Fast and stable manipulation of a charged particle in a Penning trap

We consider a spinless charged particle as spin will not play any role in the trap configuration considered below. The Schrödinger equation for this particle in an electromagnetic field is given by

$$\begin{eqnarray}&&{\rm i}\hbar \frac{\partial }{\partial t}\Psi =H(t)\Psi ,\end{eqnarray} \tag{ 1 }$$

where the Hamiltonian (expressed in some chosen gauge e.g. the Coulomb gauge) is given in coordinate representation by

$$\begin{eqnarray}\begin{array}{rcl} H(t) & = & \frac{1}{2\;m}{{\left( \frac{\hbar }{{\rm i}}\nabla -q\vec{A}(t,\vec{r}) \right)}^{2}} \\ {} & {} & +q\phi (t,\vec{r}), \\ \end{array}\end{eqnarray} \tag{ 2 }$$

with q being the charge, $\vec{A}$ the vector potential and ϕ the scalar potential (both real). We write Ψ as

$$\begin{eqnarray}&&\Psi (t,\vec{r})=\alpha (t,\vec{r}){{{\rm e}}^{{\rm i}\beta (t,\vec{r})}},\end{eqnarray} \tag{ 3 }$$

where $\alpha (t,\vec{r}),\beta (t,\vec{r})\in \mathbb{R}$. Note that $\Psi (t,\vec{r})$ corresponds to the fast-forwarded state ${{\Psi }_{{\rm FF}}}$ in [27]. $\vec{A}$ and $q\phi$ correspond to the driving potentials ${{\vec{A}}_{{\rm FF}}}$ and V_FF in [27].

Inserting the ansatz (3) into (1) and then multiplying the equation by ${{{\rm e}}^{-{\rm i}\beta (t,\vec{r})}}$, we get for the real part of the result

$$\begin{eqnarray}&&0=-\frac{{{\hbar }^{2}}}{2\;m}\Delta \alpha +\frac{1}{2\;m}{{\left( q\vec{A}-\hbar \nabla \beta \right)}^{2}}\alpha +\left( q\phi +\hbar \frac{\partial \beta }{\partial t} \right)\alpha ,\end{eqnarray} \tag{ 4 }$$

and for the imaginary part

$$\begin{eqnarray}&&\hbar \frac{\partial \alpha }{\partial t}=\frac{\hbar }{2\;m}\nabla \left( q\vec{A}-\hbar \nabla \beta \right)\alpha +\frac{\hbar }{m}\left( q\vec{A}-\hbar \nabla \beta \right)\nabla \alpha .\end{eqnarray} \tag{ 5 }$$

To write these two equations in a more compact way, let us define

$$\begin{eqnarray}&&\vec{\chi }=\vec{A}-\frac{\hbar }{q}\nabla \beta ,\ \Phi =\phi +\frac{\hbar }{q}\frac{\partial \beta }{\partial t}.\end{eqnarray} \tag{ 6 }$$

The electric and the magnetic fields are now given by

$$\begin{eqnarray}&&\vec{E}=-\nabla \Phi -\frac{\partial \vec{\chi }}{\partial t},\ \vec{B}=\nabla \times \vec{\chi }.\end{eqnarray} \tag{ 7 }$$

Using these definitions of $\vec{\chi }$ and Φ, the two equations (4) and (5) simplify to

$$\begin{eqnarray}&&\Phi =\frac{{{\hbar }^{2}}}{2\;mq\alpha }\Delta \alpha -\frac{q}{2\;m}{{\vec{\chi }}^{2}},\end{eqnarray} \tag{ 8 }$$

and

$$\begin{eqnarray}&&\frac{\partial \alpha }{\partial t}-\frac{q}{2\;m}(\nabla \vec{\chi })\alpha -\frac{q}{m}\vec{\chi }\nabla \alpha =0.\end{eqnarray} \tag{ 9 }$$

These are the two main equations.

Note that $\vec{\chi }$ and Φ as well as the main equations (8) and (9) are invariant under a gauge transformation Λ acting in the usual way

$$\begin{eqnarray}\begin{array}{rcl} {\vec{A}} & \to & \vec{A}+\nabla \Lambda ,\ \phi \to \phi -\frac{\partial \Lambda }{\partial t}\ {\rm and} \\ \Psi & \to & {{{\rm e}}^{\frac{{\rm i}}{\hbar }q\Lambda }}\Psi , \\ \end{array}\end{eqnarray} \tag{ 10 }$$

i.e., $\beta \to \beta +\frac{q}{\hbar }\Lambda$ and α is unchanged.

Let the initial state of the system ${{\psi }_{0}}(\vec{r})\equiv {{\alpha }_{0}}(\vec{r}){{{\rm e}}^{{\rm i}{{\beta }_{0}}(\vec{r})}}$ be an eigenstate of the initial time independent Hamiltonian

$$\begin{eqnarray}\begin{array}{rcl} {{H}_{0}} & = & \frac{1}{2\;m}{{\left( \frac{\hbar }{{\rm i}}\nabla -q{{{\vec{A}}}_{0}} \right)}^{2}} \\ {} & {} & +q{{\phi }_{0}}, \\ \end{array}\end{eqnarray} \tag{ 11 }$$

with eigenvalue ${{\mathcal{E}}_{0}}$. (The eigenstates of the Penning trap are reviewed in the following section.) The goal is to design a scheme ($\vec{A}(t,\vec{r})$ and $\phi (t,\vec{r})$) such that the final state (at t = T) of the system, ${{\psi }_{T}}(\vec{r})\equiv {{\alpha }_{T}}(\vec{r}){{{\rm e}}^{{\rm i}{{\beta }_{T}}(\vec{r})}}$, is an eigenstate of the final (time independent) Hamiltonian

$$\begin{eqnarray}\begin{array}{rcl} {{H}_{T}} & = & \frac{1}{2\;m}{{\left( \frac{\hbar }{{\rm i}}\nabla -q{{{\vec{A}}}_{T}} \right)}^{2}} \\ {} & {} & +q{{\phi }_{T}}, \\ \end{array}\end{eqnarray} \tag{ 12 }$$

with eigenvalue ${{\mathcal{E}}_{T}}$. The Hamiltonian should be continuous at initial and final time, i.e., $H(0)={{H}_{0}}$ and $H(T)={{H}_{T}}$.

In the inversion protocol we first design $\alpha (t,\vec{r})$ and $\beta (t,\vec{r})$ fulfilling the boundary conditions

$$\begin{eqnarray}\begin{array}{rcl} \alpha (0,\vec{r}) & = & {{\alpha }_{0}}(\vec{r}),\ \alpha (T,\vec{r})={{\alpha }_{T}}(\vec{r}), \\ \beta (0,\vec{r}) & = & {{\beta }_{0}}(\vec{r}),\ \beta (T,\vec{r})={{\beta }_{T}}(\vec{r}). \\ \end{array}\end{eqnarray} \tag{ 13 }$$

In the next step, we have to solve for $\vec{\chi }$ in (9). The function Φ is then given by (8). Because the Hamiltonian should be changing continuously at initial and final time, $\vec{\chi }$ must fulfil the following boundary conditions

$$\begin{eqnarray}\begin{array}{rcl} \vec{\chi }(0,\vec{r}) & = & {{{\vec{A}}}_{0}}-\frac{\hbar }{q}\nabla {{\beta }_{0}}, \\ \vec{\chi }(T,\vec{r}) & = & {{{\vec{A}}}_{T}}-\frac{\hbar }{q}\nabla {{\beta }_{T}}. \\ \end{array}\end{eqnarray} \tag{ 14 }$$

A consequence of these conditions can be seen by evaluating (9) at the initial and final time (see also appendix A). This leads to

$$\begin{eqnarray}&&\begin{array}{l} \frac{\partial \alpha }{\partial t}(0,\vec{r})=0, \\ \frac{\partial \alpha }{\partial t}(T,\vec{r})=0. \\ \end{array}\end{eqnarray} \tag{ 15 }$$

The boundary conditions of Φ can be seen by evaluating (8) at initial and final time leading to (see also appendix A)

$$\begin{eqnarray}\begin{array}{rcl} \Phi (0,\vec{r}) & = & {{\phi }_{0}}(\vec{r})-\frac{1}{q}{{\mathcal{E}}_{0}} \\ {} & = & {{\phi }_{0}}(\vec{r})+\frac{\hbar }{q}\frac{\partial \beta }{\partial t}(0,\vec{r}), \\ \Phi (T,\vec{r}) & = & {{\phi }_{0}}(\vec{r})-\frac{1}{q}{{\mathcal{E}}_{T}} \\ {} & = & {{\phi }_{0}}(\vec{r})+\frac{\hbar }{q}\frac{\partial \beta }{\partial t}(T,\vec{r}). \\ \end{array}\end{eqnarray} \tag{ 16 }$$

These conditions are equivalent to

$$\begin{eqnarray}\begin{array}{ll} {} & {} & \frac{\partial \beta }{\partial t}(0,\vec{r})=-\frac{1}{\hbar }{{\mathcal{E}}_{0}}, \\ {} & {} & \frac{\partial \beta }{\partial t}(T,\vec{r})=-\frac{1}{\hbar }{{\mathcal{E}}_{T}}. \\ \end{array}\end{eqnarray} \tag{ 17 }$$

Finally, the vector potential and the scalar potential in the chosen gauge are then given by

$$\begin{eqnarray}&&\vec{A}(t,\vec{r})=\vec{\chi }(t,\vec{r})+\frac{\hbar }{q}\nabla \beta (t,\vec{r}),\end{eqnarray} \tag{ 18 }$$

$$\begin{eqnarray}&&\phi (t,\vec{r})=\Phi (t,\vec{r})-\frac{\hbar }{q}\frac{\partial \beta }{\partial t},\end{eqnarray} \tag{ 19 }$$

and the electric and magnetic fields are given by (7).

The above boundary conditions guarantee that the magnetic field is continuous at the initial and final time. To make the electric field continuous at the initial and final time, we also impose

$$\begin{eqnarray}&&\frac{\partial \vec{\chi }}{\partial t}(0,\vec{r})=0,\ \frac{\partial \vec{\chi }}{\partial t}(T,\vec{r})=0.\end{eqnarray} \tag{ 20 }$$

Following the algorithm presented in section 2, we start by choosing the following ansatz for the time evolution of the wavefunction

$$\begin{eqnarray}\begin{array}{rcl} \alpha (t,r) & = & \sqrt{\frac{N!}{(N+\left| M \right|)!}}\frac{1}{\sqrt{2\pi }l(t)}{{\left( \frac{r}{\sqrt{2}l(t)} \right)}^{\left| M \right|}} \\ {} & {} & \times {\rm exp} \left( -\frac{{{r}^{2}}}{4\;l{{(t)}^{2}}} \right)L_{N}^{\left| M \right|}\left( \frac{{{r}^{2}}}{2\;l{{(t)}^{2}}} \right), \\ \end{array}\end{eqnarray} \tag{ 30 }$$

and $\beta (t,\theta )=M\theta +\zeta (t)$. For the boundary conditions (13), it follows $\alpha (0,r)={{f}_{N,M,{{l}_{0}}}}(r)$ and $\alpha (T,r)={{f}_{N,M,{{l}_{T}}}}(r)$ and so we get the condition $l(0)={{l}_{0}}$, $l(T)={{l}_{T}}$. Moreover, we get $\zeta (0)=\zeta (T)=0$.

As the next step, we have to solve the main equation (9). We assume that $\vec{\chi }$ does not depend on θ, i.e. $\vec{\chi }={{\chi }_{r}}(t,r)\hat{r}+{{\chi }_{\theta }}(t,r)\hat{\theta }$. (9) then becomes

$$\begin{eqnarray}\begin{array}{ll} {} & {} & \frac{2mr}{q}\frac{\partial \alpha }{\partial t}-{{\chi }_{r}}\left( a+2r\frac{\partial \alpha }{\partial r} \right) \\ {} & {} & \quad -r\frac{\partial {{\chi }_{r}}}{\partial r}\alpha =0, \\ \end{array}\end{eqnarray} \tag{ 31 }$$

and (8) becomes

$$\begin{eqnarray}\begin{array}{rcl} \Phi & = & -\frac{q}{2\;m}\left( \chi _{r}^{2}+\chi _{\theta }^{2} \right) \\ {} & {} & +\frac{{{\hbar }^{2}}}{2\;mq\alpha }\left( \frac{1}{r}\frac{\partial \alpha }{\partial r}+\frac{{{\partial }^{2}}\alpha }{\partial {{r}^{2}}} \right). \\ \end{array}\end{eqnarray} \tag{ 32 }$$

A solution of (9) is given by

$$\begin{eqnarray}&&{{\chi }_{r}}(t,r)=-\frac{2m}{q}\frac{1}{r\;{{\alpha }^{2}}}\int _{r}^{\infty }{\rm d}s\;s\;\alpha \frac{\partial \alpha }{\partial t}(t,s).\end{eqnarray} \tag{ 33 }$$

The solution when α and $\vec{\chi }$ depend on θ can be found in appendix B.

For α given by (30), we get from (33) that

$$\begin{eqnarray}&&{{\chi }_{r}}(t,r)=-\frac{m}{q}\frac{rl^{\prime} (t)}{l(t)},\end{eqnarray} \tag{ 34 }$$

where the prime indicates a derivative with respect to time. Note that this solution is independent of the quantum numbers N and M.

The components of the physical fields can be written as

$$\begin{eqnarray}&&{{B}_{z}}(t,r)=\frac{1}{r}\frac{\partial (r{{\chi }_{\theta }})}{\partial r},\end{eqnarray} \tag{ 35 }$$

$$\begin{eqnarray}\begin{array}{rcl} {{E}_{r}}(t,r) & = & -\frac{\partial }{\partial r}\Phi -\frac{\partial {{\chi }_{r}}}{\partial t}, \\ {} & {} & {{E}_{\theta }}(t,r)=-\frac{\partial {{\chi }_{\theta }}}{\partial t}. \\ \end{array}\end{eqnarray} \tag{ 36 }$$

We want a uniform magnetic field and a constant radial electric field $\left( {{E}_{r}}=\frac{m\omega _{z}^{2}}{2q}r \right)$ during the whole process. To achieve a uniform magnetic field ${{B}_{z}}={{B}_{z}}(t)$ we set

$$\begin{eqnarray}&&{{\chi }_{\theta }}(t,r)=\frac{r}{2}{{B}_{z}}(t)+\frac{g(t)}{r},\end{eqnarray} \tag{ 37 }$$

with an arbitrary function g. The electric field component E_r is now

$$\begin{eqnarray}\begin{array}{rcl} {{E}_{r}} & = & \frac{{{M}^{2}}{{\hbar }^{2}}-{{q}^{2}}g{{(t)}^{2}}}{mq{{r}^{3}}} \\ {} & {} & +\frac{r}{4mq}\left( {{q}^{2}}B_{z}^{2}-\frac{{{\hbar }^{2}}}{l{{(t)}^{4}}}+\frac{4\;{{m}^{2}}\;l^{\prime\prime} (t)}{l(t)} \right). \\ \end{array}\end{eqnarray} \tag{ 38 }$$

The demand ${{E}_{r}}=\frac{m\omega _{z}^{2}}{2q}r$ leads to the choice $g(t)=-M\hbar /q$ and

$$\begin{eqnarray}&&{{B}_{z}}(t)=\frac{\sqrt{{{\hbar }^{2}}-4\;{{m}^{2}}\;l{{(t)}^{3}}\;l^{\prime\prime} (t)+2\;{{m}^{2}}\omega _{z}^{2}\;l{{(t)}^{4}}}}{ql{{(t)}^{2}}},\end{eqnarray} \tag{ 39 }$$

where $q{{B}_{z}}(t)\gt 0$ is assumed. The electric field components are finally

$$\begin{eqnarray}&&{{E}_{r}}=\frac{m\omega _{z}^{2}}{2q}r,\ {{E}_{\theta }}=-\frac{r}{2}B_{z}^{\prime }(t).\end{eqnarray} \tag{ 40 }$$

Now, we have

$$\begin{eqnarray}&&\chi (t,r)=\left[ \frac{r}{2}{{B}_{z}}(t)-\frac{M\hbar }{qr} \right]\hat{\theta }-\frac{m}{q}\frac{rl^{\prime} (t)}{l(t)}\hat{r}.\end{eqnarray} \tag{ 41 }$$

Following from (14), we get the boundary conditions for $\vec{\chi }$

$$\begin{eqnarray}\begin{array}{rcl} \vec{\chi }(0,r) & = & {{{\vec{A}}}_{0}}-\frac{\hbar }{q}\nabla {{\beta }_{0}}=\left( \frac{r{{B}_{0}}}{2}-\frac{\hbar M}{qr} \right)\hat{\theta }, \\ \vec{\chi }(T,r) & = & {{{\vec{A}}}_{T}}-\frac{\hbar }{q}\nabla {{\beta }_{T}}=\left( \frac{r{{B}_{T}}}{2}-\frac{\hbar M}{qr} \right)\hat{\theta }. \\ \end{array}\end{eqnarray} \tag{ 42 }$$

With the χ given in (41), this is fulfilled if $l^{\prime} (0)=0$, $l^{\prime} (T)=0$, ${{B}_{z}}(0)={{B}_{0}}$, and ${{B}_{z}}(T)={{B}_{T}}$. To fulfil the last two conditions, we have to demand $l^{\prime\prime} (0)=0$, $l^{\prime\prime} (T)=0$.

The boundary conditions of Φ are fulfilled if the conditions (17) are satisfied, i.e. if

$$\begin{eqnarray}&&\frac{\partial \zeta }{\partial t}(0)=-\frac{1}{\hbar }{{\mathcal{E}}_{0}},\ \frac{\partial \zeta }{\partial t}(T)=-\frac{1}{\hbar }{{\mathcal{E}}_{T}}.\end{eqnarray} \tag{ 43 }$$

A simple choice of ζ may be a polynomial of degree 3 that obeys all of the boundary conditions on ζ. Note that the magnetic and the electric fields do not depend on the choice of the time dependent global phase $\zeta (t)$.

An additional boundary condition on l(t) can be derived by enforcing that the electric field is continuous at t = 0 and t = T, see the conditions (20). This requires $B_{z}^{\prime }(0)=0$ and $B_{z}^{\prime }(T)=0$. Differentiating the expression (39) for B_z(t) with respect to time gives

$$\begin{eqnarray}&&B_{z}^{\prime }(t)=-\frac{\left( 2\left( l^{\prime} (t)\left[ {{\hbar }^{2}}-{{m}^{2}}\;l{{(t)}^{3}}\;{{l}^{\prime\prime }}(t) \right] \right.+\left. {{m}^{2}}\;l{{(t)}^{4}}\;l(t) \right) \right)}{q{{l}^{3}}(t)\sqrt{\left( {{\hbar }^{2}}-4\;{{m}^{2}}\;l{{(t)}^{3}}\;{{l}^{\prime\prime }}(t)+2\;{{m}^{2}}\omega _{z}^{2}\;l{{(t)}^{4}} \right)}}.\end{eqnarray} \tag{ 44 }$$

Noting the boundary conditions on l already derived, this requires, in addition, that $l(0)=0$, $l(T)=0$.

In summary, the boundary conditions for l(t) are

$$\begin{eqnarray}\begin{array}{rcl} l(0) & = & {{l}_{0}}=\sqrt{\frac{\hbar }{2\;m\tilde{\omega }\left( 0 \right)}},\ \\ l^{\prime} (0) & = & l^{\prime\prime} (0)=l(0)=0, \\ l(T) & = & {{l}_{T}}=\sqrt{\frac{\hbar }{2\;m\tilde{\omega }\left( T \right)}}, \\ l^{\prime} (T) & = & l^{\prime\prime} (T)=l(T)=0. \\ \end{array}\end{eqnarray} \tag{ 45 }$$

These conditions are independent of the quantum numbers N and M.

If l(t) satisfies these boundary conditions, the corresponding magnetic field and electric field are given by equations (39) and (40), and fulfil $\nabla \cdot \vec{E}=0$ and $\nabla \times \vec{B}=0$. They are also independent of the quantum numbers N and M. If the system starts in the corresponding eigenstate and if these fields are implemented, then the system will end with fidelity 1 in the final state. We will show that these schemes which do not change the quantum number could be alternatively derived using an invariant based approach.

A Lewis–Riesenfeld invariant is a Hermitian operator I(t) fulfilling

$$\begin{eqnarray}&&\frac{\partial }{\partial t}I(t)=\frac{{\rm i}}{\hbar }{{\left[ I(t),H(t) \right]}_{-}},\end{eqnarray} \tag{ 46 }$$

where H(t) is the Hamiltonian for the system. If we disregard again the z dependent part, we find the following invariant for the Hamiltonian (24),

$$\begin{eqnarray}\begin{array}{rcl} I(t) & = & -\frac{l{{(t)}^{2}}}{{{\hbar }^{2}}}\Delta -2\;ml^{\prime} (t)l(t)\left( \frac{\hbar }{{\rm i}}\frac{\partial }{\partial r} \right)r \\ {} & {} & +\left( {{m}^{2}}\;l^{\prime} {{(t)}^{2}}+\frac{{{\hbar }^{2}}}{4\;l{{(t)}^{2}}} \right){{r}^{2}}, \\ \end{array}\end{eqnarray} \tag{ 47 }$$

where the function l(t) has to be a solution of the following Ermakov-like equation

$$\begin{eqnarray}&&4\;{{m}^{2}}\frac{l^{\prime\prime} (t)}{l(t)}+4\;{{m}^{2}}\tilde{\omega }{{(t)}^{2}}-\frac{{{\hbar }^{2}}}{l{{(t)}^{4}}}=0.\end{eqnarray} \tag{ 48 }$$

(In [28], the case $\tilde{\omega }=\omega$ was examined and an invariant was constructed. Eigenstates of this invariant which are simultaneously eigenstates of L_z were also constructed indirectly.) An explicit expression of the eigenstates of I is

$$\begin{eqnarray}\begin{array}{rcl} {{\Gamma }_{N,M}}(t,r,\theta ) & = & \frac{1}{\sqrt{2\pi }}\sqrt{\frac{N!}{(N+|M|)!}}\frac{1}{l(t)} \\ {} & {} & \times {{\left[ \frac{r}{\sqrt{2}l(t)} \right]}^{|M|}}L_{N}^{|M|}\left( \frac{{{r}^{2}}}{2\;l{{(t)}^{2}}} \right) \\ {} & {} & \times {\rm exp} \left( -\frac{{{r}^{2}}}{4\;l{{(t)}^{2}}}+\frac{{\rm i}ml^{\prime} (t)}{2\hbar l(t)}{{r}^{2}}+iM\theta \right), \\ \end{array}\end{eqnarray} \tag{ 49 }$$

where $N,M\in \mathbb{Z}$ and l(t) is a solution of (48). The corresponding eigenvalue of I is $(2N+|M|+1){{\hbar }^{2}}$ and the corresponding eigenvalue of L_z is $M\hbar$. A more general invariant (which allows for a time dependent mass) was described in [30]. An eigenstate of a Lewis–Riesenfeld invariant is a solution to the Schrödinger equation for H(t) up to a time dependent phase ${{\Pi }_{N,M}}(t)$ [28], which is here given by

$$\begin{eqnarray}&&{{\Pi }_{N,M}}(t)=\int _{0}^{t}{\rm d}t^{\prime} \left( M\omega (t^{\prime} )-\frac{(N+1)\hbar }{2\;ml{{(t^{\prime} )}^{2}}} \right).\end{eqnarray} \tag{ 50 }$$

The idea is to do inverse-engineering by demanding that the system follows the state $\Psi (t,r,\theta )={{\Gamma }_{N,M}}(t,r,\theta ){{{\rm e}}^{{\rm i}{{\Pi }_{N,M}}(t)}}$. First, we choose an auxiliary function l(t) (which has to fulfil different conditions at initial and final time, see below) and then we get $\tilde{\omega }(t)$ from (48). At initial and final time the eigenstates of the invariant should coincide with the eigenstates of the Hamiltonian, i.e. ${{[I(0),H(0)]}_{-}}=0={{[I(T),H(T)]}_{-}}$. Therefore, we have to impose the following boundary conditions for the auxiliary function l(t):

$$\begin{eqnarray}\begin{array}{rcl} l(0) & = & {{l}_{0}}=\sqrt{\frac{\hbar }{2\;m\tilde{\omega }\left( 0 \right)}},\ l^{\prime} (0)=0, \\ l(T) & = & {{l}_{T}}=\sqrt{\frac{\hbar }{2\;m\tilde{\omega }\left( T \right)}},\ l^{\prime} (T)=0. \\ \end{array}\end{eqnarray} \tag{ 51 }$$

From these boundary conditions and (48), it also follows that $l^{\prime\prime} (0)=0$ and $l^{\prime\prime} (T)=0$.

An additional boundary condition on l(t) can be derived by enforcing that the electric field is continuous at t = 0 and t = T. This leads to $B_{z}^{\prime }(0)=B_{z}^{\prime }(T)=0$ or $l(0)=l(T)=0$. The complete list of boundary conditions is equivalent to (45) above. As already mentioned, we can now design first an auxiliary function l(t) fulfilling the above boundary conditions and then calculate $\tilde{\omega }(t)$ and hence the magnetic field strength B_z(t) from (48). The resulting formula for B_z(t) is the same as (39) above.

Summarizing, the streamlined fast-forward formalism and the invariant based approach provide two ways to find the same boundary conditions for the auxiliary function l(t) in this setting. So, for varying the magnetic field strength, both formalisms are equivalent as they both require that one chooses the auxiliary function l(t) fulfilling these boundary conditions and then the corresponding physical potentials can be calculated in the same way. In the following we look at a numerical example of this procedure.

Let us first set $l(t)={{l}_{0}}\lambda (\tau )$ where $\tau =t/T$. From the above formalism, the time dependence of $\tilde{\omega }$ follows as

$$\begin{eqnarray}&&\tilde{\omega }(t)=\tilde{\omega }(0)\frac{1}{\lambda {{(\tau )}^{2}}}\sqrt{1-\frac{\lambda {{(\tau )}^{3}}\lambda ^{\prime\prime} (\tau )}{{{\mu }^{2}}}},\end{eqnarray} \tag{ 52 }$$

where $\mu =T\tilde{\omega }(0)$. μ can be seen as the final time in units of $1/\tilde{\omega }(0)$. Therefore, decreasing μ corresponds to decreasing the total time T of the process, with fixed $\tilde{\omega }(0)$ (i.e. fixed ${{\omega }_{0}}=\frac{q{{B}_{z}}(0)}{2\;m}\gt 0$ and fixed ${{\omega }_{z}}$). The limit $\mu \to \infty$ would correspond to the adiabatic limit where we get $\tilde{\omega }(t)/\tilde{\omega }(0)\to \frac{1}{{{\lambda }^{2}}(\tau )}$.

The corresponding magnetic field would then be given by (39) or in dimensionless variables

$$\begin{eqnarray}\begin{array}{rcl} {{B}_{z}}(t) & = & \frac{\hbar }{ql_{0}^{2}}\frac{1}{\lambda {{(\tau )}^{2}}}\left[ 1-\frac{\lambda {{(\tau )}^{3}}\lambda ^{\prime\prime} (\tau )}{{{\mu }^{2}}} \right. \\ {} & {} & +{{\left. \frac{{{\nu }^{2}}}{2-{{\nu }^{2}}}\lambda {{(\tau )}^{4}} \right]}^{1/2}}, \\ \end{array}\end{eqnarray} \tag{ 53 }$$

where $\nu =\frac{{{\omega }_{z}}}{{{\omega }_{0}}}$ is the ratio between the two initial frequencies. This parameter ν is independent of the total time T. We want to have a trap setting at initial and final time, i.e. $\tilde{\omega }{{(0)}^{2}}$ and $\tilde{\omega }{{(T)}^{2}}$ should be positive. From this, ν must be in the range $0\leqslant \nu \lt \sqrt{2}{\rm min} \{1,{{\omega }_{T}}/{{\omega }_{0}}\}$ where ${{\omega }_{T}}=\frac{q{{B}_{z}}(T)}{2\;m}\gt 0$.

Assuming a polynomial form of $\lambda (\tau )$ and using the above conditions, $\lambda (\tau )$ can be expressed as

$$\begin{eqnarray}\begin{array}{rcl} \lambda (\tau ) & = & 1-20\left( {{l}_{T}}/{{l}_{0}}-1 \right){{\tau }^{7}}+70\left( {{l}_{T}}/{{l}_{0}}-1 \right){{\tau }^{6}} \\ {} & {} & -84\left( {{l}_{T}}/{{l}_{0}}-1 \right){{\tau }^{5}}+35\left( {{l}_{T}}/{{l}_{0}}-1 \right){{\tau }^{4}}. \\ \end{array}\end{eqnarray} \tag{ 54 }$$

The final value of the magnetic field is chosen in this example such that $\tilde{\omega }(T)/\tilde{\omega }(0)=c=10$ (i.e. ${{l}_{T}}={{l}_{0}}/\sqrt{10}$). The ratio between initial and final magnetic field is then

$$\begin{eqnarray*}&&\frac{{{B}_{z}}(T)}{{{B}_{z}}(0)}=\sqrt{{{c}^{2}}\left( 1-\frac{{{\nu }^{2}}}{2} \right)+\frac{{{\nu }^{2}}}{2}}.\end{eqnarray*}$$

Figure 1 is the corresponding plot of l(t).

Zoom In Zoom Out Reset image size

Figure 2 shows $\tilde{\omega }(t)$ for different values of μ. For $\mu \approx 0.672$ (green, thick, solid line) $\tilde{\omega }{{(t)}^{2}}\gt 0$ is no longer fulfilled for all times. The requirement that ${{B}_{z}}(t)\in \mathbb{R}$ for all times results in a type of quantum speed limit of the form

$$\begin{eqnarray}&&\mu \geqslant \mathop{{\rm max} }\limits_{\tau \in [0,1]}{{\left[ \frac{\lambda {{(\tau )}^{3}}\lambda ^{\prime\prime} (\tau )}{1+{{\nu }^{2}}{{(2-{{\nu }^{2}})}^{-1}}\lambda {{(\tau )}^{4}}} \right]}^{1/2}}.\end{eqnarray} \tag{ 55 }$$

Recently the topic of quantum speed limits have gained renewed interest in the literature. For example recent work has been done on extending the usual Mandelstam–Tamm(MT) and Margolus–Levitin(ML) bounds for open system dynamics [31, 32]. It has also proven useful when using optimal control techniques [33]. The quantum speed limit of our paper is clearly more restrictive than the MT and ML bounds i.e. the minimal time of our speed limit is larger or equal than the time limit provided by those bounds.

Zoom In Zoom Out Reset image size

As an example, the wavefunction at different times with $N=M=0$ can be seen in figure 3. The shown time evolution is independent of $\mu ,\nu$ and depends only on the chosen form of l(t).

Zoom In Zoom Out Reset image size

The electric and magnetic fields derived in the previous subsection are independent of the quantum numbers N and M. Therefore, the fields can be also applied to a superposition of different eigenstates with initial magnetic field B₀ and they will produce a superposition of eigenstates with final magnetic field B_T with the same populations as initially. Let us assume an initial wavefunction of the form

$$\begin{eqnarray}&&\Psi (0,r,\theta )=\mathop{\sum }\limits_{N,M}{{c}_{N,M}}{{\Gamma }_{N,M}}(0,r,\theta ),\end{eqnarray} \tag{ 56 }$$

where ${{\Gamma }_{N,M}}(t,r,\theta )$ are the eigenfunctions of the invariant given in (49) and ${{c}_{N,M}}$ are (constant) complex coefficients. Then it follows that the state at final time will be

$$\begin{eqnarray}&&\Psi (T,r,\theta )=\mathop{\sum }\limits_{N,M}{{c}_{N,M}}{{{\rm e}}^{{\rm i}{{\Pi }_{N,M}}(T)}}{{\Gamma }_{N,M}}(T,r,\theta ),\end{eqnarray} \tag{ 57 }$$

where ${{\Pi }_{N,M}}$ is given in (50), so the populations in the different eigenstates will be the same as initially, i.e. ${{\left| {{c}_{N,M}}{{{\rm e}}^{{\rm i}{{\Pi }_{N,M(t)}}}} \right|}^{2}}={{\left| {{c}_{N,M}} \right|}^{2}}$.

It is important that the scheme is not only fast but also stable concerning errors in the implementation. We want to examine the stability of the protocol if there is a systematic error in the magnetic field B_z(t). We assume that the magnetic field is correctly implemented before and after the process, for $t\leqslant 0$ and $t\geqslant T$. Nevertheless, during the change of the magnetic field for for $0\lt t\lt T$, we assume that an inaccurate magnetic field ${{\mathcal{B}}_{\epsilon }}(t)={{B}_{z}}(t)(1+\epsilon )$ is implemented, where ${{B}_{z}}(t)$ is the correct one and a small relative systematic error which is unknown but constant.

We will examine the final fidelity as a function of for $N=M=0$. The initial state is still ${{\Psi }_{\epsilon }}(0)={{\psi }_{N=0,M=0,{{l}_{0}}}}$. The solution of the Schrödinger equation is then still given by ${{\Psi }_{\epsilon }}(t)={{\Gamma }_{0,0}}(t){{{\rm e}}^{{\rm i}{{\Pi }_{0,0}}(t)}}$ (see (49)) with l(t) replaced by ${{\ell }_{\epsilon }}(t)$, a solution of

$$\begin{eqnarray}&&4\;{{m}^{2}}\frac{{{\ell }_{\epsilon }}^{\prime\prime} (t)}{{{\ell }_{\epsilon }}(t)}+4\;{{m}^{2}}\left[ {{\left( \frac{q{{\mathcal{B}}_{\epsilon }}(t)}{2\;m} \right)}^{2}}-\frac{\omega _{z}^{2}}{2} \right]-\frac{{{\hbar }^{2}}}{{{\ell }_{\epsilon }}{{(t)}^{4}}}=0,\end{eqnarray} \tag{ 58 }$$

with ${{\ell }_{\epsilon }}(0)={{l}_{0}}$ and ${{\ell }_{\epsilon }}^{\prime} (0)=0$. The fidelity at t = T is now

$$\begin{eqnarray}\begin{array}{rcl} F & = & \left| \langle \Psi (T)|{{\Psi }_{\epsilon }}(T)\rangle \right| \\ {} & = & \frac{2\;l(T){{\ell }_{\epsilon }}(T)}{\sqrt{{{(l{{(T)}^{2}}+{{\ell }_{\epsilon }}{{(T)}^{2}})}^{2}}+\frac{4\;m}{{{\hbar }^{2}}}l{{(T)}^{4}}{{\ell }_{\epsilon }}{{(T)}^{2}}{{\ell }_{\epsilon }}^{\prime} {{(T)}^{2}}}}. \\ \end{array}\end{eqnarray} \tag{ 59 }$$

Let l(t) be given again as in (54). We once again fix $\tilde{\omega }(T)/\tilde{\omega }(0)=c=10$, noting that the magnetic field is assumed to be error-free at the initial and final time. With these values fixed, the fidelity F only depends on μ, ν and . Note ν must be in the range $0\leqslant \nu \lt \sqrt{2}$. The fidelity F for different combinations of μ and ν versus is shown in figure 4. One still gets a high fidelity even if there is a small, systematic error in the implementation of the magnetic field during the scheme. The scheme is, in some range, stable concerning this type of systematic error.

Zoom In Zoom Out Reset image size

A sensitivity S of the scheme versus this systematic error can be defined as the negative curvature of the fidelity at $\epsilon =0$, i.e. $S=-{{\left. \frac{{{\partial }^{2}}F}{\partial {{\epsilon }^{2}}} \right|}_{\epsilon =0}}$. This sensitivity S versus μ and ν is shown in figure 5. The sensitivity is increasing with increasing ratio ν for fixed μ. For fixed ν the sensitivity shows an oscillating behaviour with increasing μ. The sensitivity for arbitrary N and M is treated in appendix C.

Zoom In Zoom Out Reset image size

In order to have even more stability against systematic error in the magnetic field one could design a different l(t) which minimizes the sensitivity S and still fulfils the necessary boundary conditions (a similar strategy could also be applied to other types of systematic errors or random errors). As shown in appendix C, it would be sufficient to minimize S only for $N,M=0$.