XVIII Dynamics for stationary Hamiltonians

The derivation of the stationary Schrödinger equation for time-independent Hamiltonians $\widehat{H}$ is based on the trial solution

$$|{\mathrm{\Psi }}_n(t)⟩=\exp (-i{E}_{n}t/\mathrm{\hslash })|{\psi }_n⟩.$$

(357)

We insert this into the time-dependent Schrödinger equation (355), which gives

$$E_n|{\psi }_n⟩\exp (-i{E}_{n}t/\mathrm{\hslash })=\widehat{H}|{\psi }_n⟩\exp (-i{E}_{n}t/\mathrm{\hslash }).$$

(358)

The common factor $\exp (-i{E}_{n}t/\mathrm{\hslash })$ on both sides never vanishes, and thus can be divided out; what remains is the stationary Schrödinger equation (356).

In order to obtain a general solution of the time-dependent Schrödinger equation (355) we should note that this is still a linear differential equation, and thus obeys the superposition principle. Therefore, we can construct new solutions by adding different trial solutions (even if they correspond to different energies): the sum

$$|\mathrm{\Psi }(t)⟩=\sum _n{c}_n\exp (-i{E}_{n}t/\mathrm{\hslash })|{\psi }_n⟩$$

(359)

is still a solution, where the complex constants $c_n$ can be chosen arbitrarily. Here we assumed that the energies $E_n$ are discrete. For continuous energies, the sum is replaced by an integral,

$$|\mathrm{\Psi }(t)⟩=\int 𝑑Ec(E)\exp (-iEt/\mathrm{\hslash })|{\psi }_E⟩.$$

(360)

If both types of spectra coexist (like, e.g., in the hydrogen atom, where we have a discrete set of bound states for , and a continuum of extended states with $E>0$), we can write

	$|\mathrm{\Psi }(t)⟩$	$=$	${\displaystyle \int 𝑑Ec(E)\exp (-iEt/\mathrm{\hslash })|{\psi }_E⟩}$		(361)
			$+{\displaystyle \sum _n}{c}_n\exp (-i{E}_{n}t/\mathrm{\hslash })|{\psi }_n⟩.$

This expression covers all valid solutions of the time-dependent Schrödinger equation with time-independent Hamiltonian.

In practice, we often know the state $|\mathrm{\Psi }(t_0)⟩\equiv |{\mathrm{\Psi }}_0⟩$ of the quantum system at some time $t_0$. In that case, the specific solution of the Schrödinger equation follows from Eq. (361) by a particular choice of the constants $c_n$ and $c_E$. Since the collection of states ${\psi }_n$, ${\psi }_E$ form an orthonormal basis, the coefficients can be obtained by evaluating the scalar product,

	$c_n$	$=\exp (i{E}_n{t}_0/\mathrm{\hslash })⟨{\psi }_n|{\mathrm{\Psi }}_0⟩,$		(362)
	$c(E)$	$=\exp (iE{t}_0/\mathrm{\hslash })⟨{\psi }_E|{\mathrm{\Psi }}_0⟩.$		(363)

Here, we assumed that the states of the continuous spectrum fulfill the orthonormalisation condition

$$⟨{\psi }_{E^{\prime }}|{\psi }_E⟩=\delta (E^{\prime }-E),$$

(364)

where $\delta$ is the Dirac delta function, defined by ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\delta (x)f(x)𝑑x=f(0)$.

The expansion coefficients $c_n$ determine the probability to find the system in the discrete bound state with energy $E_n$, while $c(E)$ determines the probability density of the extended energy eigenstates in the continuous part of the spectrum:

$$P(E_n)={|c_n|}^2,P(E)={|c(E)|}^2.$$

(365)

These probabilities are independent of time, which is a consequence of energy conservation in stationary quantum systems. The expectation value of the energy follows from the general solution (361),

$$⟨\mathrm{\Psi }(t)|\widehat{H}\mathrm{\Psi }(t)⟩=\sum _n{E}_n{|c_n|}^2+\int 𝑑EE{|c(E)|}^2.$$

(366)

In the following, we formulate expressions only using the symbols for a discrete spectrum (quantities $c_n$, $E_n$, $|{\psi }_n⟩$, ${\psi }_n(x)$, etc.), but imply that these have to be replaced by their continuous counterparts (with the sum ${\sum }_n$ replaced by integration over energy) if the spectrum is continuous.

The approach above requires to determine all eigenstates of the system, and to match the initial condition $|\mathrm{\Psi }(t_0)⟩$ to a superposition of these states. In many cases, a more formal approach proves advantageous: We seek an operator $\widehat{U}(t-t_0)$ such that

$$|\mathrm{\Psi }(t)⟩=\widehat{U}(t-t_0)|\mathrm{\Psi }(t_0)⟩.$$

(367)

The operator $\widehat{U}(t)$ is known as the time-evolution operator. The linearity of the time-dependent Schrödinger equation guarantees that this operator is linear, too. Indeed, we can give a formal (but often also practically useful) expression which relates $\widehat{U}$ directly to the Hamiltonian:

$$\widehat{U}(t)=\exp (-i\widehat{H}t/\mathrm{\hslash }).$$

(368)

The right-hand side has to be interpreted as a Taylor expansion; for any operator $\widehat{A}$, $\exp (\widehat{A})={\sum }_{n=0}^{\mathrm{\infty }}{\widehat{A}}^n/n!$. This can be used to show

$$\frac{d}{dt}\widehat{U}(t)=-\frac{i}{\mathrm{\hslash }}\widehat{H}\widehat{U}(t),$$

(369)

where the derivative of an operator $\widehat{A}(t)$ is defined as $\frac{d}{dt}\widehat{A}(t)={lim}_{\epsilon \to 0}{\epsilon }^{-1}[\widehat{A}(t+\epsilon )-\widehat{A}(t)]$. The validity of Eq. (368) then follows by inserting Eq. (367) into the time-dependent Schrödinger equation (355). The initial condition is verified by observing

$$\widehat{U}(0)=\widehat{I},$$

(370)

where $\widehat{I}$ is the identity operator ($\widehat{I}|\psi ⟩=|\psi ⟩$ for all $|\psi ⟩$).

It directly follows from Eq. (368) that $\widehat{U}$ is unitary:

$${\widehat{U}}^{-1}(t)=\widehat{U}(-t)=\exp (i\widehat{H}t/\mathrm{\hslash })={\widehat{U}}^{†}(t),$$

(371)

where we used in the last step that $\widehat{H}$ is hermitian. Importantly, this guarantees that an initially normalised state $|\mathrm{\Psi }(t)⟩$ remains normalised during the time evolution.

As a first example, we study the case of a free particle in one dimension, for which the stationary Schrödinger equation

$$E\psi (x)=-\frac{{\mathrm{\hslash }}^2}{2m}\frac{d^2}{d{x}^2}\psi (x)$$

(372)

is solved by the momentum eigenfunctions

$${\psi }_p(x)=\frac{1}{\sqrt{2\pi \mathrm{\hslash }}}\exp (ipx/\mathrm{\hslash }),$$

(373)

with energy given by $E_p=p^2/2m$. Therefore, the partial solutions of the time-dependent Schrödinger equation are plane propagating waves

$${\mathrm{\Psi }}_p(x,t)=\frac{1}{\sqrt{2\pi \mathrm{\hslash }}}\exp (ipx/\mathrm{\hslash }-i{E}_{p}t/\mathrm{\hslash }).$$

(374)

Assume that at time $t_0=0$ the system is in a state $\psi (x)$, expressed as a superposition of plane waves

$$\psi (x)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}𝑑p\tilde{\psi }(p){\psi }_p(x).$$

(375)

The expansion coefficients

$$\tilde{\psi }(p)=⟨{\psi }_p|\psi ⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}𝑑x{\psi }_p^{*}(x)\psi (x)$$

(376)

follow from the Fourier transformation of the initial state, and are identical to the momentum wave function introduced in section VII.1. The time-dependent state is then given by

$$\mathrm{\Psi }(x,t)=\frac{1}{\sqrt{2\pi \mathrm{\hslash }}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}𝑑p\tilde{\psi }(p)\exp (ipx/\mathrm{\hslash }-i{E}_{p}t/\mathrm{\hslash }).$$

(377)

We can also write this as

$$\mathrm{\Psi }(x,t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}𝑑p\tilde{\mathrm{\Psi }}(p,t){\psi }_p(x),$$

(378)

i.e., as a Fourier transformation with time-dependent coefficients

$$\tilde{\mathrm{\Psi }}(p,t)=\tilde{\psi }(p)\exp (-i{E}_{p}t/\mathrm{\hslash }).$$

(379)

In terms of this time-dependent momentum wave function, the probability density for momentum is given by $P(p,t)={|\tilde{\mathrm{\Psi }}(p,t)|}^2$. For the free particle, the probability density is stationary, $P(p,t)={|\tilde{\psi }(p)|}^2$, which is a consequence of momentum conservation in absence of forces acting on the particle.

In order to get some insight into the time dependence of the position wave function, let us consider the case of a particle initially described by a wave packet centred at $x_0=0$,

$$\psi (x)={(2\pi {\sigma }^2)}^{-1/4}\exp (-x^2/4{\sigma }^2+i{p}_{0}x/\mathrm{\hslash }),$$

(380)

where $\sigma$ and $p_0$ are constants. As shown in section VII.5, this wave packet has expectation values $⟨x⟩=0$ and $⟨p⟩=p_0$, as well as uncertainties $\mathrm{\Delta }x=\sigma$ and $\mathrm{\Delta }p=\mathrm{\hslash }/2\sigma$. We have also calculated the corresponding momentum wave function

$$\tilde{\psi }(p)={(2{\sigma }^2/{\mathrm{\hslash }}^2\pi )}^{1/4}\exp [-{\sigma }^2{(p-p_0)}^2/{\mathrm{\hslash }}^2].$$

(381)

Equation (377) for the time-dependent wave function then reduces to a (complex) Gaussian integral, which can be solved analytically. The results can be written as a generalised wave packet of the form

	$\mathrm{\Psi }(x,t)$	$=$	${(2\pi {\sigma }_t^2)}^{-1/4}\exp \left[-{\displaystyle \frac{{(x-p_{0}t/m)}^2}{4{|{\sigma }_t|}^2}}\right]$
			$\times \exp \left[i{\displaystyle \frac{{\sigma }^2}{{|{\sigma }_t|}^2}}\left({\displaystyle \frac{p_{0}x}{\mathrm{\hslash }}}-{\displaystyle \frac{p_0^{2}t}{2m\mathrm{\hslash }}}+{\displaystyle \frac{x^2\mathrm{\hslash }t}{8m{\sigma }^4}}\right)\right],$

where ${\sigma }_t=\sigma +(i\mathrm{\hslash }/2m\sigma )t$. The position probability density

$$P(x,t)={(2\pi {|{\sigma }_t|}^2)}^{-1/2}\exp \left[-\frac{{(x-p_{0}t/m)}^2}{2{|{\sigma }_t|}^2}\right]$$

(383)

is still a Gaussian, with a linearly drifting expectation value $⟨x⟩=p_{0}t/m$, and an uncertainty $\mathrm{\Delta }x=|{\sigma }_t|=\sigma \sqrt{1+({\mathrm{\hslash }}^2/4m^2{\sigma }^4)t^2}$ which increases slowly as long as $t\ll m{\sigma }^2/\mathrm{\hslash }$.

Classically, the momentum of a free particle can be measured in a time-of-flight experiment, where one observes the particle’s positions $x_0$ and $x$ at times $t_0$ and $t$, and evaluates $p=m(x-x_0)/(t-t_0)$. Quantum mechanically, we cannot know position and momentum at the time. As we now demonstrate, it is still instructive to carry out such a momentum measurement. The main requirement for its accuracy is to choose $t-t_0$ sufficiently large, so that the position probability spreads out over a large region in space (therefore, the measurement remains in accordance with the uncertainty principle).

In order to see how this works, let us set $t_0=0$ and assume that the particle is initially localised around position $x_0=0$, which can be enforced in practice by confining the particle in a suitable potential. We denote the corresponding wave function by $\psi (x)$, but don’t require this to be a Gaussian wave packet. Because $\psi (x)$ is well localised, we can assume (again in accordance with the uncertainty principle) that the associated momentum wave function $\tilde{\psi }(p)$ is rather smooth.

Next, we let the particle move freely (in the practical setting above, we switch off the confining potential). For large times, the time-evolved wave function (377) can be approximated using an advanced mathematical method, known as the stationary phase approximation. The idea behind this approximation is to observe that for large $t$, the function $S(x)\equiv px-p^{2}t/2m$ in the exponent of the integrand changes rapidly as a function of the integration variable $p$. Therefore, the integrand oscillates wildly, and tends to average out. The only exceptions are the vicinities of values $p_s$ of $p$ where the function $S(x)$ is stationary, i.e., has a vanishing derivative. Here, this is the case for $p_s=mx/t$. In the vicinity of this stationary point the integral can be approximated by a complex Gaussian. The result of this approximation is

$$\mathrm{\Psi }(x,t)\approx \sqrt{m/t}\tilde{\psi }(mx/t)\exp \left(i\frac{x^{2}m}{2t\mathrm{\hslash }}-i\frac{\pi }{4}\right).$$

(384)

Therefore, the position probability density at time $t$ is

$$P(x,t)\approx \frac{m}{t}{|\tilde{\psi }(mx/t)|}^2.$$

(385)

This can be translated into a probability distribution of the momentum $p=mx/t$:

$$P(p)=P(x,t)\frac{dx}{dp}=|\tilde{\psi }(p)|.$$

(386)

In this way, we recover our original interpretation of the momentum wave function! Note that at large times, the initial wave packet has spread out over a large region in space; this counteracts the increasing accuracy of this momentum measurement, in accordance with the uncertainty principle.

Interestingly, with suitable initial conditions the quantum dynamics of the harmonic oscillator turns out to be somewhat simpler than that of the free particle. This is realised when the initial wavefunction corresponds to the displaced ground state wave function (a so-called coherent state)

$$\psi (x)={\psi }_0(x-X_0)={(2\pi {\sigma }^2)}^{-1/4}\exp [-{(x-X_0)}^2/4{\sigma }^2],$$

(387)

where ${\sigma }^2=\mathrm{\hslash }/2m\omega$, and $X_0$ is a constant. The time-dependent solution can then be found from the ansatz

$$\mathrm{\Psi }(x,t)={\psi }_0(x-X(t))\exp [ixP(t)/\mathrm{\hslash }+iS(t)/\mathrm{\hslash }],$$

(388)

where $X(t)$, $P(t)$, and $S(t)$ are functions of time. We insert this into the time-dependent Schrödinger equation and compare both sides to obtain

			$[-im\omega X\dot{X}-\dot{S}]+x\{-\dot{P}+im\omega \dot{X}\}$
		$=$	$\left[{\displaystyle \frac{\mathrm{\hslash }\omega }{2}}+{\displaystyle \frac{P^2}{2m}}-{\displaystyle \frac{m{\omega }^2}{2}}{X}^2-iPX\omega \right]+x\{m{\omega }^{2}X+iP\omega \},$

where we suppressed the argument $t$ of $X$ and $P$, and used a dot to denote time derivatives. Comparing the real and imaginary parts in the curly brackets we find

$$\dot{P}=-m{\omega }^{2}X,\dot{X}=P/m,$$

(390)

which are just the classical equations of motion for the oscillator. The solution for initial conditions $X(0)=X_0$, $P(0)=0$ is given by

$$X(t)=X_0\cos (\omega t),P(t)=-m\omega X_0\sin (\omega t).$$

(391)

We now can obtain $S$ by comparing the bracketed terms in Eq. (XVIII.5), which demands

$$\dot{S}=-\frac{\mathrm{\hslash }\omega +\dot{X}P+\dot{P}X}{2}\Rightarrow S(t)=-\frac{\mathrm{\hslash }\omega t+P(t)X(t)}{2}.$$

(392)

The time-dependent solution $\mathrm{\Psi }(x,y)$ follows by inserting these expressions in Eq. (388). This allows to determine the associated expectation values $⟨x⟩=X(t)$ of position and $⟨p⟩=P(t)$ of momentum, which therefore follow the classical motion (this is a consequence of the Ehrenfest theorem, which establishes a similar relation for arbitrary quantum systems). Furthermore, the uncertainties $\mathrm{\Delta }x=\sigma$ and $\mathrm{\Delta }p=\mathrm{\hslash }/2\sigma$ are time-independent. Therefore, the motion of this wave packet mimics the classical motion as closely as it is possible under the constraints of the uncertainty principle.

The expectation value of energy

	$⟨\widehat{H}⟩$	$=$	$\mathrm{\hslash }\omega /2+P^2(t)/2m+m{\omega }^2{X}^2(t)/2$		(393)
		$=$	$\mathrm{\hslash }\omega /2+m{\omega }^2{X}_0^2/2$		(394)

is also time-independent, but this is simply a consequence of energy conservation in stationary problems. Compared to the classical expression, we encounter an additional positive contribution $\mathrm{\hslash }\omega /2$, the ground state energy, which we associated with the zero-point motion enforced by the uncertainty principle.

A large range of quantum problems only involves a pair of quantum states, either exactly as for the spin of an electron, or approximately because other states are energetically unaccessible and therefore can be neglected. Examples of the latter situation are the dynamics in the ground and first excited state of an atom at low energies, or the low-energy dynamics of a particle in a symmetric double-well potential. In all these cases, we can reduce the wave function to a two-component vector $\psi =\left(\genfrac{}{}{0pt}{}{\alpha }{\beta }\right)$, where the two components determine the probabilities $P_1={|\alpha |}^2$, $P_2={|\beta |}^2$ to find the system in state $\left(\genfrac{}{}{0pt}{}{1}{0}\right)$ or $\left(\genfrac{}{}{0pt}{}{0}{1}\right)$, respectively.

We remind ourselves of the 3-dimensional real polarisation vector

	$\overrightarrow{P}$	$=$	$(⟨{\sigma }_x⟩,⟨{\sigma }_y⟩,⟨{\sigma }_z⟩)$		(395)
		$=$	$(2\mathrm{Re}{\alpha }^{*}\beta ,2\mathrm{Im}{\alpha }^{*}\beta ,{|\alpha |}^2-{|\beta |}^2),$		(396)

where ${\sigma }_x$, ${\sigma }_y$, and ${\sigma }_z$ are the three Pauli matrices which we introduced for the description of the electronic spin. If $\psi$ is normalised then the polarisation vector is of unit length, i.e., it is restricted to the surface of a sphere, the so-called Bloch sphere.

Linear operators $\widehat{A}$ acting on two-component states $\psi$ are of the form of a $2\times 2$-dimensional matrix. These can be expressed as a sum

$$\widehat{A}=a_{0}I+a_x{\sigma }_x+a_y{\sigma }_y+a_z{\sigma }_z,$$

(397)

where $I$ is the $2\times 2$-dimensional identity matrix. For hermitian operators, the coefficients $a_{0,x,y,z}$ are all real. Furthermore, any unitary operator can also be written formally as

	$\widehat{U}$	$=$	$\exp [-i\phi (n_x{\sigma }_x+n_y{\sigma }_x+n_z{\sigma }_x)/2]e^{i\chi }$
		$=$	$\cos (\phi /2)e^{i\chi }I-i\sin (\phi /2)e^{i\chi }(n_x{\sigma }_x+n_y{\sigma }_y+n_z{\sigma }_z),$

where we assume $n_x^2+n_y^2+n_z^2=1$. (The expression on the second line follows directly from the Taylor-series definition of the exponential function.) By examining the polarisation vector of the state $\widehat{U}\psi$ one finds that $\widehat{U}$ induces a rigid rotation of the Bloch sphere, by an angle $\phi$ about the axis $𝐧=n_x𝐢+n_y𝐣+n_z𝐤$.

We now use these general features of two-component vectors and matrices to examine the dynamics of a two-state system, based on a time-independent Hamiltonian of the form

$$H=a_x{\sigma }_x+a_y{\sigma }_y+a_z{\sigma }_z$$

(400)

where the three coefficients are all real and time-independent. (We set $a_0=0$ since this simply shifts the energy. For an electronic spin, this Hamiltonian can be realised by applying a magnetic field of suitable strength $B$ into the direction $𝐧$ of the vector $𝐚=a_x𝐢+a_y𝐣+a_z𝐤=a𝐧$, which then is given by $𝐚=\frac{g_{e}e\mathrm{\hslash }}{4m_e}𝐁$ where $g_e\approx 2$ is the g-factor.) A general solution of this problem is provided by the time-evolution operator

	$\widehat{U}(t)$	$=$	$\exp (-i\widehat{H}t/\mathrm{\hslash })$		(401)
		$=$	$\exp [-i\omega t(n_x{\sigma }_x+n_y{\sigma }_y+n_z{\sigma }_z)/2],$		(402)

where $\omega =(2a/\mathrm{\hslash })$ is known as the Larmor frequency. (For an electronic spin in a magnetic field, $\omega =\frac{g_{e}eB}{2m_e}$.) Being unitary, the time-evolution operator can be cast into the form Eq. (XVIII.6), which here simply amounts to equating $\phi =\omega t$ and $\chi =0$, while $𝐧$ remains fixed. Therefore, the dynamics corresponds to a permanent rotation of the Bloch sphere about the axis $𝐧$, with angular frequency $\omega$. This motion, known as Larmor precession, is analogous to the precession of a rotating top under the influence of gravity.

In order to get further insight into the dynamics (and illustrate the solution method based on the eigenstates of the stationary Schrödinger equation), we now specialise to the case of a Hamiltonian

(403)

where we have expressed the coefficient $a_x$ in terms of a suitable constant $\mathrm{\Delta }$ whose meaning will become clear shortly. For a spin, this Hamiltonian can be realised by applying a magnetic field into the $x$ direction. For the double well, $\mathrm{\Delta }$ is associated with the tunneling between the troughs.

The stationary Schrödinger equation $E\psi =H\psi$ is solved by the eigenvectors ${\psi }_{+}=2^{-1/2}\left(\genfrac{}{}{0pt}{}{1}{1}\right)$ (the symmetric state, with energy $-\mathrm{\Delta }/2$) and ${\psi }_{-}=2^{-1/2}\left(\genfrac{}{}{0pt}{}{1}{-1}\right)$ (the antisymmetric state, with energy $\mathrm{\Delta }/2$). Therefore, $\mathrm{\Delta }$ amounts to the energy difference between the two states (in the context of the double well, this is also known as the tunnel splitting).

We now can use Eq. (359) to construct the time-dependent state

$$\mathrm{\Psi }(t)=a_{+}\exp (i\mathrm{\Delta }t/2\mathrm{\hslash }){\psi }_{+}+a_{-}\exp (-i\mathrm{\Delta }t/2\mathrm{\hslash }){\psi }_{-},$$

(404)

where $a_{-}$ and $a_{+}$ are determined by the initial conditions $\psi (0)$. The state is normalised for ${|a_{+}|}^2+{|a_{-}|}^2=1$.

Let us assume that the system is initially in the state $\psi (0)=\left(\genfrac{}{}{0pt}{}{1}{0}\right)$. Then the solution of the Schrödinger equation is

$$\psi (t)=\left(\genfrac{}{}{0pt}{}{\cos (\mathrm{\Delta }t/2\mathrm{\hslash })}{i\sin (\mathrm{\Delta }t/2\mathrm{\hslash })}\right).$$

(405)

This gives the probability $P_1={\cos }^2(\mathrm{\Delta }t/2\mathrm{\hslash })=\frac{1}{2}[1+\cos (\mathrm{\Delta }t/\mathrm{\hslash })]$ that the system occupies the state $\left(\genfrac{}{}{0pt}{}{1}{0}\right)$ and $P_2={\sin }^2(\mathrm{\Delta }t/2\mathrm{\hslash })=\frac{1}{2}[1-\cos (\mathrm{\Delta }t/\mathrm{\hslash })]$ that the system occupies the state $\left(\genfrac{}{}{0pt}{}{0}{1}\right)$.

Hence the quantum system oscillates between the states $\left(\genfrac{}{}{0pt}{}{1}{0}\right)$ and $\left(\genfrac{}{}{0pt}{}{0}{1}\right)$. The oscillation is periodic, $P_1(t+T)=P_1(t)$, with period $T=2\pi \mathrm{\hslash }/\mathrm{\Delta }$ (corresponding to Larmor frequency $\omega =\mathrm{\Delta }/\mathrm{\hslash }$). If we wait for half a period, $t=T/2$, the state changes from $\left(\genfrac{}{}{0pt}{}{1}{0}\right)$ to $\left(\genfrac{}{}{0pt}{}{0}{1}\right)$.

For spin, these oscillations are again interpreted as a precession in the magnetic field. For the double well, the oscillations correspond to tunnelling back and forth between the two troughs.