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Persamaan schroedinger bebas waktu
 

Persamaan schroedinger bebas waktu

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    Persamaan schroedinger bebas waktu Persamaan schroedinger bebas waktu Document Transcript

    • Persamaan SchroedingerPostulat-postulat dasar Mekanika KuantumPostulat I: Setiap sistem fisis dinyatakan dengan fungsi gelombang atau fungsikeadaan, , yang secara implisit memuat informasi lengkapmengenai observabel-observabel yang dapat diketahui pada sistemtersebut.Postulat II: Setiap observabel dinyatakan atau diwakili oleh suatu operatorlinear hermitan.OperatorOperator adalah suatu instruksi matematis yang bila dikenakan ataudioperasikan pada suatu fungsi maka akan mengubah fungsi tersebut menjadifungsi lain. Untuk operator dapat ditulis sebagai[Tanda aksen ‘ bukan berarti diferensial atau turunan, tapi hanya untukmembedakan dengan fungsi asalnya].Beberapa Operator Observabel No. Observabel Operator 1. Posisi 2. Momentum Linier 3. Momentum Sudut 4. Energi Kinetik
    • Persamaan Schroedinger 5. Energi Potensial 6. Energi TotalPersamaan schroedinger bebas waktuJika fungsi potensial tidak bergantung waktu, bagaimanakah bentuk persamaanSchroedinger untuk kasus dengan potensial bebas waktu V(x)?Untuk kasus seperti itu persamaan gelombang SchroedingerBila dilakukan separasi variable (pemisahan peubah) dalam solusi persamaan di atassehingga lalu substitusikan dalam persamaan Schroedinger bebaswaktu menghasilkan :dan dapat ditulis pula kedalam bentuk :Dari persamaan di atas jelas terlihat bahwa ruas kiri dari persamaan tersebut hanyamengandung variable x, dan ruas tengah hanya mengandung variable t. Sedangkanpersamaan itu berlaku untuk semua harga x maupun t. Hal ini hanya berlaku jika ruas kiridan ruas tengah selalu bernilai tetap, misalkan sama dengan G.Dengan demikian dapat diperoleh dua persamaan berikut :
    • Persamaan SchroedingerSolusi dari persamaan adalah , dengan G = E, yangmerupakan energy total partikel yang direpresentasikan oleh fungsi gelombang .Berikut penjelasannya :Perhatikan persamaan dan lalubandingkandengan persamaan maka didapat ungkapan :sehingga otomatis nilai G sama besarnya dengan energi total partikel E. Dengan demikianuntuk kasus dengan fungsi potensial tidak bergantung waktu, diperoleh persamaanSchroedinger bebas waktu (PSBW) :dengan fungsi gelombang total:persamaan , yang dapat ditulis sebagai dinamakan persamaan harga eigen, dan harga tetap E yangmerupakan solusi yang dikenal sebagai nama persamaan karakteristik, suatu topik pentingdalam pembelajaran tentang persamaan diferensial.Sumur Potensial Persegi Tak Terhingga
    • Persamaan SchroedingerAndaikanlah suatu elektron dalam pengaruh potensial berbentuk sumur tak terhinggaberdimensi-1 seperti berikutElektron terperangkap dalam daerah – , dan sama sekali tak dapat ke luar daerahitu. Dengan perkata lain peluang elektron berada di dan di sama dengan nol.Oleh sebab itu, jika adalah fungsi gelombangnya, maka :Karena dalam daerah – , maka persamaan Schrödinger bagi electrontersebut adalah:Solusinya adalah dan . dengan syarat batas didiperoleh :Harga C dan D dihitung melalui normalisasi fungsi, yakni:Hasilnya adalah , sehingga fungsi-fungsi eigen adalah :
    • Persamaan SchroedingerFungsi-fungsi ini membentuk set ortonormal; artinya:Selanjutnya, diperoleh harga eigen energi:Energi ini berharga diskrit (tidak kontinu, tapi bertingkat-tingkat) ditandai oleh bilangankuantum n.Applications of the Schrödinger equation in nonperiodic semiconductor structuresThe infinite square-shaped quantum wellThe infinite square-shaped well potential is the simplest of all possible potential wells. Theinfinite square well potential is illustrated in Fig. 7.1(a) and is defined asU(x)=0 (-1/2 x ½L )……………………………….(7.1)U(x)= ( 1/2 L)……………………………………(7.2)
    • Persamaan SchroedingerTo find the stationary solutions for ψn(x) and En we must find functions for ψn(x) whichsatisfy the Schrödinger equation. The time-independent Schrödinger equation, containsonly the differential operator d/dx, whose eigenfunctions are exponential or sinusoidalfunctions. Since the Schrödinger equation has the form of an eigenvalue equation, it isreasonable to try only eigenfunctions of the differential operator. Furthermore, we assumethat ψn(x)=0 for L/2 , because the potential energy is infinitely high in the barrierregions. Since the 3rd Postulate requires that the wave function be continuous, the wavefunction must have zero amplitude at thetwo potential discontinuities, that is ψn(x= ± L/2)=0 We therefore employ sinusoidalfunctions and differentiate between states of even and odd symmetry. We write for evensymmetry statesψn(x)= (n = 0,2,4,…and L/2 )……………………………(7.3)and for odd-symmetry statesψn(x)= (n =1,3,5…and L/2 )……………………………(7.4)Both functions have a finite amplitude in the well-region ( L/2 ) and they have zeroamplitude in the barriers, that is :ψn(x)=0 (n = 0,1,2,…and L/2 )……………………………(7.3)
    • Persamaan SchroedingerThe shapes of the three lowest wave functions (n = 0, 1, 2,...) are shown in Fig . 7.1(b). Inorder to normalize the wave functions, the constant A must be determined. The condition = 1 yields A= ……………………………………………………….(7.6)One can verify that Eqs. (7.3) and (7.4) are solutions of the infinite square well by insertingthe normalized wave functions into the Schrödinger equation. Insertion of the ground-statewave function (n = 0) into the Schrödinger equation yieldsCalculating the derivative on the left-hand side of the equation yields the ground sateenergy of the infinite square wellThe excited state energies (n = 1, 2, 3, …) can be evaluated analogously. One obtains theeigenstate energies in the infinite square well according toThe spacing between two adjacent energy levels, that is En – En-1, is proportional to n. Thus,the energetic spacing between states increases with energy. The energy levels areschematicallyshown in Fig . 7.1(b) for the infinite square well. The probability density of a particledescribed by the wave function ψ is given by ψ* ψ (2nd Postulate). The probabilitydensities of the three lowest states are shown in Fig . 7.1(c). The eigenstate energies are, asalready mentioned, expectation values of the total energy of the respective state. It istherefore interesting to know if the eigenstate energies are purely kinetic, purely potential,or a mixture of both. The expectation value of the kinetic energy of the ground state iscalculated according to the 5th Postulate:
    • Persamaan SchroedingerUsing the momentum operator p = (ħ / i) (d / dx) one obtains the expectation value of thekinetic energy of the ground statewhich is identical to the total energy given in Eq. (7.8). Evaluation of kinetic energies of allother states yieldsThe kinetic energy coincides with the total energy given in Eq. (7.9). Thus, the energy of aparticle in an infinite square well is purely kinetic. The particle has no potential energy.We next turn to a second, more intuitive method to obtain the wave functions of the infinitepotential well. This second method is based on the de Broglie wave concept. Recall that thede . Broglie wave is defined for a constant momentum p, that is for a particle in a constantpotential.The energy of the wave is purely kinetic. In order to find solutions of the infinite squarewell, wematch the de Broglie wavelength to the width of the quantum well according to theconditionIn this equation, multiples of half of the de Broglie wavelength are matched to the width ofthe quantum well. Expressing the kinetic energy in terms of the de Broglie wavelength, thatisand inserting Eq. (7.13) into Eq. (7.14) yields
    • Persamaan SchroedingerThis equation is identical to Eq. (7.12) which was obtained by the solution of theSchrödinger equation. The de Broglie wave concept yields the correct solution of theinfinite potential well,because (i) the particle is confined to the constant potential of the well region, (ii) theenergy ofparticle is purely kinetic, and (iii) the wave function is sinusoidal.The infinite square shaped quantum well is the simplest of all potential wells. The wavefunctions (eigenfunctions) and energies (eigenvalues) in an infinite square well arerelativelysimple. There is a large number of potential wells with other shapes, for example thesquare wellwith finite barriers, parabolic well, triangular well, or V-shaped well. The exact solutions ofthese wells are more complicated. Several methods have been developed to calculateapproximate solutions for arbitrary shaped potential wells. These methods will bediscussed in the Chapter on quantum wells in this book.The asymmetric and symmetric finite square-shaped quantum wellIn contrast to the infinite square well, the finite square well has barriers of finite height. Thepotential of a finite square well is shown in Fig . 7.2. The two barriers of the well have adifferent height and therefore, the structure is denoted asymmetric square well. Thepotential energy is constant within the three regions I, II, and III, as shown in Fig . 7.2. Inorder to obtain the solutions to the Schrödinger equation for the square well potential, thesolutions in a constant potential will be considered first.
    • Persamaan SchroedingerAssume that a particle with energy E is in a constant potential U. Then two cases can bedistinguished, namely E > U and E < U. In the first case (E > U) the general solution to thetimeindependent one-dimensional Schrödinger equation is given by ψ(x) = A cos kx + B sin kx……………………………………..(7.16)where A and B are constants andInsertion of the solution into the Schrödinger equation proves that it is indeed a correctsolution.Thus the wave function is an oscillatory sinusoidal function in a constant potential with E >U. In the second case (E < U), the solution of the time-independent one-dimensionalSchrödingerequation is given by ψ(x) = C ekx + De-kx……………………………………..(7.18)where C and D are constants and ………………………(7.19)Again, the insertion of the solution into the Schrödinger equation proves that it is indeed acorrect solution. Thus the wave function is an exponentially growing or decaying function inaconstant potential with E < U.Next, the solutions of an asymmetric and symmetric square well will be calculated. The
    • Persamaan Schroedingerpotential energy of the well is piecewise constant, as shown in Fig . 7.2. Having shown thatthewave functions in a constant potential are either sinusoidal or exponential, the wavefunctions in the three regions I (x ≤ 0) II (0 < x < L), and III (x ≥ L), can be written as …………………………………………..(7.20) ………………….. ………(7.21) ……………….(7.22)In this solution, the first boundary condition of the 3rd Postulate, i. e. ψI’(0) = ψIII’ and ψII’(L) and ψIII’(L) , the following two equations are obtained : ……………………………………(7.23) …..(7.24)This homogeneous system of equations has solutions, only if the determinant of the systemvanishes. From this condition, one obtains ………………………………………..(7.25)which is the eigenvalue equation of the finite asymmetric square well. For the finitesymmetric square well, which is of great practical relevance, the eigenvalue equation isgiven by : ………………………………………….(7.26)Where K = KI = KIII. If K is expressed as a function of k (see Eq. 7.19), then Eq. (7.26) dependsonly on a single variable, i. e., k. Solving the eigenvalue equation yields the eigenvalues of kand, by using Eqs. (7.17) and (7.19), the allowed energies E and decay constants K,respectively. The allowed energies are also called the eigenstate energies of the potential.Inspection of Eq. (7.26) yields that the eigenvalue equation has a trivial solution kL = 0(and thus E = 0) which possesses no practical relevance. Non-trivial solutions of theeigenvalue equation can be obtained by a graphical method. Figure 7.3 shows the graph ofthe left-hand and right-hand side of the eigenvalue equation. The dashed curve represents
    • Persamaan Schroedingerthe right-hand side of the eigenvalue equation. The intersections of the dashed curve withthe periodic tangent function are the solutions of the eigenvalue equation. The quantumstate with the lowest non-trivial solution is called the ground state of the well. States ofhigher energy are referred to as excited states .The dashed curve shown in Fig . 7.3 has two significant points, namely a pole and an endpoint. The dashed curve has a pole when the denominator of the right-hand side of theeigenvalue equation vanishes, i. e., when kL = KL Using Eq. (7.19), it is given by :Pole: …………………………………(7.27)The dashed curve ends when k = (2mU/ 2)1/2 If k exceeds this value, the square root in Eq.(7.19) becomes imaginary. The end point of the dashed curve is thus given by :And point : …………………………………(7.28)There are no further bound state solutions to the eigenvalue equation beyond the endpoint. Now that the eigenvalues of k and K are known they are inserted into Eqs. (7.23) and(7.24) which allows for the determination of the constants A and B and the wave functions.Thus the allowed energies and the wave functions of the square well have beendetermined.It is possible to show that all states with even quantum numbers (n = 0, 1, 2, ...) are of evensymmetry with respect to the center of the well, i. e. ψ (x) = ψ (-x) All states with oddquantum numbers (n = 1, 3, 5, ...) are of odd symmetry with respect to the center of thewell, i. e.ψ (x) = ψ (-x) The even and odd state wave functions in the well are thus of the form (for n=0,2,4,…) ……………………..(7.29)And (for n=1,3,5,…) ……………………..(7.30)
    • Persamaan SchroedingerThe proof of these equations is left to the reader. The three lowest wave functions of asymmetricsquare well are shown in Fig . 7.4.Elektron Berada Dalam Sumur PotensialSumur potensial adalah daerah yang tidak mendapat pengaruh potensial sedangkandaerah mendapat pengaruh potensial. Hal ini berarti bahwa jelektron, selama ia beradaberada dalam sumur potensial, merupakan elektron-bebas. Kita katakana bahwa elektronterjebak di sumur potensial, dan kita anggap bahwa dinding potensial sangat tinggi menuju∞ sedangkan di daerah II, yaitu antara 0 dan L,V = 0. Kita katakan bahwa lebar sumur potensial ini adalah L.Gb. Elektron dalam sumur potensial (daerah II).Pada sumur potensial yang dalam, daerah I dan III adalah daerah dimanaemungkinan keberadaan elektron bisa dianggap nol, ψ (x) =0 dan ψ (x) =0 . Solusipersamaan Schrödinger untuk daerah II adalah solusi untuk elektronbebas ………………………………(3.21)Persyaratan kekontinyuan di x = 0 mengharuskan
    • Persamaan Schroedinger BI + B2 = ψ1 (0) = 0 B1=B2dan persyaratan kekontinyuan di L mengharuskan sehingga …………………………..(3.22)Persamaan (3.22) mengharuskan k2L= nπ atau k2 = (dengan n bilangan bulat), sehinggafungsi gelombang di daerah II menjadiΨ*(x) ψ2(x)=4B22 ……………….(3.23)Untuk n = 1, fungsi ini bernilai nol di x =0 dan x =L , dan maksimum di x =L/ 2 . Untuk n= 2, nilai nol terjadi di x = 0, L/2, dan L. Untuk n = 3, nilai nol terjadi di x = 0, L/3, 2L/3,dan L; dan seterusnya, seperti terlihat pada Gb.3.3. Selain di x = 0, jumlah titiksimpul gelombang, yaitu titik di mana fungsinya bernilai nol, sama dengan nilai n.Karena di daerah II V = 0, maka atau . Denganmemasukkan nilai k2 kita peroleh energi elektron: ……………………………………(3.25)Kita lihat di sini bahwa energi elektron mempunyai nilainilai tertentu yang diskrit, yangditentukan oleh bilangan bulat n. Nilai diskrit ini terjadi karena pembatasan yangharus dialami oleh ψ2, yaitu bahwa ia harus berada dalam sumur potensial. Ia harusbernilai nol di batasbatas dinding potensial dan hal itu akan terjadi bila lebar sumurpotensial L sama dengan bilangan bulat kali setengah panjang gelombang. Jika tingkatenergi untuk n = 1 kita sebut tingkat energi yang pertama, maka tingkat energi yang kedua
    • Persamaan Schroedingerpada n = 2, tingkat energi yang ketiga pada n = 3 dan seterusnya.Jika kita kaitkan dengan bentuk gelombangnya, dapat kita katakan bahwa tingkattingkat energi tersebut sesuai dengan jumlah titik simpul gelombang.Dengan demikian maka diskritisasi energi elektron terjadi secara wajar melaluipemecahan persamaan Schödinger. Hal ini berbeda dari pendekatan Bohr yang harusmembuat postulat mengenai momentum sudut yang harus diskrit agar kuantisasienergi terjadi.Persamaan (3.25) memperlihatkan bahwa selisih energi antara satu tingkat dengantingkat berikutnya, misalnya antara n = 1 dan n = 2, berbanding terbalik dengankwadrat lebar sumur potensial. Makin lebar sumur ini, makin kecil selisih energitersebut, artinya tingkattingkat energi semakin rapat. Untuk L sama dengan satusatuan misalnya, selisih energi untuk n=2 dan n=1 adalah E2 – E1 =3h2/8m dan jikaL 10 kali lebih lebar maka selisih ini menjadi E2 – E1 =0,0h32/8m. gb.3.4Di x = 0 dan x = L amplitudo gelombang tidak lagi nol dan demikian jugaprobabilitas keberadaan elektronnya. Selain itu penurunan amplitudo akan makinlambat jika sumur potensial makin dangkal. Hal ini berarti bahwa makin dangkal