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Show that there is no acceptable solution to the Schrodinger equation for the infinite square well with E=0orE<0(This is a special case of the general theorem in Problem 2.2, but this time do it by explicitly solving the Schrodinger equation, and showing that you cannot meet the boundary conditions.)

Short Answer

Expert verified

Ecannot be less than zero because nis squared, also, it cannot be zero because if E=0then this will violate the uncertainty principle.

Step by step solution

01

Significance of the Schrodinger equation

The position of a particular electron is described in space and time using Schrodinger equation in the form of a mathematical expression. It considered the wave nature of the electron.

02

Identification of solution of the Schrodinger equation for infinite square wall

Write equation (2.27).

En=n2222ma2

Encannot be negative because all the constants are positive or squared (that is., if we take n=1,2, the square will kill the negative signs, so Encannot be less than zero).

The other case in this problem when E=0(i.e.,n=0 ) is not accepted because Eis the eigenvalue of the total energy operator (we can think about it as the total energy of the wave function) which cannot be zero because this will violate the uncertainty principle.

Since if E=0this corresponds to zero kinetic energy (because E in the potential energy in the infinite potential well is zero), and zero kinetic energy corresponds to zero momentum, and the particle will rest inside the potential.

The calculation of the minimum kinetic energy by restricting the particle to move in a region of width xa(i.e., the length of the well) can be done, then using the uncertainty principle.

The calculation of the minimum uncertainty in the momentum can be done, so the result can bep/(2a) , so the minimum kinetic energy is2/(8ma2)where this result qualitatively agrees with the exact value of the ground state.

Thus, Ecannot be less than zero because nis squared, also, it cannot be zero because if E=0then this will violate the uncertainty principle.

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Most popular questions from this chapter

In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials.

a. The Rodrigues formula says thatHn()=(1)ne2(d诲尉)ne2

Use it to derive H3 and H4 .

b. The following recursion relation gives you Hn+1 in terms of the two preceding Hermite polynomials: Hn+1()=2尉贬n()2nHn1()

Use it, together with your answer in (a), to obtain H5 and H6 .

(c) If you differentiate an nth-order polynomial, you get a polynomial of

Order (n-1). For the Hermite polynomials, in fact,

dHn诲尉=2nHn1()

Check this, by differentiatingH5and H6.

d. Hn()is the nth z-derivative, at z = 0, of the generating function exp(z2+2z)or, to put it another way, it is the coefficient ofznn! in the Taylor series expansion for this function: ez2+2尉锄=n=0znn!Hn()

Use this to obtain H0,H1and H2.

The transfer matrix. The S- matrix (Problem 2.52) tells you the outgoing amplitudes (B and F)in terms of the incoming amplitudes (A and G) -Equation 2.175For some purposes it is more convenient to work with the transfer matrix, M, which gives you the amplitudes to the right of the potential (F and G)in terms of those to the left (A and b):

(FG)=(M11M12M21M22)(AB)[2.178]

(a) Find the four elements of the M-matrix, in terms of the elements of theS-matrix, and vice versa. ExpressRI,TI,RrandTr(Equations 2.176and 2.177) in terms of elements of the M-matrix.,

(b) Suppose you have a potential consisting of two isolated pieces (Figure 2.23 ). Show that the M-matrix for the combination is the product of the twoM-matrices for each section separately: M=M2M1[2.179]

(This obviously generalizes to any number of pieces, and accounts for the usefulness of the M-matrix.)

FIGURE : A potential consisting of two isolated pieces (Problem 2.53 ).

(c) Construct the -matrix for scattering from a single delta-function potential at point V(x)=-(x-a) :

(d) By the method of part , find the M-matrix for scattering from the double delta functionV(x)=-[(x+a)+(X-a)] .What is the transmission coefficient for this potential?

Solve the time-independent Schrodinger equation with appropriate boundary conditions for the 鈥渃entered鈥 infinite square well: V(x)=0(for-a<x<+a), V(x)=(otherwise). Check that your allowed energies are consistent with mine (Equation 2.30), and confirm that your 'scan be obtained from mine (Equation 2.31) by the substitution x 鈫 (x + a)/2 (and appropriate renormalization). Sketch your first three solutions, and compare Figure 2.2. Note that the width of the well is now 2a.

In the ground state of the harmonic oscillator, what is the probability (correct to three significant digits) of finding the particle outside the classically allowed region?

Delta functions live under integral signs, and two expressions (D1xandD2x)involving delta functions are said to be equal if

-+f(x)D1(x)dx=-+f(x)D2(x)dxfor every (ordinary) function f(x).

(a) Show that

(cx)=1|c|(x)(2.145)

where c is a real constant. (Be sure to check the case where c is negative.)

(b) Let (x) be the step function:

(x){1,x>00,x>0(2.146).

(In the rare case where it actually matters, we define (0) to be 1/2.) Show that dldx=

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