Question

What are the most likely locations of a particle in a box of length L in the state n = 2 and where are the nodes for this wavefunction? Express your ansers in terms (fractions) of the length L.

Answer 1

The second-order differential equation

−ℏ22md2ψdx2+V(x)ψ=Eψ−ℏ22md2ψdx2+V(x)ψ=Eψ

has two solutions (because it is a second order equation) in the region between x=0x=0 and x=Lx=L where V(x)=0V(x)=0:

ψ=sin(kx)ψ=sin⁡(kx)

and

ψ=cos(kx),ψ=cos⁡(kx),

where kk is defined as

k=2mE/ℏ2−−−−−−−√.k=2mE/ℏ2.

Hence, the most general solution is some combination of these two:

ψ=Asin(kx)+Bcos(kx).ψ=Asin⁡(kx)+Bcos⁡(kx).

We could, alternatively use exp(ikx)exp⁡(ikx) and exp(−ikx)exp⁡(−ikx) as the two independent solutions (we do so later in Sec. 1.4 to illustrate) because sin(kx)sin⁡(kx) and cos(kx)cos⁡(kx) can be rewritten in terms of exp(ikx)exp⁡(ikx) and exp(−ikx)exp⁡(−ikx); that is, they span exactly the same space.

The fact that ψψ must vanish at x=0x=0 (n.b., ψψ vanishes for x<0x<0 because V(x)V(x) is infinite there and ψψ is continuous, so it must vanish at the point x=0x=0) means that the weighting amplitude of the cos(kx)cos⁡(kx) term must vanish because cos(kx)=1cos⁡(kx)=1 at x=0x=0. That is,

B=0.B=0.

The amplitude of the sin(kx)sin⁡(kx) term is not affected by the condition that ψψ vanish at x=0x=0, since sin(kx)sin⁡(kx) itself vanishes at x=0x=0. So, now we know that ψψ is really of the form:

ψ(x)=Asin(kx).ψ(x)=Asin⁡(kx).

The condition that ψψ also vanish at x=Lx=L (because it vanishes for x<0x<0 where V(x)V(x) again is infinite) has two possible implications. Either A=0A=0 or kk must be such that sin(kL)=0sin⁡(kL)=0. The option A=0A=0 would lead to an answer ψψ that vanishes at all values of xx and thus a probability that vanishes everywhere. This is unacceptable because it would imply that the particle is never observed anywhere.

            The other possibility is that sin(kL)=0sin⁡(kL)=0. Let’s explore this answer because it offers the first example of energy quantization that you have probably encountered. As you know, the sin function vanishes at integral multiples of pp. Hence kLkL must be some multiple of ππ; let’s call the integer nn and write Lk=nπLk=nπ (using the definition of kk) in the form:

L2mEℏ2−−−−−√=nπ.L2mEℏ2=nπ.

Solving this equation for the energy EE, we obtain:

E=n2π2ℏ22mL2E=n2π2ℏ22mL2

This result says that the only energy values that are capable of giving a wave function ψ(x)ψ(x) that will obey the above conditions are these specific EE values. In other words, not all energy values are allowed in the sense that they can produce ψψ functions that are continuous and vanish in regions where V(x)V(x) is infinite. If one uses an energy EE that is not one of the allowed values and substitutes this EE into sin(kx)sin⁡(kx), the resultant function will not vanish at x=Lx=L. I hope the solution to this problem reminds you of the violin string that we discussed earlier. Recall that the violin string being tied down at x=0x=0 and at x=Lx=L gave rise to quantization of the wavelength just as the conditions that ψψ be continuous at x=0x=0 and x=Lx=L gave energy quantization.

Substituting k=nπ/Lk=nπ/L into ψ=Asin(kx)ψ=Asin⁡(kx) gives

ψ(x)=Asin(npxL).ψ(x)=Asin⁡(npxL).

The value of A can be found by remembering that |ψ|2|ψ|2  is supposed to represent the probability density for finding the particle at xx. Such probability densities are supposed to be normalized, meaning that their integral over all xx values should amount to unity. So, we can find A by requiring that

1=∫|ψ(x)|2dx=|A|2∫sin2(npxL)​dx1=∫|ψ(x)|2dx=|A|2∫sin2⁡(npxL)​dx

​

where the integral ranges from x=0x=0 to x=Lx=L. Looking up the integral of sin2(ax)sin2⁡(ax) and solving the above equation for the so-called normalization constant AA gives

A=2L−−√A=2L  and so

ψ(x)=2L−−√sin(npxL).ψ(x)=2Lsin⁡(npxL).

The values that nn can take on are n=1,2,3,⋯n=1,2,3,⋯; the choice n=0n=0 is unacceptable because it would produce a wave function ψ(x)ψ(x) that vanishes at all xx.

The full x- and t- dependent wave functions are then given as

Ψ(x,t)=2L−−√sin(npxL)exp[−itℏn2π2ℏ22mL2].Ψ(x,t)=2Lsin⁡(npxL)exp⁡[−itℏn2π2ℏ22mL2].

Notice that the spatial probability density |Ψ(x,t)|2|Ψ(x,t)|2 is not dependent on time and is equal to |ψ(x)|2|ψ(x)|2 because the complex exponential disappears when Ψ∗ΨΨ∗Ψ is formed. This means that the probability of finding the particle at various values of xx is time-independent.

Another thing I want you to notice is that, unlike the classical dynamics case, not all energy values EE are allowed. In the Newtonian dynamics situation, EE could be specified and the particle’s momentum at any xx value was then determined to within a sign. In contrast, in quantum mechanics, one must determine, by solving the Schrödinger equation, what the allowed values of EE are. These EE values are quantized, meaning that they occur only for discrete values E=n2π2h22mL2E=n2π2h22mL2 determined by a quantum number nn, by the mass of the particle m, and by characteristics of the potential (LL in this case).