(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 6.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 55122, 1493] NotebookOptionsPosition[ 51115, 1369] NotebookOutlinePosition[ 51713, 1392] CellTagsIndexPosition[ 51670, 1389] WindowFrame->Normal ContainsDynamic->False*) (* Beginning of Notebook Content *) Notebook[{ Cell[CellGroupData[{ Cell["\<\ Quantum Wells -- Eigenvalues of the Schr\[ODoubleDot]dinger Equation\ \>", "Title", FontSize->24], Cell[CellGroupData[{ Cell["Introduction", "Section"], Cell[TextData[{ "We investigate the bound states of a particle in a quantum well in one \ dimension with potential V(x), where V(x) \[LessEqual] 0 and V(x) \[Rule] 0 \ for |x| \[Rule] \[Infinity]. This notebook shows one way to do this using ", StyleBox["Mathematica. ", FontSlant->"Italic"], "In units where \[HBar] = m = 1 (where m is the mass of the particle), Schr\ \[ODoubleDot]dinger's equation is " }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{ FractionBox[ RowBox[{ SuperscriptBox["d", "2"], "\[Psi]"}], RowBox[{"d", " ", SuperscriptBox["x", "2"]}]], " ", "+", " ", RowBox[{"2", " ", RowBox[{"(", RowBox[{"E", " ", "-", " ", RowBox[{"V", RowBox[{"(", "x", ")"}]}]}], ")"}], " ", "\[Psi]"}]}], " ", "=", " ", "0"}]], "DisplayFormula"], Cell[TextData[{ "We take the potential to be an even function of x, so the wave functions \ are either even or odd functions of x, \[Psi](x) = \[PlusMinus]\[Psi](-x). \ The sign, +1 or -1, is called the parity of the state. It is shown in the \ textbooks on quantum mechanics that (i) the ground state is symmetric \ \[Psi](x) = \[Psi](-x) and has no zeroes (nodes), (ii) the first excited \ state is odd and has one node, and (iii) for each higher energy eigenvalue \ the the number of nodes increases by one and the parity alternates. \n\nWe \ will find the lowest energy level and corresponding wavefunction for each \ parity. We will set up a numerical method, but, in this notebook, only apply \ it to a simple case where the solution can also be obtained exactly, namely \ the rectangular well of width L and depth ", Cell[BoxData[ FormBox[ SubscriptBox["V", "0"], TraditionalForm]]], "(< 0), i.e. where V(x) = 0 for |x| > L/2 and V(x) = ", Cell[BoxData[ FormBox[ SubscriptBox["V", "0"], TraditionalForm]]], " for |x| \[LessEqual] L/2. We take L = 1, and initially take ", Cell[BoxData[ FormBox[ SubscriptBox["V", "0"], TraditionalForm]]], "= -8. \n\nFirst, therefore, we set up the potential and plot it. 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BaseStyle->{15, FontFamily -> "Times", Bold}, Epilog->{ InsetBox[ FormBox[ StyleBox["\"Region 1\"", FontSize -> 10, StripOnInput -> False], TraditionalForm], {-0.75, -5}], InsetBox[ FormBox[ StyleBox["\"Region 3\"", FontSize -> 10, StripOnInput -> False], TraditionalForm], {0.75, -5}], InsetBox[ FormBox[ StyleBox["\"Region 2\"", FontSize -> 10, StripOnInput -> False], TraditionalForm], {0, -5}]}, FrameStyle->Thickness[Large], PlotRange->{{-1, 1}, {-8., 0.}}, PlotRangeClipping->True, PlotRangePadding->{ Scaled[0.02], Scaled[0.02]}]], "Output", CellChangeTimes->{ 3.421373926210504*^9, 3.4213739997283173`*^9, 3.421506481589897*^9, 3.482974559349204*^9, 3.515287571721611*^9, {3.515293570407526*^9, 3.515293605204743*^9}, {3.515293659288105*^9, 3.515293675177878*^9}, 3.515293712533917*^9, {3.515293755995674*^9, 3.5152937858393497`*^9}, 3.515293821477844*^9}] }, Open ]], Cell["\<\ Although we only consider the rectangular well here, the numerical method can \ be applied to any potential which is an even function of x vanishes outside \ some range of x (at least to a sufficiently good approximation) simply by \ changing the expression for V(x). \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Setting up the Problem", "Section"], Cell[TextData[{ "In region 1, ", Cell[BoxData[ FormBox[ SubscriptBox["\[Psi]", "1"], TraditionalForm]]], "= A ", Cell[BoxData[ FormBox[ SuperscriptBox["e", "\[Kappa]x"], TraditionalForm]]], "where \[Kappa] = ", Cell[BoxData[ FormBox[ RowBox[{ SqrtBox[ RowBox[{"2", "|", "E", "|"}]], ","}], TraditionalForm]]], "and in region 3, ", Cell[BoxData[ FormBox[ SubscriptBox["\[Psi]", "3"], TraditionalForm]]], "= B ", Cell[BoxData[ FormBox[ SuperscriptBox["e", RowBox[{"-", "\[Kappa]x"}]], TraditionalForm]]], ", so we set up these functions:" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"B", " ", "=", " ", "1"}], ";"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"\[Psi]1", "[", "x_", "]"}], " ", ":=", " ", RowBox[{"A", " ", RowBox[{"Exp", "[", " ", RowBox[{ RowBox[{ RowBox[{"Abs", "[", RowBox[{"2", " ", "en"}], "]"}], "^", RowBox[{"(", RowBox[{"1", "/", "2"}], ")"}]}], " ", "x"}], "]"}]}]}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"\[Psi]3", "[", "x_", "]"}], " ", ":=", " ", RowBox[{"B", " ", RowBox[{"Exp", "[", RowBox[{ RowBox[{"-", " ", RowBox[{ RowBox[{"Abs", "[", RowBox[{"2", " ", "en"}], "]"}], "^", RowBox[{"(", RowBox[{"1", "/", "2"}], ")"}]}]}], " ", "x"}], "]"}]}]}]], "Input"], Cell[TextData[{ "The variable ", StyleBox["en", FontWeight->"Bold"], " is the energy.\nWe also define the Schr\[ODoubleDot]dinger equation for \ region 2, using a delayed assignment, \":=\", since we will only use it \ later:" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"eqn", "[", "en_", "]"}], " ", ":=", " ", RowBox[{ RowBox[{ RowBox[{"\[Psi]2", "''"}], "[", "x", "]"}], " ", "+", " ", RowBox[{"2", RowBox[{"(", RowBox[{"en", " ", "-", " ", RowBox[{"v", "[", "x", "]"}]}], ")"}], RowBox[{"\[Psi]2", "[", "x", "]"}]}]}]}]], "Input"], Cell[TextData[{ "We also set up the calculation of the wavefunction in region 2 matching the \ function and its derivative to ", Cell[BoxData[ FormBox[ SubscriptBox["\[Psi]", "1"], TraditionalForm]]], "at x=-L/2. We will adjust the energy so that either the derivative of \ \[Psi] vanishes at x=0 (for even eigenfunctions) or \[Psi] vanishes (for odd \ eigenfunctions). This value for the energy will be an eigenvalue. Since the \ energy will be determined by a boundary condition at x = 0 we only need to \ integrate from x = -L/2 up to x = 0." }], "Text", CellChangeTimes->{3.421508353115506*^9}], Cell[BoxData[ RowBox[{ RowBox[{"wavefunc2", "[", "energy_", "]"}], " ", ":=", RowBox[{"(", RowBox[{ RowBox[{"en", " ", "=", " ", "energy"}], ";", " ", RowBox[{"NDSolve", "[", RowBox[{ RowBox[{"{", " ", RowBox[{ RowBox[{ RowBox[{"eqn", "[", "energy", "]"}], " ", "==", " ", "0"}], ",", " ", RowBox[{ RowBox[{"\[Psi]2", "[", RowBox[{ RowBox[{"-", "L"}], "/", "2"}], "]"}], " ", "\[Equal]", " ", RowBox[{"\[Psi]1", "[", RowBox[{ RowBox[{"-", "L"}], "/", "2"}], "]"}]}], ",", " ", "\n", "\t\t\t\t\t", RowBox[{ RowBox[{ RowBox[{"\[Psi]2", "'"}], "[", RowBox[{ RowBox[{"-", "L"}], "/", "2"}], "]"}], " ", "\[Equal]", " ", RowBox[{ RowBox[{"\[Psi]1", "'"}], "[", RowBox[{ RowBox[{"-", "L"}], "/", "2"}], "]"}]}]}], "}"}], ",", " ", "\n", "\t\t", "\[Psi]2", ",", " ", RowBox[{"{", RowBox[{"x", ",", " ", RowBox[{ RowBox[{"-", "L"}], "/", "2"}], ",", " ", "0"}], "}"}]}], " ", "]"}]}], ")"}]}]], "Input"], Cell[TextData[{ "Note that it is necessary to set the energy variable ", StyleBox["en", FontWeight->"Bold"], ", used in the computation of ", Cell[BoxData[ FormBox[ SubscriptBox["\[Psi]", "1"], TraditionalForm]]], " and ", Cell[BoxData[ FormBox[ SubscriptBox["\[Psi]", "3"], TraditionalForm]]], ", to the current value of the variable ", StyleBox["energy ", FontWeight->"Bold"], " used in the function ", StyleBox["wavefunc2.", FontWeight->"Bold"], " This is done first and then ", StyleBox["NDSolve", FontWeight->"Bold"], " is called to solve the ", "Schr\[ODoubleDot]dinger equation.", " We define ", StyleBox["wavefunc2[en_]", "Input"], " with a delayed assignment, so that each time it is called, the \ differential equation is integrated again. This is what we want in order to \ calculate the eigenvalue. Note that ", StyleBox["wavefunc2[en_] ", "Input"], "will be given as a replacement rule in the form ", StyleBox["{{\[Psi]2\[Rule]InterpolatingFunction[{{-0.5,0.5}},\"<>\"]}}", "Input"], ". However, it is more convenient to obtain the wavefunction in region 2 \ directly as a function of x. To do so we define ", StyleBox["sol2[x, en]", "Input"], " by" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"sol2", "[", RowBox[{ RowBox[{"x_", "?", "NumericQ"}], ",", " ", RowBox[{"en_", "?", "NumericQ"}]}], "]"}], " ", ":=", " ", RowBox[{ RowBox[{"\[Psi]2", "[", "x", "]"}], " ", "/.", " ", RowBox[{ RowBox[{"wavefunc2", "[", "en", "]"}], "[", RowBox[{"[", "1", "]"}], "]"}]}]}]], "Input"], Cell[TextData[{ "which applies the replacement rule and removes the outer set of curly \ brackets by taking the first element of the list. Note too that we have added \ the hieroglyphics\n\t", StyleBox["?NumericQ", FontWeight->"Bold"], "\nto the arguments of sol2. This is necessary in ", StyleBox["Mathematica", FontSlant->"Italic"], " version 5 and later when the solution is put into ", StyleBox["FindRoot", FontWeight->"Bold"], " below to determine the energy eigenvalue.(", StyleBox["?NumericQ", FontWeight->"Bold"], " imposes that the function is only evaluated if the arguments are \ numerical.) It is also convenient to define a function for the derivative of \ the wave function (since we will require that this is zero at x=0 for the \ even parity solutions):" }], "Text", CellChangeTimes->{{3.421508382705098*^9, 3.421508384048872*^9}, { 3.4526191197849407`*^9, 3.452619121384219*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"sol2prime", "[", RowBox[{ RowBox[{"x_", "?", "NumericQ"}], ",", RowBox[{"en_", "?", "NumericQ"}]}], "]"}], ":=", RowBox[{ RowBox[{ RowBox[{"\[Psi]2", "'"}], "[", "x", "]"}], "/.", RowBox[{ RowBox[{"wavefunc2", "[", "en", "]"}], "[", RowBox[{"[", "1", "]"}], "]"}]}]}]], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["Even Parity Solution", "Section"], Cell["\<\ We are now in a position to start calculating. First of all we look for an \ even eigenfuction by setting B = A and requiring that the derivative of the \ wavefunction vanishes at x = 0. We give two initial starting guesses for the \ eigenvalue, -5 and -8. \ \>", "Text"], Cell[BoxData[ RowBox[{ RowBox[{"A", " ", "=", " ", "B"}], ";"}]], "Input"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"eval", " ", "=", " ", RowBox[{"energy", " ", "/.", " ", RowBox[{"FindRoot", " ", "[", " ", RowBox[{ RowBox[{"sol2prime", "[", RowBox[{"0", ",", " ", "energy"}], "]"}], ",", RowBox[{"{", RowBox[{"energy", ",", " ", RowBox[{"-", "5"}], ",", " ", RowBox[{"-", "8"}]}], "}"}]}], " ", "]"}]}]}]], "Input"], Cell[BoxData[ RowBox[{"-", "5.878749912503076`"}]], "Output", CellChangeTimes->{ 3.421373926546163*^9, 3.421373999941866*^9, 3.421506482101499*^9, 3.4829745596375847`*^9, 3.515287572004737*^9, 3.515293570601911*^9, { 3.5152936594338284`*^9, 3.515293675343548*^9}, 3.5152937127036943`*^9, { 3.515293756161433*^9, 3.515293786007101*^9}, 3.5152938216279907`*^9}] }, Open ]], Cell["\<\ We conclude that E =-5.87875 is an eigenvalue corresponding to an even \ eigenfunction of the Hamiltonian. For this simple example of the rectangular \ well we can explicitly verify that it is an eigenvalue by working out the \ exact solution. As the textbooks show, even-parity eigenvalues are given \ by the solutions of\ \>", "Text"], Cell[BoxData[ RowBox[{"\[Kappa]", " ", "=", " ", RowBox[{"k", " ", "tan", RowBox[{"(", RowBox[{"kL", "/", "2"}], ")"}]}]}]], "DisplayFormula"], Cell[TextData[{ "where k = ", Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"2", RowBox[{"(", RowBox[{"|", SubscriptBox["V", "0"], "|", " ", RowBox[{"-", " ", RowBox[{"|", "E", "|"}]}]}], ")"}]}]], TraditionalForm]]], ". We also solve this equation with ", StyleBox["FindRoot", "Input"], ", starting with the numerical value found above, ", StyleBox["eval", FontWeight->"Bold"], ",", " as the initial guess:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{" ", RowBox[{ RowBox[{ RowBox[{"kappa", " ", ":=", RowBox[{"Sqrt", "[", " ", RowBox[{"Abs", "[", RowBox[{"2", "analen"}], "]"}], " ", "]"}]}], ";", " ", RowBox[{"k", " ", ":=", " ", RowBox[{"Sqrt", "[", RowBox[{"2", " ", RowBox[{"(", RowBox[{ RowBox[{"Abs", "[", "v0", "]"}], " ", "-", " ", RowBox[{"Abs", "[", "analen", "]"}]}], ")"}]}], "]"}]}], ";"}], "\n", " ", RowBox[{"evalanal", " ", "=", RowBox[{"analen", " ", "/.", RowBox[{"FindRoot", " ", "[", " ", RowBox[{ RowBox[{"kappa", " ", "==", " ", RowBox[{"k", " ", RowBox[{"Tan", " ", "[", RowBox[{"k", " ", RowBox[{"L", " ", "/", " ", "2"}]}], "]"}]}]}], " ", ",", " ", RowBox[{"{", RowBox[{"analen", ",", " ", "eval"}], "}"}]}], " ", "]"}]}]}]}]}]], "Input"], Cell[BoxData[ RowBox[{"-", "5.87874986356345`"}]], "Output", CellChangeTimes->{ 3.4213739266020317`*^9, 3.421373999997266*^9, 3.421506482325324*^9, 3.482974559685679*^9, 3.515287572071817*^9, 3.5152935706563883`*^9, { 3.5152936594872293`*^9, 3.5152936753949614`*^9}, 3.5152937127503757`*^9, { 3.515293756215149*^9, 3.515293786058676*^9}, 3.515293821662888*^9}] }, Open ]], Cell[TextData[{ "The numbers for ", StyleBox["eval", FontWeight->"Bold"], " and ", StyleBox["evalanal", FontWeight->"Bold"], " agree showning that our numerical method correctly found an even-parity \ eigenvalue.\n\nNow we want the eigenfunction coresponding to our eigenvalue. \ Since we know the eigenvalue, we do not want to keep recalculating the \ wavefunction so we define a function ", StyleBox["efunc2", "Input"], " with ", StyleBox["immediate", FontSlant->"Italic"], " assignment, and input the eigenvalue for the energy " }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"efunc2", "[", "x_", "]"}], "=", RowBox[{ RowBox[{"\[Psi]2", "[", "x", "]"}], "/.", RowBox[{ RowBox[{"wavefunc2", "[", "eval", "]"}], "[", RowBox[{"[", "1", "]"}], "]"}]}]}], ";"}]], "Input"], Cell[TextData[{ "(where", StyleBox["[[1]] ", "Input"], "removes one set of curly bracket as above). (Calling ", StyleBox["wavefunct2", FontWeight->"Bold"], " not only computes the eigenfunction in Region 2, but also sets the value \ of the energy to be used in ", StyleBox["\[Psi]1", FontWeight->"Bold"], " and ", StyleBox["\[Psi]3", FontWeight->"Bold"], " to the desired eigenvalue.) We have now obtained the eigenfunction in all \ three regions, so let's collect these into a single (not normalized) function \ ", StyleBox["\[Psi]nn[x_]", "Input"], ", which can then easily be plotted (remember that we only computed ", StyleBox["efunc2[x]", FontWeight->"Bold"], " for -L/2 \[LessEqual] x < 0, and so we also have to specify it for 0 \ \[LessEqual] x \[LessEqual] L/2, noting that the eigenfunction is even):" }], "Text", CellChangeTimes->{{3.4215084013066597`*^9, 3.421508431885395*^9}, 3.515289108328967*^9}], Cell[BoxData[ RowBox[{ RowBox[{"\[Psi]nn", " ", "[", "x_", "]"}], " ", ":=", " ", RowBox[{ RowBox[{"efunc2", "[", RowBox[{"-", "x"}], "]"}], " ", "/;", " ", RowBox[{"0", " ", "\[LessEqual]", " ", "x", " ", "\[LessEqual]", " ", RowBox[{"L", "/", "2"}]}]}]}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"\[Psi]nn", "[", "x_", "]"}], " ", ":=", " ", RowBox[{ RowBox[{"efunc2", "[", "x", "]"}], " ", "/;", " ", RowBox[{ RowBox[{ RowBox[{"-", "L"}], "/", "2"}], " ", "\[LessEqual]", " ", "x", " ", "<", " ", "0"}]}]}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"\[Psi]nn", "[", "x_", "]"}], " ", ":=", " ", RowBox[{ RowBox[{"\[Psi]1", "[", "x", "]"}], " ", "/;", " ", RowBox[{"x", " ", "<", " ", RowBox[{ RowBox[{"-", "L"}], "/", "2"}]}]}]}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"\[Psi]nn", " ", "[", "x_", "]"}], " ", ":=", " ", RowBox[{ RowBox[{"\[Psi]3", "[", "x", "]"}], " ", "/;", " ", RowBox[{"x", " ", ">", " ", RowBox[{"L", "/", "2"}]}]}]}]], "Input"], Cell[TextData[{ "Note the use of the symbol ", StyleBox["/;", FontWeight->"Bold"], " , which means \"provided that\".\n\nWe first normalize the wavefunction, \ using immediate assignment so that the normalization constant is not \ recomputed each time a value for the normalized wave function is needed for \ the plot." }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{"normconst", " ", "=", " ", RowBox[{"Sqrt", "[", " ", RowBox[{"NIntegrate", " ", "[", " ", RowBox[{ RowBox[{ RowBox[{"\[Psi]nn", "[", "x", "]"}], "^", "2"}], ",", " ", RowBox[{"{", RowBox[{"x", ",", " ", RowBox[{"-", "Infinity"}], ",", " ", "Infinity"}], "}"}]}], " ", "]"}], " ", "]"}]}], ";"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"\[Psi]", "[", "x_", "]"}], " ", ":=", " ", RowBox[{ RowBox[{"\[Psi]nn", "[", "x", "]"}], " ", "/", " ", "normconst"}]}], ";"}]], "Input"], Cell[TextData[{ "and then plot it. We first define the plot for \[Psi]norm (but do not yet \ display it because of the semicolon after the command). Then we add the \ labels with the S", StyleBox["how[Graphics[... ]] ", "Input"], "command. This looks a bit complicated but has the advantage that the \ current values of the well depth and width, and the eigenvalue are \ automatically printed without needing to modify this command each time." }], "Text", CellChangeTimes->{{3.421373947863536*^9, 3.421373960411188*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"fig", " ", "=", " ", RowBox[{"Plot", "[", RowBox[{ RowBox[{"\[Psi]", "[", "x", "]"}], ",", " ", RowBox[{"{", RowBox[{"x", ",", " ", RowBox[{ RowBox[{"-", "1.2"}], "L"}], ",", " ", RowBox[{"1.2", "L"}]}], "}"}], ",", " ", RowBox[{"PlotStyle", " ", "\[Rule]", " ", RowBox[{"{", RowBox[{ RowBox[{"AbsoluteThickness", "[", "2", "]"}], ",", " ", "Red"}], "}"}]}], ",", RowBox[{"AxesLabel", " ", "->", " ", RowBox[{"{", RowBox[{"\"\\"", ",", " ", "\"\<\[Psi]\>\""}], "}"}]}]}], " ", "]"}]}], ";"}]], "Input", CellChangeTimes->{ 3.421373969316154*^9, {3.515293636727295*^9, 3.5152936702368717`*^9}, { 3.51529373104038*^9, 3.515293739116338*^9}, {3.515293782206849*^9, 3.515293782357408*^9}}], 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]], Cell["\<\ We see that the wavefunction matches up smoothly at x = \[PlusMinus]L/2 (i.e. \ \[PlusMinus]1/2) as expected. Also there are no nodes (zeroes) in the \ wavefunction which means, since we are in one dimension, that it is the \ ground state.\ \>", "Text", CellChangeTimes->{{3.42150845157757*^9, 3.4215084652902117`*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["Odd Parity Solution", "Section"], Cell["Now we look at odd-parity solutions.by requiring", "Text"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"A", " ", "=", " ", RowBox[{"-", "B"}]}], ";"}], " "}]], "Input"], Cell["\<\ and looking for a solution where \[Psi] vanishes at x = 0. We give two \ starting estimates for the eigenvalue of -0.5 and -2.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"eval", " ", "=", " ", RowBox[{"en", " ", "/.", " ", RowBox[{"FindRoot", " ", "[", RowBox[{ RowBox[{"sol2", "[", RowBox[{"0", ",", " ", "en"}], "]"}], " ", ",", RowBox[{"{", RowBox[{"en", ",", " ", RowBox[{"-", "0.5"}], ",", " ", RowBox[{"-", "2"}]}], "}"}]}], " ", "]"}]}]}]], "Input"], Cell[BoxData[ RowBox[{"-", "0.8142029405843095`"}]], "Output", CellChangeTimes->{ 3.4213739269357557`*^9, 3.42137400028139*^9, 3.421506482879223*^9, 3.482974560468131*^9, 3.515287572744134*^9, 3.515293571272848*^9, { 3.515293659749983*^9, 3.5152936756877213`*^9}, 3.515293712983899*^9, { 3.515293756431756*^9, 3.5152937863753853`*^9}, 3.515293821883922*^9}] }, Open ]], Cell["\<\ We check that the eigenvalue agrees with the analytical eigenvalue, which is \ a solution of \[Kappa] = \[Dash]k cot (kL / 2).\ \>", "Text", CellChangeTimes->{{3.452619262255045*^9, 3.452619265743215*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{" ", RowBox[{"evalanal", " ", "=", " ", RowBox[{"analen", " ", "/.", RowBox[{"FindRoot", " ", "[", " ", RowBox[{ RowBox[{"kappa", " ", "==", " ", RowBox[{ RowBox[{"-", " ", "k"}], " ", RowBox[{"Cot", " ", "[", RowBox[{"k", " ", RowBox[{"L", " ", "/", " ", "2"}]}], "]"}]}]}], " ", ",", " ", RowBox[{"{", RowBox[{"analen", ",", " ", "eval"}], "}"}]}], " ", "]"}]}]}]}]], "Input"], Cell[BoxData[ RowBox[{"-", "0.8142029672826213`"}]], "Output", CellChangeTimes->{ 3.421373926985882*^9, 3.4213740003489122`*^9, 3.421506483021553*^9, 3.4829745605214577`*^9, 3.515287572779653*^9, 3.515293571306571*^9, { 3.515293659789938*^9, 3.515293675727974*^9}, 3.515293713017693*^9, { 3.5152937564649057`*^9, 3.515293786408902*^9}, 3.5152938219110117`*^9}] }, Open ]], Cell["\<\ Indeed it agrees. Next we calculate the eigenfunction for -L/2 \[LessEqual] x \[LessEqual] 0 \ using immediate assignment and the energy set to the eigenvalue.\ \>", "Text", CellChangeTimes->{3.4526192833555727`*^9}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"efunc2", "[", "x_", "]"}], "=", RowBox[{ RowBox[{"\[Psi]2", "[", "x", "]"}], "/.", RowBox[{ RowBox[{"wavefunc2", "[", "eval", "]"}], "[", RowBox[{"[", "1", "]"}], "]"}]}]}], ";"}]], "Input"], Cell["\<\ We also need to redefine eigenfunction for 0 \[LessEqual] x \[LessEqual] L/2 \ (where we didn't explicitly compute it) since it is now odd (not even)\ \>", "Text"], Cell[BoxData[ RowBox[{ RowBox[{"\[Psi]nn", " ", "[", "x_", "]"}], " ", ":=", " ", RowBox[{ RowBox[{"-", RowBox[{"efunc2", "[", RowBox[{"-", "x"}], "]"}]}], " ", "/;", " ", RowBox[{"0", " ", "\[LessEqual]", " ", "x", " ", "\[LessEqual]", " ", RowBox[{"L", "/", "2"}]}]}]}]], "Input"], Cell["and to compute the normalization factor for this state", "Text"], Cell[BoxData[ RowBox[{ RowBox[{"normconst", " ", "=", " ", RowBox[{"Sqrt", "[", RowBox[{"NIntegrate", " ", "[", " ", RowBox[{ RowBox[{ RowBox[{"\[Psi]nn", "[", "x", "]"}], "^", "2"}], ",", " ", RowBox[{"{", RowBox[{"x", ",", " ", RowBox[{"-", "Infinity"}], ",", "Infinity"}], "}"}]}], " ", "]"}], " ", "]"}]}], ";"}]], "Input"], Cell["\<\ Everything else is the same as for the even-parity eigenstate and uses a \ delayed assignment. 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The wavefunction has only one node, \ showing that it is indeed the lowest energy odd-parity eigenstate.\ \>", "Text", CellChangeTimes->{{3.51529385915067*^9, 3.5152938844043083`*^9}, { 3.5152939186435633`*^9, 3.5152939578770657`*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["Conclusions", "Section"], Cell["\<\ The method can now be used to determine eigenvalues and eigenvectors for \ other potential wells where analytical solutions don't exist provided V(x) = \ 0 for |x| > L/2. Two examples which you could consider are: (a) A single well\ \>", "Text"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"V", RowBox[{"(", "x", ")"}]}], " ", "=", " ", RowBox[{"-", " ", RowBox[{ FractionBox[ SubscriptBox["V", "0"], "2"], " ", "[", RowBox[{"1", " ", "+", " ", RowBox[{"cos", " ", RowBox[{"(", RowBox[{"2", "\[Pi]", " ", RowBox[{"x", " ", "/", " ", "L"}]}], " ", ")"}]}]}], " ", "]"}]}]}], ",", " ", RowBox[{"(", RowBox[{ RowBox[{ RowBox[{"-", "L"}], "/", "2"}], " ", "<", " ", "x", " ", "<", " ", RowBox[{"L", "/", "2"}]}], ")"}]}]], "DisplayFormula"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"and", " ", "V", RowBox[{"(", "x", ")"}]}], " ", "=", " ", "0"}], ",", " ", RowBox[{ RowBox[{ RowBox[{ RowBox[{"otherwise", ".", " ", "This"}], " ", "has", " ", "depth"}], " ", "-", RowBox[{ SubscriptBox["V", "0"], "at", " ", "x"}]}], " ", "=", " ", RowBox[{ RowBox[{ RowBox[{"0", " ", "and", " ", "vanishes", " ", "as"}], " ", "|", "x", "|"}], " ", "\[Rule]", " ", RowBox[{"L", "/", "2."}]}]}]}], TextForm]], "Text"], Cell["(b) A double well", "Text"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"V", RowBox[{"(", "x", ")"}]}], " ", "=", " ", RowBox[{"-", " ", RowBox[{ FractionBox[ SubscriptBox["V", "0"], "2"], " ", "[", RowBox[{"1", " ", "-", " ", RowBox[{"cos", " ", RowBox[{"(", RowBox[{"4", "\[Pi]", " ", RowBox[{"x", " ", "/", " ", "L"}]}], " ", ")"}]}]}], " ", "]"}]}]}], ",", " ", RowBox[{"(", RowBox[{ RowBox[{ RowBox[{"-", "L"}], "/", "2"}], " ", "<", " ", "x", " ", "<", " ", RowBox[{"L", "/", "2"}]}], ")"}]}]], "DisplayFormula"], Cell[TextData[{ "which vanishes at x = 0 and \[PlusMinus] L/2, and has minima of depth ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"-", SubscriptBox["V", "0"]}], " "}], TraditionalForm]]], "at x = \[PlusMinus] L/4. Increase the depth of the potential and notice how \ the lowest even and odd energies become close to each other. Explain this in \ terms of quantum mechanical tunneling. \n\nYou can also obtain higher energy \ bound states of either parity by giving different starting values to the ", StyleBox["FindRoot", FontWeight->"Bold"], " command. For a given value of the well depth it is of interest to \ determine how many bound states there are. You will find that there is always \ one bound state (this is a feature of one-dimension; it is not true in three \ dimensions) and this is of even parity and has no nodes. However, the well \ needs to be sufficiently deep for a second bound state to appear.\n\nThe \ method used here relies on the potential vanishing for |x| > L/2. More \ general potential wells can be studied using a similar but somewhat more \ complicated method called the \"shooting method\", see e.g. my handout on \ this, and Kinzel, ", StyleBox["Physics by Computer", FontSlant->"Italic"], ", p. 130-135." }], "Text", CellChangeTimes->{{3.421508485749483*^9, 3.421508494124935*^9}}] }, Open ]] }, Open ]] }, WindowSize->{928, 1003}, WindowMargins->{{Automatic, 466}, {Automatic, 6}}, PrintingCopies->1, PrintingPageRange->{1, Automatic}, PrintingOptions->{"Magnification"->1., "PaperOrientation"->"Portrait", "PaperSize"->{611.25, 789.5625}, "PostScriptOutputFile"->"/home/peter/courses/115/mathematica/quantumwell.nb.\ ps"}, Magnification->1.5, FrontEndVersion->"6.0 for Mac OS X x86 (32-bit) (May 21, 2008)", StyleDefinitions->"Default.nb" ] (* End of Notebook Content *) (* Internal cache information *) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[CellGroupData[{ Cell[590, 23, 109, 3, 149, "Title"], Cell[CellGroupData[{ Cell[724, 30, 31, 0, 100, "Section"], Cell[758, 32, 424, 8, 84, "Text"], Cell[1185, 42, 380, 13, 59, "DisplayFormula"], Cell[1568, 57, 1317, 25, 291, "Text"], Cell[2888, 84, 73, 1, 40, "Input"], Cell[2964, 87, 77, 2, 40, "Input"], Cell[3044, 91, 97, 3, 40, "Input"], Cell[3144, 96, 216, 6, 42, "Input"], Cell[3363, 104, 214, 6, 42, "Input"], Cell[CellGroupData[{ Cell[3602, 114, 1795, 46, 133, "Input"], Cell[5400, 162, 2767, 60, 369, "Output"] }, Open ]], Cell[8182, 225, 293, 6, 107, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[8512, 236, 41, 0, 100, "Section"], Cell[8556, 238, 599, 25, 47, "Text"], Cell[9158, 265, 77, 2, 40, "Input"], Cell[9238, 269, 323, 10, 42, "Input"], Cell[9564, 281, 349, 11, 42, "Input"], Cell[9916, 294, 243, 7, 84, "Text"], Cell[10162, 303, 335, 10, 42, "Input"], Cell[10500, 315, 607, 12, 129, "Text"], Cell[11110, 329, 1181, 35, 111, "Input"], Cell[12294, 366, 1221, 36, 177, "Text"], Cell[13518, 404, 349, 10, 42, "Input"], Cell[13870, 416, 917, 21, 174, "Text"], Cell[14790, 439, 349, 11, 42, "Input"] }, Open ]], Cell[CellGroupData[{ Cell[15176, 455, 39, 0, 100, "Section"], Cell[15218, 457, 283, 5, 84, "Text"], Cell[15504, 464, 77, 2, 40, "Input"], Cell[CellGroupData[{ Cell[15606, 470, 377, 10, 40, "Input"], Cell[15986, 482, 373, 6, 40, "Output"] }, Open ]], Cell[16374, 491, 350, 6, 107, "Text"], Cell[16727, 499, 155, 4, 30, "DisplayFormula"], Cell[16885, 505, 472, 18, 77, "Text"], Cell[CellGroupData[{ Cell[17382, 527, 924, 27, 64, "Input"], Cell[18309, 556, 374, 6, 40, "Output"] }, Open ]], Cell[18698, 565, 560, 16, 152, "Text"], Cell[19261, 583, 261, 8, 42, "Input"], Cell[19525, 593, 940, 24, 152, "Text"], Cell[20468, 619, 293, 7, 42, "Input"], Cell[20764, 628, 282, 8, 42, "Input"], Cell[21049, 638, 245, 7, 42, "Input"], Cell[21297, 647, 229, 6, 42, "Input"], Cell[21529, 655, 336, 8, 107, "Text"], Cell[21868, 665, 398, 11, 40, "Input"], Cell[22269, 678, 195, 6, 42, "Input"], Cell[22467, 686, 524, 9, 107, "Text"], Cell[22994, 697, 847, 23, 64, "Input"], Cell[CellGroupData[{ Cell[23866, 724, 1910, 50, 179, "Input"], Cell[25779, 776, 6452, 114, 392, "Output"] }, Open ]], Cell[32246, 893, 330, 6, 62, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[32613, 904, 38, 0, 100, "Section"], Cell[32654, 906, 64, 0, 39, "Text"], Cell[32721, 908, 116, 4, 40, "Input"], Cell[32840, 914, 150, 3, 62, "Text"], Cell[CellGroupData[{ Cell[33015, 921, 362, 10, 40, "Input"], Cell[33380, 933, 371, 6, 40, "Output"] }, Open ]], Cell[33766, 942, 216, 4, 62, "Text"], Cell[CellGroupData[{ Cell[34007, 950, 492, 14, 64, "Input"], Cell[34502, 966, 374, 6, 40, "Output"] }, Open ]], Cell[34891, 975, 228, 6, 107, "Text"], Cell[35122, 983, 261, 8, 42, "Input"], Cell[35386, 993, 173, 3, 62, "Text"], Cell[35562, 998, 314, 8, 42, "Input"], Cell[35879, 1008, 70, 0, 39, "Text"], Cell[35952, 1010, 387, 11, 40, "Input"], Cell[36342, 1023, 187, 4, 62, "Text"], Cell[36532, 1029, 789, 22, 64, "Input"], Cell[CellGroupData[{ Cell[37346, 1055, 2142, 58, 179, "Input"], Cell[39491, 1115, 7783, 136, 413, "Output"] }, Open ]], Cell[47289, 1254, 364, 7, 84, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[47690, 1266, 30, 0, 100, "Section"], Cell[47723, 1268, 259, 6, 129, "Text"], Cell[47985, 1276, 583, 19, 51, "DisplayFormula"], Cell[48571, 1297, 548, 17, 41, "Text"], Cell[49122, 1316, 33, 0, 39, "Text"], Cell[49158, 1318, 583, 19, 51, "DisplayFormula"], Cell[49744, 1339, 1343, 26, 311, "Text"] }, Open ]] }, Open ]] } ] *) (* End of internal cache information *)