Solving for C1 and C2: A Wave Function Boundary Condition

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zhillyz
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Homework Statement



A one-dimensional wave function associated with a localized particle can be written as

[itex]\varphi (x) = \begin{cases}<br /> 1- \frac{x^2}{8}, & \text{if } 0<x<4, \\<br /> C_1 - \frac{C_2}{x^2}, & \text{if} \,x \geq 4.<br /> \end{cases}[/itex]

Determine [itex]C_1[/itex] and [itex]C_2[/itex] for which this wave function will obey the boundary condition of continuity at x = 4.

Homework Equations



N\A

The Attempt at a Solution



So I am thinking the boundary condition is to make sure both equations hold at x = 4, and fed into the first equation it equals -1 so equate the second to -1 also and find values for [itex]C_1 \text{and} C_2[/itex] which would be 1 and 32 respectively? Is this correct because the question is worth 6marks which seems like a lot.
 
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Is [itex]C_1=1[/itex] and [itex]C_2=32[/itex] the only solution to [itex]-1=C_1-\frac{C_2}{16}[/itex]?

You have two unknowns and one equation, so if you want a unique solution you will need one more independent equation for [itex]C_1[/itex] and [itex]C_2[/itex]. What can you say about [itex]\varphi'(x)[/itex]?
 


[itex]16C_1+16 = C_2[/itex] So for values of C_1 = 1,2,3,4 C_2 will = 32,48,64,80 respectively.

or

[itex]C_2(n) = C_2(n-1) +16[/itex]

The first order differential of [itex]\varphi[/itex]? Em that it would be part of the shrodinger equation?
 


zhillyz said:
[itex]16C_1+16 = C_2[/itex] So for values of C_1 = 1,2,3,4 C_2 will = 32,48,64,80 respectively.

or

[itex]C_2(n) = C_2(n-1) +16[/itex]

Who says that the constants have to be integers? There are an infinite number of solutions.

The first order differential of [itex]\varphi[/itex]? Em that it would be part of the shrodinger equation?

You need to review your notes/textbook on the boundary conditions of the wavefunction. For a finite potential/barrier, the first derivative of the wavefunction must be continuous.