- #1

Irishdoug

- 102

- 16

- Homework Statement
- Self studying Quantum Mechanics. I've a finite square well. Inside the well the V(x) = 0. Thus the problem becomes that of the free particle. I'm aware the solution to the SE is Asinkx + Bcoskx however I can't figure out how.

- Relevant Equations
- ##\psi ##'' = ##-k^2 \psi ## were k = ##(2mE)^{0.5}## / ##\hbar##

I've tried to carry out the solution to this as a normal 2nd order Differential Equation

##\psi ##'' - ##-k^2 \psi ## = 0

Assume solution has form ##e^{\gamma x}##

sub this in form ##\psi## and get

##\gamma ^2## ##e^{\gamma x} ## + ##k^2 e^{\gamma x}## = 0

Solution is ##\gamma## = 0 or ##k^2##

Now have two real roots that are not equal thus have

c1##e^{\gamma_1 x}## + c2##e^{\gamma_2 x}##

I presume this is wrong, as I cannot figure out how to turn it into c1sinkx +c2coskx.

I'm aware of Eulers equation but I've no imaginary number and just using the real part means I've no sin term.

##\psi ##'' - ##-k^2 \psi ## = 0

Assume solution has form ##e^{\gamma x}##

sub this in form ##\psi## and get

##\gamma ^2## ##e^{\gamma x} ## + ##k^2 e^{\gamma x}## = 0

Solution is ##\gamma## = 0 or ##k^2##

Now have two real roots that are not equal thus have

c1##e^{\gamma_1 x}## + c2##e^{\gamma_2 x}##

I presume this is wrong, as I cannot figure out how to turn it into c1sinkx +c2coskx.

I'm aware of Eulers equation but I've no imaginary number and just using the real part means I've no sin term.

Last edited: