For the one-dimensional case on the
x-axis, the
time-independent Schrödinger equation can be written as: {{NumBlk|| -\frac{\hbar^2}{2 m} \frac{d^2 \psi}{d x^2} + V(x) \psi = E \psi |}} where • \hbar is the
reduced Planck constant, • m is the
mass of the particle, • V(x) is the
potential energy at each point
x, • \psi is the (complex valued)
wavefunction, or "
eigenfunction", and • E is the
energy, a
real number, sometimes called eigenenergy. For the case of the particle in a one-dimensional box of length
L, the potential is V_0 outside the box, and zero for
x between -L/2 and L/2. The wavefunction is composed of different wavefunctions; depending on whether
x is inside or outside of the box, such that: \psi = \begin{cases} \psi_1, & \text{if }xL/2\text{ (the region outside the well)} \end{cases}
Inside the box For the region inside the box,
V(
x) = 0 and Equation 1 reduces to -\frac{\hbar^2}{2 m} \frac{d^2 \psi_2}{d x^2} = E \psi_2,
resembling the time-independent
free schrödinger equation, hence E = \frac{k^2 \hbar^2}{2m} . Letting k = \frac{\sqrt{2mE}}{\hbar}, the equation becomes \frac{d^2 \psi_2}{d x^2} = -k^2 \psi_2 . with a general solution of \psi_2 = A \sin(kx) + B \cos(kx)\, . where
A and
B can be any
complex numbers, and
k can be any real number.
Outside the box For the region outside of the box, since the potential is constant, V(x) = V_0 and equation becomes: -\frac{\hbar^2}{2 m} \frac{d^2 \psi_1}{d x^2} = ( E - V_0) \psi_1 There are two possible families of solutions, depending on whether
E is less than V_0 (the particle is in a bound state) or
E is greater than V_0 (the particle is in an unbounded state). If we solve the time-independent Schrödinger equation for an energy E > V_0, letting k' = \frac{\sqrt{2m(E - V_0)}}{\hbar} such that \frac{d^2 \psi_1}{d x^2} = -k'^2 \psi_1 then the solution has the same form as the inside-well case: \psi_1 = C \sin(k' x) + D \cos(k' x) and, hence, will be oscillatory both inside and outside the well. Thus, the solution is never square integrable; that is, it is always a non-normalizable state. This does not mean, however, that it is impossible for a quantum particle to have energy greater than V_0, it merely means that the system has continuous spectrum above V_0, i.e., the non-normalizable states still contribute to the continuous part of the spectrum as
generalized eigenfunctions of an
unbounded operator. This analysis will focus on the bound state, where E . Letting \alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar} produces \frac{d^2 \psi_1}{d x^2} = \alpha^2 \psi_1 where the general solution is a real exponential: \psi_1 = Fe^{- \alpha x}+ Ge^{ \alpha x} Similarly, for the other region outside the box: \psi_3 = He^{- \alpha x}+ Ie^{ \alpha x} Now in order to find the specific solution for the problem at hand, we must specify the appropriate boundary conditions and find the values for
A,
B,
F,
G,
H and
I that satisfy those conditions.
Finding wavefunctions for the bound state Solutions to the Schrödinger equation must be continuous, and continuously differentiable. These requirements are
boundary conditions on the differential equations previously derived, that is, the matching conditions between the solutions inside and outside the well. In this case, the finite potential well is symmetrical, so symmetry can be exploited to reduce the necessary calculations. Summarizing the previous sections: \psi = \begin{cases} \psi_1, & \text{if }x L/2\text{ (the region outside the box)} \end{cases} where we found \psi_1, \psi_2 , and \psi_3 to be: \begin{align} \psi_1 &= Fe^{- \alpha x}+ Ge^{ \alpha x} \\ \psi_2 &= A \sin(k x) + B \cos(k x) \\ \psi_3 &= He^{- \alpha x}+ Ie^{ \alpha x} \end{align} We see that as x goes to -\infty, the F term goes to infinity. Likewise, as x goes to +\infty, the I term goes to infinity. In order for the wave function to be square integrable, we must set F = I = 0, and we have: \psi_1 = Ge^{ \alpha x} and \psi_3 = He^{- \alpha x} Next, we know that the overall \psi function must be continuous and differentiable. In other words, the values of the functions and their derivatives must match up at the dividing points: These equations have two sorts of solutions, symmetric, for which A = 0 and G = H, and antisymmetric, for which B = 0 and G=-H. For the symmetric case we get He^{- \alpha L/2} = B \cos(k L/2) - \alpha He^{- \alpha L/2} = - k B \sin(k L/2) so taking the ratio gives \alpha=k \tan(k L/2) . Similarly for the antisymmetric case we get \alpha=-k \cot(k L/2) . Recall that both \alpha and k depend on the energy. What we have found is that the continuity conditions
cannot be satisfied for an arbitrary value of the energy; because that is a result of the infinite potential well case. Thus, only certain energy values, which are solutions to one or either of these two equations, are allowed. Hence we find that the energy levels of the system below V_0 are discrete; the corresponding eigenfunctions are
bound states. (By contrast, for the energy levels above V_0 are continuous.) The energy equations cannot be solved analytically. Nevertheless, we will see that in the symmetric case, there always exists at least one bound state, even if the well is very shallow. Graphical or numerical solutions to the energy equations are aided by rewriting them a little and it should be mentioned that a nice approximation method has been found by Lima which works for any pair of parameters L and V_0. If we introduce the dimensionless variables u=\alpha L/2 and v=k L/2 , and note from the definitions of \alpha and k that u^2 = u_0^2-v^2, where u_0^2 = \frac{1}{2}m L^2 V_0 \hbar^{-2} , the master equations read \sqrt{u_0^2-v^2} = \begin{cases} v \tan v, & \text{(symmetric case) } \\ -v \cot v, & \text{(antisymmetric case) } \end{cases} In the plot to the right, for u_0^2=20, solutions exist where the blue semicircle intersects the purple or grey curves (v \tan v and -v \cot v). Each purple or grey curve represents a possible solution, v_j within the range \frac{\pi}{2}(j-1) \leq v_i . The total number of solutions, N, (i.e., the number of purple/grey curves that are intersected by the blue circle) is therefore determined by dividing the radius of the blue circle, u_0, by the range of each solution \pi/2 and using the floor or ceiling functions: N = \left\lfloor\frac{2u_0}{\pi}\right\rfloor+1=\left\lceil\frac{2u_0}{\pi}\right\rceil In this case there are exactly three solutions, since N = \lceil 2\sqrt{20}/\pi\rceil = \lceil 2.85 \rceil = 3. v_1 =1.28, v_2=2.54 and v_3=3.73, with the corresponding energies E_n={2\hbar^2 v_n^2\over m L^2} . If we want, we can go back and find the values of the constants A, B, G, H in the equations now (we also need to impose the normalisation condition). On the right we show the energy levels and wave functions in this case (where x_0\equiv\hbar/\sqrt{2m V_0}). We note that however small u_0 is (however shallow or narrow the well), there is always at least one bound state. Two special cases are worth noting. As the height of the potential becomes large, V_0\to\infty, the radius of the semicircle gets larger and the roots get closer and closer to the values v_n=n\pi/2, and we recover the case of the
infinite square well. The other case is that of a very narrow, deep well - specifically the case V_0\to\infty and L\to 0 with V_0 L fixed. As u_0\propto \sqrt{V_0} L it will tend to zero, and so there will only be one bound state. The approximate solution is then v^2 = u_0^2 - u_0^4, and the energy tends to E=-m L^2 V_0^2/2\hbar^2. But this is just the energy of the bound state of a
Delta function potential of strength V_0 L, as it should be. A simpler graphical solution for the energy levels can be obtained by normalizing the potential and the energy through multiplication by {8 m}{L^2} / h^2 . The normalized quantities are \tilde{V}_0= V_0 \frac{8 m}{h^2} L^2 \qquad \tilde{E}= E \frac{8 m}{h^2} L^2 giving directly the relation between the allowed couples (V_0, E) as \sqrt{\tilde{V}_0}={\sqrt{\tilde{E}}}\, \left|{\sec(\sqrt{\tilde{E}} \, {\pi}/{2})}\right|, \qquad \sqrt{\tilde{V}_0} = {\sqrt{\tilde{E}}}\, \left|{\csc(\sqrt{\tilde{E}} \, {\pi}/{2})}\right| for the even and odd parity wave functions, respectively. In the previous equations only the positive derivative parts of the functions have to be considered. The chart giving directly the allowed couples (V_0, E) is reported in the figure.
Asymmetric well Consider a one-dimensional asymmetric potential well given by the potential V(x) = \begin{cases} V_1, & \text{if }-\infty with V_2 > V_1. The corresponding solution for the wave function with E is found to be \psi(x) = \begin{cases} c_1 e^{k_1 x}, & \text{for }xa, \text{where } k_2 = \sqrt{(2m/\hbar^2)(V_2-E)} \end{cases} and \sin\delta = \frac{k\hbar}{\sqrt{2mV_1}}. The energy levels E=k^2\hbar^2/(2m) are determined once k is solved as a root of the following transcendental equation ka = n\pi - \sin^{-1}\left(\frac{k\hbar}{\sqrt{2mV_1}}\right) - \sin^{-1}\left(\frac{k\hbar}{\sqrt{2mV_2}}\right) where n=1,2,3,\dots Existence of root to above equation is not always guaranteed, for example, one can always find a value of a so small, that for given values of V_1 and V_2, there exists no discrete energy level. The results of symmetrical well is obtained from above equation by setting V_1 = V_2 = V_o. == Particle in a spherical potential well ==