Question 11.39: We can use the technique of Problem 11.38 to investigate tim...

We can use the technique of Problem 11.38 to investigate time evolution when the Hamiltonian does depend on time, as long as we choose Δt small enough. Evaluating H at the midpoint of each time step we simply replace Equation 11.154 with

\left(1+\frac{1}{2} i \frac{\Delta t}{\hbar} H \right) \Psi (t+\Delta t)=\left(1-\frac{1}{2} i \frac{\Delta t}{\hbar} H \right) \Psi (t)            (11.154).

\left[1+\frac{1}{2} i \frac{\Delta t}{\hbar} H \left(t+\frac{\Delta t}{2}\right)\right] \Psi(t+\Delta t)=\left[1-\frac{1}{2} i \frac{\Delta t}{\hbar} H \left(t+\frac{\Delta t}{2}\right)\right] \Psi(t)            (11.155).

Consider the driven harmonic oscillator of Problem 11.36 with

f(t)=A \sin (\Omega t)                (11.156).

where A is a constant with the units of length and Ω is the driving frequency.
In the following we will set m=\omega=\hbar=A=1 and look at the effect of varying Ω. Use the same parameters for the spatial discretization as in Problem 11.38, but set N_{t}=1000 . For a particle that starts off in the ground state at t = 0 , create a movie showing the numerical and exact solutions as well as the instantaneous ground state from t=0 \text { to } t=2 \pi / \Omega for

(a) \Omega=\omega / 5 . In line with the adiabatic theorem, you should see that the numerical solution is close (up to a phase) to the instantaneous ground state.

(b) \Omega=5\omega  . In line with what you’ve learned about sudden perturbations, you should see that the numerical solution is barely affected by the driving force.

(c) \Omega=6\omega /5  .

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

The exact solution for the driven oscillator which we’ll compare to

\operatorname{In}[1]:=\operatorname{Clear}[ m , \omega, \hbar, \Omega, A ] ;

\ln [2]=f\left[t_{-}\right]=A \operatorname{Sin}[\Omega t] ;

\operatorname{In}[3]=\operatorname{xc}\left[ t _{-}\right]=\text {Simplify }[\omega \text { Integrate }[ f [t p] \sin [\omega(t-t p)],\{t p, 0, t\}]] ;

\operatorname{In}[4]=\operatorname{psi}\left[n_{-}, x_{-}\right]:=\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} \frac{1}{\sqrt{2^{n} n !}} \text { HermiteH }\left[n, \sqrt{\frac{m \omega}{\hbar}} x\right] \operatorname{Exp}\left[-\frac{1}{2}\left(\sqrt{\frac{m \omega}{\hbar}} x\right)^{2}\right] ;

\ln [5]:=\Psi\left[n_{-}, x_{-}, t_{-}\right]= \text { Simplify }\left[\text { psi } [ n , x – x c [ t ] ] \operatorname { E x p } \left[\frac { I } { \hbar } \left(-\left(n+\frac{1}{2}\right) \hbar \omega t+m x c^{\prime}[t]\left(x-\frac{x c[t]}{2}\right)+\right.\right.\right.

\left.\left.\frac{m \omega^{2}}{2} \text { Integrate }[f[t p] \times c[t p],\{t p, 0, t\}]\right]\right] ;

\text { Numerics } \Omega=\omega / 5 .

Physical constants

\operatorname{In}[6]= m =1 ; \omega=1 ; \hbar=1 ; \Omega=.2 ; \quad A =1 ;

Numerical constants

\ln [7]=N x=100 ; \text { xmin }=-10 . ; \quad \operatorname{xmax}=10 . ; dx =\frac{x \max -x m i n}{N x+1} ;

Nt =1000 ; d t =\frac{2 \pi / \Omega}{ Nt } ;

Construct H.

The Identity Matrix

\ln [8]=\text { Openone }=\text { SparseArray }\left[\left\{i_{-}, i_{-}\right\} \rightarrow 1 .,\{N x, N x\}\right] ;

The second derivative operator

\operatorname{In}[9]=D 2=\frac{1.0}{d x^{2}} \text { SparseArray }\left[\left\{\left\{i_{-}, i_{-}\right\} \rightarrow-2 .,\left\{i_{-}, j_{-}\right\} / ; A b s[i-j]=1 \rightarrow 1 .\right\},\{N x, N x\}\right] ;

The position operator

\ln [10]=X=\text { SparseArray }\left[\left\{i_{-}, i_{-}\right\} \rightarrow x m i n+d x i,\{N x, N x\}\right] ;

The Hamiltonian

I n[11]:=H\left[t_{-}\right]=-\frac{\hbar^{2}}{2 m} D 2+\frac{1}{2} m \omega^{2} X \cdot X-m \omega^{2} X f[t] ;

Construct the time-evolution operator

The numerator in Cayley’s form

\operatorname{In}[12]=\text { Uplus }\left[t_{-}\right]=\text {Openone }+I H\left[t+\frac{d t}{2}\right] \frac{d t}{2} ;

The denominator in Cayley’s form

\ln [13]==\text { Uminus }\left[t_{-}\right]=\text {Openone }-I H\left[t+\frac{d t}{2}\right] \frac{d t}{2} ;

Solve the t-dep Schrödinger equation

\ln [14]==\operatorname{Psi0}=\text { Eigenvectors }[H[\theta]][[N x]] / \operatorname{Sqrt}[d x] ;

\operatorname{In}[15]:=\operatorname{Psi}=\operatorname{Psi 0} ;

Data =

Table [

(* Step the wave function forward in time *)

\text { Psi }=\text { Linear Solve[Uplus }[(i-1) d t], \text { Uminus }[(i-1) d t] . \text { Psi }] ;

If [

(* Only Plot 100 frames regardless of number of time steps *)

\operatorname{Mod}[i, N t / 100]=0 ,

(* The exact solution *)

g\left[x_{-}\right]=\Phi[0, x, i d t] ;

(* The instantaneous ground state *)

h\left[x_{-}\right]=p s i[0, x-f[i d t]] ;

(* Plotting commands *)

Show [

\{ Plot [\{\operatorname{Re}[g[x]], \quad \operatorname{Im}[g[x]], \quad A b s[g[x]], \quad A b s[h[x]]\} ,

\{x,-3,3\}, \text { PlotRange } \rightarrow\{-1,1\}, \text { AxesLabel } \rightarrow\left\{" x ", "|\Psi|^{\prime \prime}\right\} ,

\text { Plotstyle } \rightarrow \text { \{Blue, Red, Black, \{Black, Dashed\}\}] } ,

\text { ListPlot[ }\{R e[\text { Psi], Im[Psi], Abs [Psi] }\}, \text { PlotRange } \rightarrow\{-1,1\} ,

\text { DataRange } \rightarrow\{ xmin + dx , xmax – dx \} ,

\text { Plotstyle } \rightarrow \text { PointSize[Medium]] } ,

\}

],

Nothing

],

\{\dot{i}, 1, Nt \}

];

\text { In[17]:= ListAnimate [Data] } .

(b)\Omega=5 \omega:

\text { Numerics } \Omega=5 \omega .

Physical constants

\ln [18]:= m =1 ; \omega=1 ; \quad \hbar=1 ; \Omega=5 ; \quad A =1 ;

Numerical constants

\ln [19]=N x=100 ; \text { xmin }=-10 . ; \text { xmax }=10 . ; d x=\frac{x \max -\operatorname{xmin}}{N x+1} ;

Nt =1000 ; d t =\frac{2 \pi / \Omega}{ Nt } ;

Construct H

The Identity Matrix

\operatorname{In}[20]=\text { Openone }=\text { SparseArray }\left[\left\{i_{-}, i_{-}\right\} \rightarrow 1 .,\{N x, N X\}\right] ;

The second derivative operator

\operatorname{In}[21]= D 2=\frac{1.0}{ dx ^{2}} \text { SparseArray }\left[\left\{\left\{ i _{-}, i_{-}\right\} \rightarrow-2 .,\left\{i_{-}, j_{-}\right\} / ; \text {Abs }[i-j]=1 \rightarrow 1 .\right\},\{ Nx , N X \}\right] ;

The position operator

\operatorname{In}[22]==X=\text { SparseArray }\left[\left\{i_{-}, i_{-}\right\} \rightarrow \operatorname{xmin}+d x i,\{N x, N x\}\right] ;

The Hamiltonian

\text { In }[23]=H\left[t_{-}\right]=-\frac{\hbar^{2}}{2 m} D 2+\frac{1}{2} m \omega^{2} X \cdot x-m \omega^{2} X f[t] ;

Construct the time-evolution operator

The numerator in Cayley’s form

\operatorname{In}[24]=\text { Uplus }\left[t_{-}\right]=\text {Openone }+I H\left[t+\frac{d t}{2}\right] \frac{d t}{2} ;

The denominator in Cayley’s form

\operatorname{In}[25]=\text { Uminus }\left[t_{-}\right]=\text {Openone }-I H\left[t+\frac{d t}{2}\right] \frac{d t}{2} ;

Solve the t-dep Schrödinger equation

\operatorname{In}[26]=\operatorname{Psi} 0=\text { Eigenvectors }[H[\theta]][[N x]] / \operatorname{Sqrt}[d x] ;

\operatorname{In}[27]:=\operatorname{Psi}= Psi 0 ;

Data =

Table [

(* Step the wave function forward in time *)

\text { Psi }=\text { Linear Solve[Uplus }[(i-1) d t], \text { Uminus }[(i-1) d t] . \text { Psi] } ;

If [

(* Only Plot 100 frames regardless of number of time steps *)

\operatorname{Mod}[i, N t / 100]=0 ,

(* The exact solution *)

g\left[x_{-}\right]=\Phi[0, x, i d t] ;

(* The instantaneous ground state *)

h\left[x_{-}\right]=\operatorname{psi}[0, x-f[i d t]] ;

(* Plotting commands *)

Show[

\{\operatorname{Plot}[\{\operatorname{Re}[g[x]], \operatorname{Im}[g[x]], \quad \operatorname{Abs}[g[x]], \quad A b s[h[x]]\} ,

\{x,-3,3\}, \text { PlotRange } \rightarrow\{-1,1\}, \text { AxesLabel } \rightarrow\{" x ", "|\Psi| "\} ,

\text { PlotStyle } \rightarrow \text { \{Blue, Red, Black, \{Black, Dashed\} \}] } ,

\text { ListPlot [\{Re[Psi], Im[Psi], Abs[Psi] \}, PlotRange } \rightarrow\{-1,1\} ,

\text { DataRange } \rightarrow\{ xmin + dx , xmax – dx \} ,

\text { Plotstyle } \rightarrow \text { PointSize[Medium]] } .

\}

],

Nothing

],

\{i, 1, \text { Nt }\}

];

\text { In[29]: ListAnimate [Data] } .

\text { Numerics } \Omega=6 / 5 \omega .

Physical constants

\operatorname{In}[30]:= m =1 ; \omega=1 ; \hbar=1 ; \Omega=6 / 5 ; A =1 ;

Numerical constants

\ln [3]:=N x=100 ; \operatorname{xmin}=-10 . j \operatorname{xmax}=10 . j dx =\frac{\operatorname{xmax}- xmin }{ Nx +1} ;

Nt =1000 ; dt =\frac{2 \pi / \Omega}{ Nt } ;

Construct H

The Identity Matrix

\ln [32]=0 \text { penone }=\text { SparseArray }\left[\left\{i_{-}, i_{-}\right\} \rightarrow 1 .,\{N x, N x\}\right] ;

The second derivative operator

\ln [33]: D 2=\frac{1.0}{ d x^{2}} \text { SparseArray }\left[\left\{\left\{i_{-}, i_{-}\right\} \rightarrow-2 .,\left\{i_{-}, j_{-}\right\} / ; \text {Abs }[i-j]=1 \rightarrow 1 .\right\},\{ Nx , Nx \}\right] ;

The position operator

\operatorname{In}[34]:=X=\text { SparseArray }\left[\left\{i_{-}, i_{-}\right\} \rightarrow \operatorname{xmin}+d x i,\{N x, N x\}\right] ;

The Hamiltonian

\operatorname{In}[35]=H\left[t_{-}\right]=-\frac{\hbar^{2}}{2 m} D 2+\frac{1}{2} m \omega^{2} X \cdot X-m \omega^{2} \times f[t] ;

Construct the time-evolution operator

The numerator in Cayley’s form

\ln [36]:=\text { Uplus }\left[t_{-}\right]=\text {OpenOne }+I H\left[t+\frac{d t}{2}\right] \frac{d t}{2} ;

The denominator in Cayley’s form

\operatorname{In}[37]=\text { Uminus }\left[t_{-}\right]=\text {Openone }-I H\left[t+\frac{d t}{2}\right] \frac{d t}{2} ;

Solve the t-dep Schrödinger equation

\ln [38]:=\operatorname{Psi0}=\text { Eigenvectors }[H[\theta]][[N x]] / \operatorname{Sqrt}[d x] ;

\operatorname{In}[39]:=\operatorname{Psi}=\operatorname{Psi} 0 ;

Data =

Table [

(* Step the wave function forward in time *)

\text { Psi }=\text { LinearSolve[Uplus }[(i-1) d t], \text { Uminus }[(i-1) d t] . \text { Psi] } ;

If [

(* Only Plot 100 frames regardless of number of time steps *)

\operatorname{Mod}[i, Nt / 100]=0 ;

(* The exact solution *)

g\left[x_{-}\right]=Ψ[0, x, i d t] ,

(* The instantaneous ground state *)

h\left[x_{-}\right]=p s i[0, x-f[i d t]] ;

(* Plotting commands *)

Show[

\{P \operatorname{lot}[\{\operatorname{Re}[g[x]], \quad \operatorname{Im}[g[x]], \quad A b s[g[x]], \quad A b s[h[x]]\} ,

\{ x ,-3,3\}, \text { PlotRange } \rightarrow\{-1,1\}, \text { AxesLabel } \rightarrow\left\{" x^{\prime \prime}, "|\Psi| "\right\} ,

\text { Plotstyle } \rightarrow \text { \{Blue, Red, Black, \{Black, Dashed\}\}] } ,

\text { ListPlot[\{Re[Psi], Im[Psi], Abs[Psi] }\}, \text { PlotRange } \rightarrow\{-1,1\} ,

\text { DataRange } \rightarrow\{x \min +d x, x \max -d x\} ,

\text { PlotStyle } \rightarrow \text { PointSize[Medium]] } .

\}

],

Nothing

],

\{i, 1, Nt \}

];

ListAnimate[Data]

257
258
259

Related Answered Questions