligned}$$ -To find $\vb{A}_n$ for a particular choice of $f$, +To find $$\vb{A}_n$$ for a particular choice of $$f$$, we need to evaluate the inner product -$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$: +$$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$$: $$\begin{aligned} \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n} @@ -131,16 +131,16 @@ $$\begin{aligned} &= i \nabla_\vb{R} f + \inprod{\psi_n}{\nabla_\vb{R} \psi_n} \end{aligned}$$ -Unfortunately, $f$ does not vanish as we would have liked, -so $\vb{A}_n$ depends on our choice of $f$. +Unfortunately, $$f$$ does not vanish as we would have liked, +so $$\vb{A}_n$$ depends on our choice of $$f$$. However, the curl of a gradient is always zero, -so although $\vb{A}_n$ is not unique, -its curl $\nabla_\vb{R} \cross \vb{A}_n$ is guaranteed to be. -Conveniently, we can introduce a curl in the definition of $\gamma_n(C)$ +so although $$\vb{A}_n$$ is not unique, +its curl $$\nabla_\vb{R} \cross \vb{A}_n$$ is guaranteed to be. +Conveniently, we can introduce a curl in the definition of $$\gamma_n(C)$$ by applying Stokes' theorem, under the assumption -that $\vb{A}_n$ has no singularities in the area enclosed by $C$ -(fortunately, $\vb{A}_n$ can always be chosen to satisfy this): +that $$\vb{A}_n$$ has no singularities in the area enclosed by $$C$$ +(fortunately, $$\vb{A}_n$$ can always be chosen to satisfy this): $$\begin{aligned} \boxed{ @@ -149,10 +149,10 @@ $$\begin{aligned} } \end{aligned}$$ -Where we defined $\vb{B}_n$ as the curl of $\vb{A}_n$. -Now $\gamma_n(C)$ is guaranteed to be unique. -Note that $\vb{B}_n$ is analogous to a magnetic field, -and $\vb{A}_n$ to a magnetic vector potential: +Where we defined $$\vb{B}_n$$ as the curl of $$\vb{A}_n$$. +Now $$\gamma_n(C)$$ is guaranteed to be unique. +Note that $$\vb{B}_n$$ is analogous to a magnetic field, +and $$\vb{A}_n$$ to a magnetic vector potential: $$\begin{aligned} \vb{B}_n(\vb{R}) @@ -160,9 +160,9 @@ $$\begin{aligned} = \Imag\!\Big\{ \nabla_\vb{R} \cross \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\} \end{aligned}$$ -Unfortunately, $\nabla_\vb{R} \psi_n$ is difficult to evaluate explicitly, -so we would like to rewrite $\vb{B}_n$ such that it does not enter. -We do this as follows, inserting $1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$ along the way: +Unfortunately, $$\nabla_\vb{R} \psi_n$$ is difficult to evaluate explicitly, +so we would like to rewrite $$\vb{B}_n$$ such that it does not enter. +We do this as follows, inserting $$1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$$ along the way: $$\begin{aligned} i \vb{B}_n @@ -172,9 +172,9 @@ $$\begin{aligned} &= \sum_{m} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n} \end{aligned}$$ -The fact that $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary +The fact that $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary means it is parallel to its complex conjugate, -and thus the cross product vanishes, so we exclude $n$ from the sum: +and thus the cross product vanishes, so we exclude $$n$$ from the sum: $$\begin{aligned} \vb{B}_n @@ -200,8 +200,8 @@ $$\begin{aligned} } \end{aligned}$$ -Which only involves $\nabla_\vb{R} \hat{H}$, -and is therefore easier to evaluate than any $\Ket{\nabla_\vb{R} \psi_n}$. +Which only involves $$\nabla_\vb{R} \hat{H}$$, +and is therefore easier to evaluate than any $$\Ket{\nabla_\vb{R} \psi_n}$$. |