Schrodinger ground state wave function

In summary, the problem is to find the ground state wave function for a 1-D particle in a box with V=0 between x=-a/2 and x=a/2 and V=\infty outside of that range. Using Schrodinger's time-independent equation, the equation can be simplified to \frac{-\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}=E\psi. By using a trick from the textbook, the equation can be rewritten as \frac{d^{2}\psi}{dx^{2}}+k^{2}\psi=0, where k^{2}=\frac{2mE}{\hbar^{2}}. The general
  • #1
wrongusername
61
0

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 = [tex]\infty[/tex]

Homework Equations



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

The Attempt at a Solution



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

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

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

[tex]E\psi+\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}=0[/tex]

[tex]\frac{2mE}{\hbar^{2}}\psi+\frac{d^{2}\psi}{dx^{2}}=0[/tex]

Now I use a trick from my book:

[tex]k^{2}=\frac{2mE}{\hbar^{2}}[/tex]

[tex]\frac{d^{2}\psi}{dx^{2}}+k^{2}\psi=0[/tex]

And now there's 2 solutions to this differential equation: [tex]\psi=A\cos\left(kx\right)[/tex] and [tex]\psi=A\sin\left(kx\right)[/tex]. Where do i go from here? I am quite lost unfortunately :frown:
 
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  • #2
Hi wrongusername,

wrongusername said:

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 = [tex]\infty[/tex]

Homework Equations



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

The Attempt at a Solution



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

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

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

[tex]E\psi+\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}=0[/tex]

[tex]\frac{2mE}{\hbar^{2}}\psi+\frac{d^{2}\psi}{dx^{2}}=0[/tex]

Now I use a trick from my book:

[tex]k^{2}=\frac{2mE}{\hbar^{2}}[/tex]

[tex]\frac{d^{2}\psi}{dx^{2}}+k^{2}\psi=0[/tex]

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

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?
 
  • #3
alphysicist said:
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, [tex]V\left(x\right)=\infty[/tex] for all x not between -a/2 and a/2, though if I plug infinity in for V in the Schrodinger equation, I would get [tex]\infty=E\psi[/tex]?

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...
 
  • #4
wrongusername said:
Thank you for the reply!

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

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?
 
  • #5




I would like to commend you for attempting to solve the problem using Schrodinger's time-independent equation. Your approach is correct up to the point where you obtained the differential equation \frac{d^{2}\psi}{dx^{2}}+k^{2}\psi=0. This is known as the eigenvalue equation, and it is a common problem in quantum mechanics to solve for the eigenvalues and eigenfunctions of a given potential.

To find the ground state wave function, we need to consider the boundary conditions. In this case, the potential is infinite outside of the given box, which means that the wave function must be zero at the boundaries (x = -a/2 and x = a/2). This means that the sine solution is not valid, as it does not satisfy the boundary conditions. Therefore, the ground state wave function is given by \psi=A\cos\left(kx\right), where k is determined by the boundary condition.

To find the value of k, we can use the fact that the wave function must be continuous at the boundaries. This means that \psi\left(-\frac{a}{2}\right)=\psi\left(\frac{a}{2}\right)=0. Substituting this into the wave function, we get A\cos\left(-\frac{ka}{2}\right)=A\cos\left(\frac{ka}{2}\right)=0. This is only possible if k=\frac{n\pi}{a}, where n is an integer. Therefore, the ground state wave function is given by \psi=A\cos\left(\frac{n\pi x}{a}\right).

I hope this helps to guide you in the right direction. Keep in mind that solving for eigenvalues and eigenfunctions is a common problem in quantum mechanics, and there are many resources available that can aid in understanding this concept. Good luck with your studies!
 

Related to Schrodinger ground state wave function

1. What is the Schrodinger ground state wave function?

The Schrodinger ground state wave function is a mathematical representation of the lowest possible energy state of a quantum mechanical system, based on the Schrodinger equation. It describes the probability distribution of finding a particle in a given location at a specific time.

2. How is the Schrodinger ground state wave function calculated?

The Schrodinger ground state wave function is calculated by solving the Schrodinger equation, which takes into account the potential energy of the system and the properties of the particles within it. This equation results in a wave function that describes the behavior of the system.

3. What is the significance of the Schrodinger ground state wave function?

The Schrodinger ground state wave function is significant because it provides a fundamental understanding of the behavior of particles on a quantum level. It allows scientists to make predictions about the behavior of particles, such as their position and energy, which can then be tested and verified experimentally.

4. How does the Schrodinger ground state wave function relate to the uncertainty principle?

The Schrodinger ground state wave function is related to the uncertainty principle in that it represents the probability of finding a particle in a particular location. This means that the more precisely we know the position of a particle, the less certain we are about its momentum, and vice versa.

5. Can the Schrodinger ground state wave function be applied to all systems?

The Schrodinger ground state wave function can be applied to any quantum mechanical system, as long as it follows the principles of quantum mechanics. This includes systems such as atoms, molecules, and even larger structures like crystals. However, it cannot be applied to macroscopic objects, as they behave according to classical mechanics.

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