Solutions to schrodinger equation with potential V(x)=V(-x)

In summary, the conversation discusses the use of the Schrodinger equation to show that a wavefunction can be purely odd or even, and the potential for it to be a combination of both. It is also mentioned that any function can be written as a sum of even and odd functions. The conversation concludes with the possibility of an eigenfunction being neither odd nor even if the eigenspace is of higher dimension.
  • #1
partyday
3
0
Homework Statement
Write down the time independent one dimensional Schrodinger
equation for a particle of mass m in a potential V (x). Show that if V (x) = V (−x) and real, then the
solutions ψn(x) have the property that ψn(x) = ±ψn(−x).
Relevant Equations
V(x) = V(-x)

##C \psi''(x) + V(x)\psi(x)=E\psi(x)##
C is just the constant by ##\psi''##

My initial attempt was to write out the schrodinger equation in the case that x>0 and x<0, so that

$$ \frac {\psi'' (x)} {\psi (x)} = C(E-V(x))$$
and

$$ \frac {\psi'' (-x)} {\psi (-x)} = C(E-V(-x))$$

And since V(-x) = V(x) I equated them and rearranged them so

$$\frac {\psi'' (x)} {\psi (x)} = \frac {\psi '' (-x)} {\psi(-x)} $$ I feel like this is where I'm supposed to be in the problem, but I'm struggling to connect this result to ψn(x) = ±ψn(−x). Any suggestions?
 
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  • #2
fyi i no nothing about qm but this just seems to be about even odd functions
so the question is asking to show wavefunction is purely odd or purely even
i think you might you want to use the fact that any function can be written as a sum of even and odd functions
also if a function is purely odd or even what can you tell about the second derivative.
if this wrong please forgive me.
 
  • #3
timetraveller123 said:
fyi i no nothing about qm but this just seems to be about even odd functions
so the question is asking to show wavefunction is purely odd or purely even
i think you might you want to use the fact that any function can be written as a sum of even and odd functions
also if a function is purely odd or even what can you tell about the second derivative.
if this wrong please forgive me.
I think I see the solution now. Bit of a brainfart for me. The ratios of the first and second derivative are the same for an odd or even function at x and -x, and that means that by that fact it's either ##\psi(x) = \psi(-x) ## (even) or ##\psi(x) = - \psi(-x)## (odd) right?
 
  • #4
i think you might also need to show that any wave function is necessarily of that form

i am slightly confused why can't the wavefunction be like zero for negative x and x=zero. and non-zero for positive values of x or the other way around.
 
  • #5
partyday said:
I think I see the solution now. Bit of a brainfart for me. The ratios of the first and second derivative are the same for an odd or even function at x and -x, and that means that by that fact it's either ##\psi(x) = \psi(-x) ## (even) or ##\psi(x) = - \psi(-x)## (odd) right?

I think you may be trying to prove something that is not necessarily true. What is generally true is as follows:

If ##V## is an even function, then for each eigenvalue you can find an eigenfunction that is either even or odd.

If you make the further assumption that the eigenspace associated with an eigenvalue is one-dimensional, then any eigenfunction must be odd or even - this follows from the above result.

But, if you have an eigenspace corresponding to a single eigenvalue that is of higher dimension, then you can combine odd and even eigenfunctions to get an eigenfunction that is neither odd nor even.
 
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FAQ: Solutions to schrodinger equation with potential V(x)=V(-x)

1. What is the Schrodinger equation with potential V(x)=V(-x)?

The Schrodinger equation with potential V(x)=V(-x) is a mathematical equation that describes the behavior of a quantum mechanical system. It is used to determine the wave function of a particle in a given potential, where the potential is symmetric about the origin (V(x)=V(-x)).

2. What does the potential V(x)=V(-x) represent?

The potential V(x)=V(-x) represents a symmetric potential, where the potential energy of the system remains the same when the position of the particle is reflected about the origin. This can occur in systems with symmetrical boundaries or in systems with an even potential function.

3. How does the Schrodinger equation with potential V(x)=V(-x) differ from the regular Schrodinger equation?

The only difference between the Schrodinger equation with potential V(x)=V(-x) and the regular Schrodinger equation is the symmetry of the potential. The rest of the equation remains the same, with the same operators and mathematical principles used to solve it.

4. What are some applications of the Schrodinger equation with potential V(x)=V(-x)?

The Schrodinger equation with potential V(x)=V(-x) has many applications in quantum mechanics, such as in the study of bound states in potential wells, quantum tunneling, and the behavior of particles in symmetric potentials. It is also used in the development of materials and technologies, such as semiconductors and superconductors.

5. How do you solve the Schrodinger equation with potential V(x)=V(-x)?

The Schrodinger equation with potential V(x)=V(-x) can be solved using various methods, such as the separation of variables, the variational method, or numerical methods. The exact method used depends on the specific system and potential being studied.

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