Quantum Mechanics using Matrix Methods

Transcription

1 Quantum Mechanics using Matrix Methods Introduction and the simple harmonic oscillator In this notebook we study some problems in quantum mechanics using matrix methods. We know that we can solve quantum mechanics in any complete set of basis functions. If we choose a particular basis, the Hamiltonian will not, in general, be diagonal, so the task is to diagonalize it to find the eigenvalues (which are the possible results of a measurement of the energy) and the eigenvectors. In many cases this can not be done exactly and some numerical approximation is needed. A common approach, which is the basis of a lot of quantum chemistry is to take a finite basis set and diagonalize it numerically. The ground state of this reduced basis set will not be the exact ground state, but by increasing the size of the basis we can improve the accuracy and check if the energy converges as we increase the basis size. We will apply this approach here for an anharmonic oscillator. The reference for this material is Kinzel and Reents, p We first discuss the exactly solvable case of the simple harmonic oscillator. The Hamiltonian is given by H 0 p m m Ω x where p is the momentum, x the position, m the mass and Ω the angular frequency of the classical oscillator. x and p satisfiy the commutation relation [x, p] = i ħ. The Hamiltonian can be written in dimensionless form as H 0 ħω p p 0 x x 0 where p 0 = ħm Ω and x 0 = ħ m Ω, are the basic momentum and length scales. From now on, we will give the energy in units of ħ Ω, x in units of x 0, and p in units of p 0, so the reduced Hamiltonian is H 0 p x. In these units the commutation relation is [x, p] = i. In any textbook on quantum mechanics, it is shown that the energy levels are given by E n n, n 0,,,.... and the wavefunctions are given by Ψ n x n n Π e x H n x, where H n (x) is a Hermite polynomial. The books also show that it is easier to determine the energy levels using operator methods rather than the Schr dinger equation, and that is the approach that we will take here. In this approach, one introduces so-called raising and lowering operators, a and a, related to x and p by x i a a, p a a,

2 matrix_qm.nb which have the commutation relations a, a. and in terms of which the Hamiltonian is written H 0 a a. Denoting eigenstates of the Hamiltonian by n, then one finds a n n n, a n n n and so a a n n n, from which it follows that the energy E n is equal to (n + /) as stated above. The matrix elements of x in the basis n are given by X nm n x m m n,m m n,m n m nm,. Hence we can conveniently define the matrix elements for x as follows: In[]:= Clear"Global " In[]:= xn, m : Sqrtn m ; Absn m In[3]:= xn, m : 0 ; Absn m where we have used the symbol /;, which means "assuming that". From these we can generate a matrix for x keeping a basis of "basissize" states: In[4]:= xbasissize : Tablexn, m, n, 0, basissize, m, 0, basissize Let s see what we get for basissize = 4 In[5]:= x4 MatrixForm Out[5]//MatrixForm= It is trivial to generate the Hamiltonian matrix of the simple harmonic oscillator, since it is diagonal, i.e. In[6]:= h0basissize : DiagonalMatrix Tablen, n, 0, basissize

3 matrix_qm.nb 3 In[7]:= Out[7]//MatrixForm= h04 MatrixForm The anharmonic oscillator Now we make the problem non-trivial by adding an anharmonic term. We will take it to be proportional to x 4, i.e. H H 0 Λ x 4 It is easy to generate the matrix for H using the matrix obtained above for x and the convenient "dot" notation in Mathematica for performing matrix products: In[8]:= hbasissize, Λ : h0basissize Λ xbasissize. xbasissize. xbasissize. xbasissize For example, with a basis size of 4 we get In[9]:= h4, Λ MatrixForm Out[9]//MatrixForm= 3 Λ 4 3 Λ Λ Λ 3 Λ Λ Λ 0 5 Λ 4 The eigenvalues can also be obtained numerically and then sorted. Here we give a function (with delayed assignment) for doing this: In[0]:= evalsbasissize, Λ : Sort Eigenvalues N hbasissize, Λ Now we get some numbers. First consider the trivial example of the simple harmonic oscillator, Λ = 0, and a basis size of 0. In[]:= Out[]= evals0, 0 0.5,.5,.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5 Naturally we get the right answer, since the matrix is diagonal in this case. Next we consider the anharmonic oscillator, Λ =. In[]:= Out[]= evals0, ,.756, , 7.599, 4.5, , 46.87, 46.93, 46.84, The energy levels are no longer uniformily spaced. We expect that the lowest few levels will be fairly accurate, but the higher levels will be inaccurate because they feel more severely the truncation of the basis states. To test this, we plot the eigenvalues for a range of Λ, starting with a basis size of 5.

4 4 matrix_qm.nb In[3]:= basissize 5; In[4]:= p Plot Evaluateevalsbasissize, Λ, Λ, 0,, PlotRange 0,, PlotStyle AbsoluteThickness, AxesLabel "Λ", "E" 0 8 Out[4]= 6 4 To see how accurate these are we compare the results for the size-5 basis with those for a size-0 basis. In[5]:= basissize 0; In[6]:= p Plot Evaluate evalsbasissize, Λ, Λ, 0,, PlotRange 0, 0, PlotStyle AbsoluteThickness, Hue0, Dashing0.0, 0.0, BaseStyle FontSize 4, FontWeight "Bold", FontColor Hue0 ; In[7]:= Showp, p, PlotLabel "basissize 5 solid and 0 dashed" 0 8 E Out[7]= 6 4 The lowest levels appear to have converged well, but the higher levels have not. This is not surprising because the highest of the plotted levels are some of the highest levels possible for the size-0 basis, and so the effects of the finite basis size are naturally large in that region. Now we focus on the ground state, which is given by the first element of evals, e.g. for basis size 0 and Λ = 0. we have In[8]:= basissize 0;

5 matrix_qm.nb 5 In[9]:= evalsbasissize, 0. Out[9]= For a given basis size, we can plot the lowest energy level versus Λ, In[0]:= Plot evalsbasissize, Λ, Λ, 0,, PlotRange 0.5, 0.8, AxesLabel "Λ", "E 0 " Out[0]= We see that the energy level starts off, as it must, at the value / for Λ = 0, and then increases with increasing Λ. Note that to plot all the eigenvalues as a function of Λ it is necessary to Evaluate the evals function inside the Plot function, whereas to plot just the lowest eigenvalue it is important not to use Evaluate otherwise the wrong eigenvalue is plotted (For some reason, Plot seems to ignore Sort in this case.) The details of the functioning of the Plot command, and when one should Evaluate the function to be plotted, remain rather mysterious to me. In fact,in versions prior to version 5, I don t think one could get Mathematica to correctly plot the ground state energy at all, even if one omitted the Evaluate command. You could get round this by obtaining a list of the desired numerical values for a set of values of Λ, and then plotting them with ListPlot, using PlotJoined -> True if you want a continuous line rather than dots. The routines given at the back of Kinzel and Reents show how to do this (though it is not discussed in the text). We really wish to check that our calculation of the ground state energy used a large enough basis size in order to be accurate. To check this, we plot the ground state energy versus the inverse of the size of the basis set. Here are some results for Λ = 0.:

6 6 matrix_qm.nb In[]:= ListPlotTable basissize, evalsbasissize, 0., basissize, 9, 30, Frame True, Axes False, PlotStyle PointSize0.0, Hue0, Epilog Line 30, , 8, , PlotRange All, FrameLabel "basis size", "E 0 ", RotateLabel False, PlotLabel "Λ 0." Out[]= E We see that for basis size greater than about 6, the energy has converged well. As you might expect, and you can verify yourself, a larger basis set is needed if Λ is larger. The eigenvectors can also be found, and these give the linear coefficients of the Hermite polynomials which make up the coordinate space wavefunction. Hence the wavefunction of the anharmonic oscillator can also be obtained by this method. Double Well Potential Next we use matrix methods to calculate the lowest energy levels in a double well potential. The Hamiltonian is given by p H x V x, where V x x 4 Λ 4. Note that the coeficient of x is negative. The minima of the potential energy are at x 0 ±, V min V x 0 Λ 4 Λ. We plot the potential for the case of Λ = 0..

7 matrix_qm.nb 7 In[]:= Plotx^ Λ x^4 4. Λ 0., x, 4, 4 Out[]= 4 4 The new physics in this example is the possibility of tunneling between the two minima. The reader is referred to any quantum mechanics book for more details on tunneling. Setup To obtain the Hamiltonian, it is convenient to use the matrix elements of p, given in any quantum mechanics textbook, (as well those of of x which we already have). (To avoid needing p we could alternatively write H = H 0 x x 4 4 where H 0 is the simple harmonic oscillator Hamiltonian which is diagonal). In[3]:= pj, k : I Sqrtj k ; j k In[4]:= pj, k : I Sqrtj k ; j k In[5]:= pj, k : 0 ; Absj k In[6]:= pbasissize : Tablepn, m, n, 0, basissize, m, 0, basissize In[7]:= hn, Λ : pn. pn xn. xn Λ xn. xn. xn. xn 4 The function evals is the same as before. Λ = 0. Consider first Λ = 0.. We can find where the two minima are either analytically, which gives x 0 = / Λ, and E min 4 Λ, as stated above, or use Mathematica: In[8]:= Out[8]= FindMinimum x^ Λ x^4 4. Λ 0., x,.5, x.3607 We extract the energy levels for n (the basis size) equal to 0 and 0 In[9]:= Out[9]= evals0, , , , ,.08039,.5807,.56578, , 5.9,

8 8 matrix_qm.nb In[30]:= evals0, 0. Out[30]= , , , ,.5557,.4794, 3.378, 4.408, 5.569, , 7.73, , 9.378, ,.9696, 3.53, 5.883, , , 9.86 Next we plot the lowest energy level versus /n. It seems to be converging. In[3]:= ListPlot Table size, evalssize, 0., size, 0, 30, 4, PlotStyle Hue, PointSize0.0, Axes False, Frame True, FrameLabel "basis size", "E 0 ", RotateLabel False, PlotLabel "Λ 0." Out[3]= E Now we want to plot more than one energy level for different values of n. We construct evalslist, a two-dimensional different values of n between 0 and 30, list containing all the eigenvalues for In[3]:= evalslist Tableevalsn, 0., n,, 30 ; and from this plot the two lowest levels versus /n (note we plot each set of data twice, once to get the points without the lines, and once to get the lines without the points; this is done by the Joined -> {True, False, Tre, False} option.)

9 matrix_qm.nb 9 In[33]:= ListPlot Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Axes False, Frame True, Joined True, False, True, False, PlotStyle Thick, Red, Red, PointSizeLarge, Thick, Blue, Blue, PointSizeLarge, FrameLabel "basis size", "E 0 ", RotateLabel False, PlotLabel "Λ 0." Out[33]= E As expected, both levels converge as the basis size increases. Λ = 0. Next we consider a smaller value, Λ = 0., for which the minima are deeper. In[34]:= ClearΛ; Plotx^ Λ x^4 4. Λ 0., x, 4.7, Out[34]=

10 0 matrix_qm.nb In[35]:= minpot FindMinimum x^ Λ x^4 4. Λ 0., x, 3.5 Out[35]=.5, x 3.68 When the potential is very deep the two the particle will only tunnel very slowly from one well to the other. The two lowest energy levels are split by this small tunneling splitting, and the eigenstates are the symmetric and anti-symmetric combinations of the wavefunction for the particle localized in the ground state of the left well and the right well. If we represent the potential at the bottom of each well as a parabola then the lowest energy of each of these states is given by the simple harmonic ground state energy above V min, i.e. Ε 0 V min Ω min where Ω min is classical freqency of oscillation about the minimum, i.e. Ω min = V x 0 =. Hence Ε 0 4 Λ. We expect then, that for deep wells, the two lowest energy levels will be given, to a good approximation by E Ε 0 0, E 0 Ε 0 deep wells where 0 is the (small) tunneling splitting. For Λ = 0., Ε 0 is given by In[36]:= Out[36]= Ε0 4Λ Λ 0. Now we compare this with the actual energy of the two lowest levels. In[37]:= evalslist Tableevalsn, 0., n,, 30 ;

11 matrix_qm.nb In[38]:= ListPlot Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Axes False, Frame True, Joined True, False, True, False, FrameLabel "basis size", "E 0, E ", RotateLabel False, PlotLabel "Λ 0.".80.8 Out[38]= E 0, E We see that the two lowest levels are close together as expected (the difference being the tunneling splitting), and the energies are fairly close to Ε , though they are not precisely equal to that value. The difference arises because, for this value of Λ, the wells are not deep enough to accurately represent their behavior about the minima as parabolas. Now we also include the next two levels. These are also fairly close together, but not as close as the lowest two because, having more energy, they can tunnel more easily from one well to the other.

12 matrix_qm.nb In[39]:= ListPlot Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Table n, evalslist n3, n, 0, 30, Table n, evalslist n3, n, 0, 30, Table n, evalslist n4, n, 0, 30, Table n, evalslist n4, n, 0, 30, Axes False, Frame True, Joined True, False, True, False, True, False, True, False, FrameLabel "basis size", "E 0,E,E,E 3 ", RotateLabel False, FrameLabel "Λ 0.".0 Out[39]= E 0,E,E,E Λ = 0.05 Finally we take Λ to have the smaller value of 0.05, which makes the wells deeper still: In[40]:= ClearΛ; Plotx^ Λ x^4 4. Λ 0.05, x, 6.5, Out[40]= 3 4 We calculate Ε 0 for this value of Λ :

13 matrix_qm.nb 3 In[4]:= minpot FindMinimum x^ Λ x^4 4. Λ 0.05, x, 3.5 Out[4]= 5., x In[4]:= Out[4]= Ε0 4Λ Λ 0.05 Now we plot the lowest two levels for different basis sizes: In[43]:= In[44]:= In[45]:= In[46]:= evalslist Tableevalsn, 0.05, n,, 30 ; lw ListPlot Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Table n, evalslist n, n, 0, 30, Axes False, Frame True, Joined True, False, True, False, FrameLabel "basis size", "E 0, E ", RotateLabel False, PlotLabel "Λ 0.05"; pe0 ListPlotTable n, Ε0, n, 0, 30, PlotStyle AbsoluteThickness, Dashing0.0, 0.0, Hue0.65, Showlw, pe0 Joined True ; Out[46]= E 0, E Ε Finally we show the lowest 8 levels. They are grouped into four pairs as expected,

14 4 matrix_qm.nb Finally we show the lowest 8 levels. They are grouped into four pairs as expected, with each pair corresponding to a symmetric and anti-symmetric combination of wavefunctions localized in the left and the right well. In[47]:= In[48]:= lw8 ListPlot FlattenTable Table n, evalslist ni, n, 0, 30, i,, 8, Table Table n, evalslist ni, n, 0, 30, i,, 8,, Axes False, Frame True, Joined Flatten TableTrue, 8, TableFalse, 8, FrameLabel "basis size", "E 0,..,E 7 ", RotateLabel False, PlotLabel "Λ 0.05", PlotRange 4.8, 0.5 ; pe ListPlotTable n, Ε0, n, 0, 30, PlotStyle AbsoluteThickness, Dashing0.0, 0.0, Hue0.35, Joined True; In[49]:= pe ListPlotTable n, Ε0, n, 0, 30, PlotStyle AbsoluteThickness, Dashing0.0, 0.0, Hue0.85, Joined True; In[50]:= Showlw8, pe0, pe, pe 0 Out[50]= E 0,..,E For the case of a really deep well, the lowest levels would correspond to simple harmonic oscillator levels in each of the two wells with energy Ε n V min n Ω min where here we have V min 4 Λ and Ω min =. For each n, tunneling causes the two degenerate levels to split by a small amount n, so E n Ε n n, E n n Ε n n 0,,,... very deep wells. Ε n

15 matrix_qm.nb 5 fer progressively mroe and more, indicating that the potential cannot be accurately represented by a parabola at these energies.