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Two particles in a box have a total energy 522/2mL2.

(a) Which states are occupied?

(b) Make a sketch ofP5(x1,x2)versusx1for points along the linex2=x1.

(c) Make a similar sketch ofPA(x1,x2).

(d) Repeat parts (b) and (c) but for points on the linex2=Lx1

Short Answer

Expert verified

(a) The state of one particle is n1=1 and other particles isn2=2

(b) The sketch is shown in figure 1

(c) The sketch is shown in figure 2

(d) The sketch is shown in figure 1 with curve B for symmetric function and curve A for asymmetric function.

Step by step solution

01

Given information:

The total energy of particle is 522/2mL2.

02

Concept of probability density and energy

(a) The expression for energy of two particles in one dimension is given by,

E=n122h22mL2+n222h22mL2

(b) The expression for probability density is given by,role="math" localid="1659954573251" P(x)=[1L(sin蟺虫Lsin2蟺虫Lsin2蟺虫Lsin蟺虫L)]2

(c) The expression for probability density is given by,

role="math" localid="1659954661753" P(x)=[1L(sin蟺虫Lsin2蟺虫Lsin2蟺虫Lsin蟺虫L)]2

(d) The expression for probability density is given by,

P(x)=[1Lsin蟺虫1Lsin2蟺虫2Lsin2蟺虫1Lsin蟺虫2L]2

03

Evaluate total energy 

(a)

Suppose n1=1and n2=2.

The total energy is calculated as,

E=n12222mL2+n22222mL2=(1)2222mL2+(2)2222mL2

04

Evaluate probability density of (b)

(b)

The probability density for symmetric function is calculated as,

P(x)=1Lsin蟺虫Lsin2蟺虫L+sin2蟺虫Lsin蟺虫L2=4L2sin2蟺虫Lsin22蟺虫L

The graph between probability density and xfor symmetric and asymmetric function is shown below,

Figure 1: Curve 1 shows the symmetric function and curve 2 shown asymmetric functions.

05

Evaluate probability density of (c)

(c)

The probability density for anti-symmetric function is calculated as,

P(x)=1Lsin蟺虫Lsin2蟺虫Lsin2蟺虫Lsin蟺虫L2=0

The graph between probability density and xfor symmetric and asymmetric function is shown below,

Figure 2: Curve 1 shows the symmetric function and curve 2 shown asymmetric function

06

Evaluate probability density of (d)

(d)

The probability density is calculated as,

Because of the extra minus sign in the above equation comparison between part (b) and (c) shows P(x) for the symmetric wave function along the path is equal toP(x)for anti-symmetric case in case of part (b) and (c), Corresponding to curve B in figure 1 .

Similarly for anti-symmetric wave function along the path is equal to P(x)for symmetric case in case of part (b) and (c), corresponding to curve in figure 1.

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