What Does Continuous Derivative Mean in Quantum Mechanics?

[NEWO]

In Quantum Mechanics the one of the constraints on the wavefunction is that its derivative is continuous, what I have a problem with is that I have forgotten what this actually means in terms of an equation.

ie

$$\frac{d\psi){dx}=?[\tex]<br /> <br /> This is driving me nutty and looked on the internet but not found what I am looking for! If some one can help me then that would be great!<br /> <br /> Thank you <br /> <br /> newo$$

 

[malawi_glenn]

Suppose we have this potential shape:

a, V = +inf at x < 0
b, V = - V_0 at 0 < x < a V_0 > 0
c, V = 0 at x > a

We get one wavefuntion at region b, call it psi_1(x) and one for region c, called psi_2(x), by solving the shrödinger equation.

Now psi_1(a) = psi_2(a)
and
d(psi_1(a))/dx = d(psi_2(a))/dx

and we have:

psi_1(a) / [d(psi_1(a))/dx ] = psi_2(a) / [d(psi_2(a))/dx]

was it this you were looking for?

 

[NEWO]

Thats what i was looking for i remember now thanks!

Suppose we have this potential shape:

a, V = +inf at x < 0
b, V = - V_0 at 0 < x < a V_0 > 0
c, V = 0 at x > a

We get one wavefuntion at region b, call it psi_1(x) and one for region c, called psi_2(x), by solving the shrödinger equation.

Now psi_1(a) = psi_2(a)
and
d(psi_1(a))/dx = d(psi_2(a))/dx

and we have:

psi_1(a) / [d(psi_1(a))/dx ] = psi_2(a) / [d(psi_2(a))/dx]

was it this you were looking for?

 

[malawi_glenn]

hehe great, Good luck with the QM! =)

You can also check with the definitions and requriments of an continuous function in calcus books.