Solutions of Schrödinger's Equation: Particle in a Box, Study notes of Quantum Physics

Download Solutions of Schrödinger's Equation: Particle in a Box and more Study notes Quantum Physics in PDF only on Docsity! Physics 486 Lecture 4 Page 1 Physics 486, Spring ‘09 Lecture 4 Solutions of Schrödinger’s Equation: Particle in a Box Last time ( ) ( ) t iE t t ϕ ϕ∂ = − ∂
2 2 2 ( ) ( ) ( ) ( ) 2 d x V x x E x m dx ψ ψ ψ− + =
KE term PE term Total E term 2 2 2 ( , ) ( , ) ( ) ( , ) 2 x t x t V x x t i m x t − ∂ Ψ ∂Ψ+ Ψ = ∂ ∂

2 2 ( , )( , ) ( ) ( , ) 2 t t V t i m t − ∂Ψ∇ Ψ + Ψ = ∂ r r r r

Time- independent Schrodinger eq. This is the Schrodinger equation: In three dimensions it looks slightly different: ( , ) ( ) ( )x t x tψ ϕΨ = Here ∇2 is the Laplacian operatior, defined as ∇2=(∂/∂x)2+(∂/∂y)2+(∂/∂z)2. By separating variables we transformed Schrodinger’s equation into two ODEs: Eq. for time- dependent part (1) (2) Physics 486 Lecture 4 Page 2 Eigenstates /( ) iEtt eϕ −=
The time-dependent equation, for any given value of E, has a very simple solution: /n nEω =
But what determines the values of E? It is the time-independent Schrodinger equation, which has many solutions, each with its own energy: 1 2 3( ), ( ), ( )...x x xψ ψ ψ 1 2 3, , ...E E E ( , ) ( ) ni tn nx t x e ωψ −Ψ = Putting it all together, the separable solutions all have the general form Separable solutions comprise a very small fraction of all the possible solutions to Schrodinger’s equation. Why are they significant? (1) They are “stationary states”, or more commonly “eigenstates”, whose probability density has no time dependence: 22 2 ( , ) ( , ) ( ) ( )ni tn n n nP x t x t x e x ωψ ψ−= Ψ = = Eigenstates are “characteristic” modes of a quantum mechanical system, analogous to the normal modes of oscillation of a classical, vibrating string. Eigenstates (2) Any solution to the Schrodinger equation can be written as a linear combination of stationary states. That is, and Schrodinger wave function for a system, Ψ(x,t), can be written Where cn are just complex numbers. In mathematical terminology, the eigenfunctions ψn(x) form a “complete basis set”. We will come back to this point later. (3) They have definite total energy. If the system is in eigenstate n, a measurement of the total energy will with certainty return the value of En. As a result, solving the Schrodinger equation amounts to determining all the time-independent solutions, ψn(x). Determining ψnwill be the focus of much of this course. We will sometimes colloquially refer to the time-independent ψn(x) as the “wave function”. However one should remember that, in the end, the true wave function also contains the time-dependent part, φ(t). ( , ) ( ) ni tn n n x t c x e ωψ −Ψ =∑ Physics 486 Lecture 4 Page 5 Constraints on the form of ψ(x) (5). Inversion symmetry of the potential If the potential experienced by a particle has inversion symmetry, i.e., V(x) = -V(x), the corresponding solutions to the SEQ will have either either even or odd parity: ψ(x) = ±ψ(-x) ψ(x) x Even parity: ψ(x) = +ψ(-x) ψ(x) x Odd parity: ψ(x) = -ψ(-x) Particle in a Box: the Infinite Square Well As a specific important example, consider a quantum particle confined to a small region, 0 < x < L, by infinite potential walls. We call this a “one-dimensional (1D) box”. V(x) 0 L ∞ ∞ V = 0 for 0 < x < L V = ∞ everywhere else ψ(x) must be finite, continuous and single-valued. (because probability |ψ (x)|2 must be well defined everywhere) dψ/dx must be finite, continuous and single valued. (because dψ/dx is related to the classical momentum) In a region of finite potential: Physics 486 Lecture 4 Page 6 Particle in a Box: the Infinite Square Well Region I: When V = ∞, what is ψ(x)? For V = ∞, the SEQ can only be satisfied if: (otherwise, the energy would have to be infinite, to cancel V.) U = 0 for 0 < x < L U = ∞ everywhere else V(x) 0 L ∞ ∞ I I II ψΙ(x) = 0 ψΙ ψΙ A basic boundary-value problem. There are 2 distinct regions, (I) outside, and (II) inside the well. Note: The infinite well is an idealization. On the atomic scale there are no infinitely high and sharp barriers. ψ(x) and dψ(x)/dx go to zero continuously near a boundary. 2 2 2 ( ) ( ) ( ) 2 d x V x E x m dx ψ ψ ψ− + =
Particle in a Box: the Infinite Square Well Region II: When V = 0, what is ψ(x)? The general solution to this is: V(x) 0 L ∞ ∞ II 1 2( ) ikx ikxx A e A eψ −= + ψ 2 2 2 22 2 k h E m mλ = =
2 2 2 ( ) 2 ( ) d x mE x dx ψ ψ = −    
1 2( ) sin cosx B kx B kxψ = + Or alternatively 2 2 k mE=
where Note: this satisfies the requirement: 2 2 2 ( ) ( ) ( ) 2 d x V x E x m dx ψ ψ ψ− + =
Physics 486 Lecture 4 Page 7 Particle in a Box: the Infinite Square Well B1 and B2 are coefficients to be determined by the boundary conditions. What are the boundary conditions? Region I: ( ) 0I xψ = Region II: 1 2( ) sin cosII x B kx B kxψ = + V(x) 0 L ∞ ∞ II I I ψΙ ψΙ ψΙ Key point: The wave function must be continuous at all boundaries. Evidently the wave function in region II must go to zero at both boundaries. Let us now apply this boundary condition and see what it tells us about B1, B2, and k. Left boundary (x = 0): ( 0) ( 0)I IIx xψ ψ= = = ( ) ( )1 20 sin 0 cos 0B B= + 20 B= Particle in a Box: the Infinite Square Well Right boundary (x = L): ( ) ( )I IIx L x Lψ ψ= = = ( )10 sinB kL= B1=0 would work but that is not an interesting solution (zero everywhere). The only non-trivial solutions occur when ...3,2,1, == n L n kn π 2 Using k we find: n 2L π λ λ = = Allowed wavefunctions have an integral # of half-wavelengths that precisely “fit” in the well. This is precisely the condition one finds for an EM wave in a cavity and modes on a string fixed at both ends. Cannot have any wave vector we want … only discrete values that fit the boundary conditions. We also now only have discrete wavelengths: n λ 4 L/2 3 2L/3 2 L 1 2L Physics 486 Lecture 4 Page 10 General properties of bound states Several trends exhibited by the particle-in-box states are generic to bound state wavefunctions in any potential (even complicated ones). (2). The lowest energy bound state always has finite energy -- a “zero-point” energy Why? Even the lowest energy bound state requires some wavefunction curvature (kinetic energy) to satisfy boundary conditions (1) The overall curvature of the wavefunction increases with increasing kinetic energy, T = E-V. 2 2 2 ( ) 2 d x m dx ψ−
2 2 p like m       V=∞ ψ(x) 0 L V=∞ n=1 n=2 x n=3 V=∞ V=∞ 0 xL En n=1 n=2 n=3 T1 T2 T3 General properties of bound states (3). The -th eigenstate has (-1) zero- crossings. (e.g., the first (ground) eigenstate has 0, the second eigenstate has 1, the third eigenstate has 2, etc.,…) (4). If the potential V(x) has a center of symmetry (such as the center of the well above), the eigenstates will be (alternating) even and odd functions about that center of symmetry V=∞ ψ(x) 0 L V=∞n=1 n=2 x n=3 Physics 486 Lecture 4 Page 11 General properties of bound states (5). Effects of a change in potential: A change in potential will be reflected in a change in the curvature of the wavefunction (see rule #1), since the curvature increases with increasing T=E-V(x). But what about the amplitude of the wavefunction? V=∞ 0 L V=∞ x I II Consider the change in potential between regions I and II above. The wavefunction, and its derivative, can be written: sin( )A kxψ φ= + / cos( )d dx Ak kxψ φ= + Squaring both of these relations and adding gives: ( ) ( )22 1 2 2 2 2 2 2 2/ sin ( ) cos ( ) sin ( ) cos ( )k d dx A kx A kx A kx kxψ ψ φ φ φ φ−+ = + + + = + + + ( )22 1 2/k d dx Aψ ψ−+ = ( ) 22 1 /A k d dxψ ψ−= + Across the boundary between I and II, both ψ and dψ/dx MUST be continuous, while k can change discontinuously. So, as k=(2m(E-V)/
2)1/2 decreases across the boundary, A must increase. Classically, we would say that the probability amplitude increases w/ decreasing k because the particle moves more slowly there. Example 1: Transition energies l Calculate ground state energy and energy for a transition. An electron is trapped in a “quantum nanobox” that is L = 4 nm long. Assuming that the potential seen by the electron is approximately that of an infinite square well, estimate the ground (lowest) state energy of the electron. Allowed energies for electron in a 1D box: What photon energy is required to excite the trapped electron to the next available energy level (i.e., n = 2)? 2L/nnλ = 2 1nE n E= So, energy difference between n = 2 and n = 1 levels: ( )2 2 12 1 E E∆ = − 13E= = 0.071 eV E1 = ground state energy (n = 1) V=∞ V=∞ 0 xL En n=1 n=2 n=3 2 2 2 2 2 2 2 1 1 2 2 2 1 2 1.505 2 1.505 with 8 4 1.505 0.0235 eV 4(4 ) n n n n h eV nm E m h eV nm E E n E mL L eV nm E nm λ λ ⋅= = ⋅= = = ⋅= = Physics 486 Lecture 4 Page 12 Example 2: Bound state properties E5 V=∞ V=∞ 0 L x Uo x ψ l Let’s reinforce your intuition about the properties of bound state wavefunctions with this example: lThrough “nano-engineering,” you are able to create a “step” in the potential “seen” by an electron trapped in a 1D structure, as shown below. You’d like to estimate the wavefunction for an electron in the 5th energy level of this potential, without solving the SEQ. Qualitatively sketch the 5th wavefunction: Things to consider: (2) Wavefunction must go to zero at x = 0 and x = L. (3) Kinetic energy is lower on right side of well, so the curvature of ψ is smaller there (wavelength is longer). (4) Kinetic energy is lower on right side, so amplitude of ψ is larger there. (Classically, the particle spends more time there because it is moving more slowly). (1) 5th wavefunction has 4 zero-crossings Example 3: More bound state properties The wavefunction below describes a quantum particle in a range ∆x: ψ(x) x ∆x 8 zero crossings In what energy level is the particle? n = (a) 7 (b) 8 (c) 9