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Questions and Answers
The solution to the Schrödinger equation in the region x < a/2 is given by O1(x) = Aek1x.
The solution to the Schrödinger equation in the region x < a/2 is given by O1(x) = Aek1x.
True
The bound state eigenfunctions of symmetric one-dimensional Hamiltonians are always symmetric under space inversion.
The bound state eigenfunctions of symmetric one-dimensional Hamiltonians are always symmetric under space inversion.
False
The solution Oa(x) is symmetric under space inversion.
The solution Oa(x) is symmetric under space inversion.
False
The eigenvalues can be determined using the boundary conditions at x = 0.
The eigenvalues can be determined using the boundary conditions at x = 0.
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The solution Os(x) is antisymmetric under space inversion.
The solution Os(x) is antisymmetric under space inversion.
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The solution to the Schrödinger equation in the region x > a/2 is given by O3(x) = Dek1x.
The solution to the Schrödinger equation in the region x > a/2 is given by O3(x) = Dek1x.
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The eigenfunctions of the Schrödinger equation are always continuous at x = ±a/2.
The eigenfunctions of the Schrödinger equation are always continuous at x = ±a/2.
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The solutions of the Schrödinger equation are either odd or even under space inversion.
The solutions of the Schrödinger equation are either odd or even under space inversion.
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The radius R of the circle becomes zero when V0 approaches infinity.
The radius R of the circle becomes zero when V0 approaches infinity.
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When V0 approaches infinity, both tan(n) and cot(n) become zero.
When V0 approaches infinity, both tan(n) and cot(n) become zero.
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The function R^2 : n^2 crosses :n cot(n) and :n tan(n) at the asymptotes :n = nH/2.
The function R^2 : n^2 crosses :n cot(n) and :n tan(n) at the asymptotes :n = nH/2.
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The energy expression for the infinite well is given by :n^2 = ma^2 E_n / (2h^2).
The energy expression for the infinite well is given by :n^2 = ma^2 E_n / (2h^2).
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The equation :n^2 = ma^2 E_n / (2h^2) is derived from the equation :n^2 : n^2 = ma^2 E_n / h^2.
The equation :n^2 = ma^2 E_n / (2h^2) is derived from the equation :n^2 : n^2 = ma^2 E_n / h^2.
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The equation :n^2 = 2ma^2 E_n / h^2 is only valid for finite values of V0.
The equation :n^2 = 2ma^2 E_n / h^2 is only valid for finite values of V0.
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There is only one bound state resulting from the intersection of 4\tan^2:\0 and \tan:\0.
There is only one bound state resulting from the intersection of 4\tan^2:\0 and \tan:\0.
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The number of bound states in a finite square well potential depends on the size of R, which in turn depends on the width a of the well.
The number of bound states in a finite square well potential depends on the size of R, which in turn depends on the width a of the well.
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The harmonic oscillator is only used in classical mechanics.
The harmonic oscillator is only used in classical mechanics.
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There is always more than one bound state in a finite square well potential, regardless of the value of V0.
There is always more than one bound state in a finite square well potential, regardless of the value of V0.
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The equation :0\tan:\0 = 4*\tan^2:\0 + 4*\cos^2:\0 is derived from the numerical solutions of the corresponding equations.
The equation :0\tan:\0 = 4*\tan^2:\0 + 4*\cos^2:\0 is derived from the numerical solutions of the corresponding equations.
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The ground state of a finite square well potential is always an odd state.
The ground state of a finite square well potential is always an odd state.
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The number of bound states in a finite square well potential increases as the depth V0 of the well decreases.
The number of bound states in a finite square well potential increases as the depth V0 of the well decreases.
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The harmonic oscillator is not important in quantum mechanics.
The harmonic oscillator is not important in quantum mechanics.
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The values of :0 and :1 are exactly 0 and 1, respectively.
The values of :0 and :1 are exactly 0 and 1, respectively.
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The energy of the bound states in a finite square well potential is proportional to $\frac{1}{a^2}$.
The energy of the bound states in a finite square well potential is proportional to $\frac{1}{a^2}$.
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The equation $R = \frac{ma^2V0}{2\hbar^2}$ is used to determine the number of bound states in a finite square well potential.
The equation $R = \frac{ma^2V0}{2\hbar^2}$ is used to determine the number of bound states in a finite square well potential.
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The energy of the first bound state E1 is greater than the energy of the zeroth bound state E0.
The energy of the first bound state E1 is greater than the energy of the zeroth bound state E0.
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The harmonic oscillator is only used in theoretical physics.
The harmonic oscillator is only used in theoretical physics.
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The values of :0 and :1 are exact values.
The values of :0 and :1 are exact values.
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The power series method is used to solve the Schrödinger equation in the analytic method.
The power series method is used to solve the Schrödinger equation in the analytic method.
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The constant x0 has the dimensions of energy.
The constant x0 has the dimensions of energy.
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The solutions of the differential equation are expressed in terms of trigonometric functions.
The solutions of the differential equation are expressed in terms of trigonometric functions.
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A Gaussian type solution is suggested by the term x in the differential equation.
A Gaussian type solution is suggested by the term x in the differential equation.
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The solution O(x) = f(x) exp(-x2/2) is physically acceptable.
The solution O(x) = f(x) exp(-x2/2) is physically acceptable.
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The differential equation (4.116) is a linear equation.
The differential equation (4.116) is a linear equation.
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Study Notes
Schrödinger Equation and Solutions
- The solution to the Schrödinger equation in the regions x < a/2 and x > a/2 can be written as:
- O1(x) = Ae^(k1x) for x < a/2
- O3(x) = De^(-k1x) for x > a/2
- The bound state eigenfunctions of symmetric one-dimensional Hamiltonians are either even or odd under space inversion.
- The solutions of the Schrödinger equation can be written as:
- Odd (antisymmetric) solution: Oa(x) = C sin(k2x) for x < a/2
- Even (symmetric) solution: Os(x) = B cos(k2x) for x < a/2
Bound States and Eigenvalues
- The number of bound states depends on the size of R, which in turn depends on the depth V0 and the width a of the well.
- The deeper and broader the well, the larger the value of R, and hence the greater the number of bound states.
- There is always at least one bound state, no matter how small V0 is.
- When R < π/2, there is only one bound state corresponding to n = 0.
- When R > π/2, there are two bound states: an even state (the ground state) corresponding to n = 0 and the first odd state corresponding to n = 1.
Limiting Case
- In the limiting case V0 → ∞, the function R^2 :n^2 will cross :n cot :n and :n tan :n at the asymptotes :n = nπ/2.
- The energy expression for the infinite well is recovered: E_n = (n^2π^2)/(2ma^2).
Harmonic Oscillator
- The harmonic oscillator is a useful model for a variety of vibrational phenomena in physics.
- The Schrödinger equation for the harmonic oscillator can be reduced to a differential equation: d^2O(x)/dx^2 + (2mE/x^2 - mω^2)O(x) = 0.
- The solutions of the differential equation can be expressed in terms of Hermite polynomials.
- The occurrence of the term x^2O(x) in the differential equation suggests trying a Gaussian type solution.
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Description
This quiz covers solutions to the Schrödinger equation in different regions of x. It involves solving the equation and eliminating physically unacceptable solutions. The solution is expressed in terms of exponential functions.
