Schrodinger ground state wave function

[wrongusername]

Homework Statement

Find the ground state wave function for the 1-D particle in a box if V = 0 between x = -a/2 and x = a/2 and V = \infty

Homework Equations

I would guess -- Schrödinger's time-independent equation?
\frac{-\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}+V\left(x\right)\psi=E\psi

The Attempt at a Solution

V\left(x\right)=0 for -\frac{a}{2}\leq x\leq\frac{a}{2}

So then, we don't need the V(x) term in the equation, and it simplifies to \frac{-\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}=E\psi

So now I move the nasty stuff over to the right side:

E\psi+\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}=0

\frac{2mE}{\hbar^{2}}\psi+\frac{d^{2}\psi}{dx^{2}}=0

Now I use a trick from my book:

k^{2}=\frac{2mE}{\hbar^{2}}

\frac{d^{2}\psi}{dx^{2}}+k^{2}\psi=0

And now there's 2 solutions to this differential equation: \psi=A\cos\left(kx\right) and \psi=A\sin\left(kx\right). Where do i go from here? I am quite lost unfortunately

 

[alphysicist]

Hi wrongusername,

Homework Statement

Find the ground state wave function for the 1-D particle in a box if V = 0 between x = -a/2 and x = a/2 and V = \infty

Homework Equations

I would guess -- Schrödinger's time-independent equation?
\frac{-\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}+V\left(x\right)\psi=E\psi

The Attempt at a Solution

V\left(x\right)=0 for -\frac{a}{2}\leq x\leq\frac{a}{2}

So then, we don't need the V(x) term in the equation, and it simplifies to \frac{-\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}=E\psi

So now I move the nasty stuff over to the right side:

E\psi+\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}=0

\frac{2mE}{\hbar^{2}}\psi+\frac{d^{2}\psi}{dx^{2}}=0

Now I use a trick from my book:

k^{2}=\frac{2mE}{\hbar^{2}}

\frac{d^{2}\psi}{dx^{2}}+k^{2}\psi=0

And now there's 2 solutions to this differential equation: \psi=A\cos\left(kx\right) and \psi=A\sin\left(kx\right). Where do i go from here? I am quite lost unfortunately

The general solution for the part from -a/2 to a/2 will be a linear combination of those two solutions (A cos(kx) + B sin(kx) ).

So you've found the solution in that area, now what is the solution for those places outside the box? And what has to be true about these three separate solutions?

 

[wrongusername]

Hi wrongusername,



The general solution for the part from -a/2 to a/2 will be a linear combination of those two solutions (A cos(kx) + B sin(kx) ).

So you've found the solution in that area, now what is the solution for those places outside the box? And what has to be true about these three separate solutions?

Thank you for the reply!

Well, V\left(x\right)=\infty for all x not between -a/2 and a/2, though if I plug infinity in for V in the Schrödinger equation, I would get \infty=E\psi?

Wouldn't it be two solutions since, like you said, the two solutions would be combined into one general solution with a linear combination?

All three solutions would have to reflect the ground state energy? I'm not sure how though...

 

[alphysicist]

Thank you for the reply!

Well, V\left(x\right)=\infty for all x not between -a/2 and a/2, though if I plug infinity in for V in the Schrödinger equation, I would get \infty=E\psi?

No, I don't think that would be quite helpful here. Instead, leave it as V, but keep in mind that V>E. Find the solution (using your trick to get the equation into a similar form as before), and then let V go to infinity.

Remember that here we want the wavefunction to go to zero as x goes to plus or minus infinity.

Wouldn't it be two solutions since, like you said, the two solutions would be combined into one general solution with a linear combination?

Every region will have two mathematical solutions; but boundary conditions will set some constants equal to zero, force values for others, etc.

All three solutions would have to reflect the ground state energy? I'm not sure how though...

There is one wavefunction they are looking here, that is made up of the solutions you found in each region. But look at, for example, the point x=a/2. There the solution from the inner region (-a/2 < x < a/2) is meeting the solution from the right side (x > a/2). How do they go together? What property of the wavefunction is used at a/2?