The quantum "box" actually called a 1-dimensional infinite square well, is perhaps the simplest quantum system to which first year students in QM are exposed. It is also an excellent case to see how differential equations are applied.

One notes the box limits are from 0 to a and this is the x -direction. Inside the box the potential V = 0 and outside, V = ¥, hence the term, "infinite square well".

From these constraints, one sees that any particle cannot be at locations, x > a or less than 0, else it would have infinite energy. As we know, energies in QM are quantized, so no infinities are possible.

Therefore, outside the interval (0,a) one has the quantum wave form:

**y*y**

**= 0**

That is the product of the wave function by its complex conjugate = 0. The Schrodinger wave equation is written (for one dimension):

**H^ y = E y**

**where H^ denotes the Hamiltonian operator:**

**H^ = [-iħ d/dx]**

^{2}/2mThen the differential 1-D Schrodinger equation to solve becomes:

**dy**

^{2}/dx^{2}+ K^{2}y**= 0,**

**where K = Ö**

**[2mE] / ħ**

And this second order differential equation has the solution:

y = C sin Kx + D cos Kx

**(C, D constants)**

**We set boundary conditions thus:**

**At x = 0+, y**

**(0+) = D (since sin(0) = 0)**

**At x = 0, y**

**(0) = 0**

As a cautionary note, one must postulate that the wave function is continuous (since we want y*y to represent something physically observable).

**Then:**

**y (0+) = y**

**(0-) therefore D = 0**

**and:**

**y**

**= C sin Kx**

**Now, at x = a+, y**

**(a+) = 0**

**At x = a-, y**

**(a-) = C sin K(a)**

*For continuity*: y (a+) = y**(a-)**

Therefore: C sin K(a-) = 0

Therefore: sin K(a-) = 0

Þ sin (Ö[2mE] / ħ) a = 0, and

Therefore: C sin K(a-) = 0

__but C can't = 0__, or no particle!Therefore: sin K(a-) = 0

Þ sin (Ö[2mE] / ħ) a = 0, and

**((Ö[2mE] / ħ) a = n π**

**Therefore:**

**a = nπ/K and n =**

__+__1,__+__2, ..etc.Thus: 2mE = n

^{2}π

^{2}ħ

^{2}

E = { n

^{2}π

^{2}ħ

^{2}} / 2ma

^{2}

(which yields the energy eigenvalues.)

E = { n

Thus, E

E = { n

^{2}π

^{2}ħ

^{2}}/ 2ma

^{2}(E

_{n}= 1, 2, 3, .....n)

Thus, E

_{n }is restricted to discrete values:

**π**

^{2}ħ^{2}/ 2ma^{2},**4 π**^{2}ħ^{2}/ 2ma^{2},**9 π**^{2}ħ^{2}/ 2ma^{2}etc.Thus one sees the very basis of quantum mechanics, in the fact that energies associated with the quantum system must be quantized. Energies do not occur in unending streams but rather parceled out in "chunks" we call quanta. These differ according to the energy level in the system. One may also be concerned with where the particle is within the gien system. In this case one uses the probability density

**‖y‖**.

^{2}In general the probability P =

**‖**

**y‖**dV

^{2 }For a 1-dimensional system such as we're examining: P =

**‖**

**y‖**dx

^{2 }
The probability P

_{ab}of finding a particle between a and b:
P

_{ab }=**ò**^{b}_{a}_{ }‖y‖^{2 }dxProblems for the Math Maven:

1) A particle m is located in an infinite well, trapped between positions x = 0 and x = L. If it is found that L = 0.2nm then determine the energy levels (

**E**) for the states n=1 and n = 2. (Take m to be the same mass as an electron).

_{n}2) What is the probability the particle will be found between x = 0 and x = L/4?

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