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ฟิสิกส์โอลิมปิก วิทยาศาสตร์โอลิมปิก ข้อสอบแข่งขัน ข้อสอบชิงทุน => GRE - Physics => Topic started by: conantee on December 31, 2009, 10:05:20 AM



Title: GR8677.018
Post by: conantee on December 31, 2009, 10:05:20 AM
18. The wavefunction \psi (x) = A \mbox{ exp} [ - \frac{b^2 x^2}{2}] , where A and b are real constants, is a normalized eigenfunction of the Schrodinger equation for a particle of mass M and energy E in a one dimensional potential V(x) such that V(x) = 0 at x=0. Which of the following is correct?
(A) V = \frac{\hslash ^2 b^4}{2M}
(B) V = \frac{\hslash ^2 b^4 x^2}{2M}
(C) V = \frac{\hslash ^2 b^6 x^4}{2M}
(D) E = \hslash ^2 b^2 (1 - b^2 x^2)
(E) E = \frac{\hslash ^2 b^4}{2M}


Title: Re: GR8677.018
Post by: conantee on December 31, 2009, 10:26:29 AM
First, energy of eigenstates does not depend on x-coordinate, so (D) is wrong.
Next, we can obtain the dimension of energy from schrodinger's equation. The kinetic energy term is -\frac{\hslash ^2}{2M} \frac{\partial ^2 \psi}{\partial x^2}. So, the dimension of energy is \frac{\hslash ^2}{mass} \frac{1}{length ^2}. Well, from the wave function \psi (x) , b has dimension of 1/length. Hence, dimension of energy in (E) is wrong.
Next, we know that V(x) = 0 at x = 0, so (A) is also wrong.

Now, let's think about Schrodinger equation for eigenstates.
-\frac{\hslash ^2}{2M} \frac{\partial ^2 \psi}{\partial x^2} + V(x) \psi = E \psi.
We can plug \psi (x) and find V(x). However, full calculation is not necessary. Since the kinetic energy term involves with just second derivative, there is no x^4 coming from Gaussian function. So, (C) is wrong.

The correct is therefore (B). 30 of 100 people got the correct answer.