Why Are Allowed Energy Levels Odd at ##x=0##?

[tryingtolearn1]

Homework Statement

Find the allowed energy levels of a particle of mass m moving in the one-dimensional potential

energy well $$

V(x)=\left\{

\begin{array}{ll}

\frac{1}{2}m\omega^2x^2, \ x > 0\\

0, \ x < 0

\end{array}

\right.

$$

Relevant Equations

Harmonic oscillator

I know that there is a boundary condition at $x=0$ where the wave function becomes zero however why are the allowed energy levels odd i.e. $n=1, 3, 5 ..$

 

[LCSphysicist]

Homework Statement:: Find the allowed energy levels of a particle of mass m moving in the one-dimensional potential

energy well $$

V(x)=\left\{

\begin{array}{ll}

\frac{1}{2}m\omega^2x^2, \ x > 0\\

0, \ x < 0

\end{array}

\right.

$$
Relevant Equations:: Harmonic oscillator

I know that there is a boundary condition at $x=0$ where the wave function becomes zero however why are the allowed energy levels odd i.e. $n=1, 3, 5 ..$

I believe there is no a physical way to answer this, maybe @PeroK can know one. But, anyway, this is a math implication so even we don't know why, we know that need to be it.

Why don't you try to solve the Schrödinger equation in both sides? Remember the conditions:
$$\psi, \psi'$$ is continuous.

 

[etotheipi]

I know that there is a boundary condition at $x=0$ where the wave function becomes zero

Why?

 

[vela]

$$
V(x)=\left\{
\begin{array}{ll}
\frac{1}{2}m\omega^2x^2, \ x > 0\\
0, \ x < 0
\end{array}
\right.
$$

Is this the correct potential?

 

[tryingtolearn1]

Is this the correct potential?

Opps the correct potential is:

$$

V(x)=\left\{
\begin{array}{ll}
\frac{1}{2}m\omega^2x^2, \ x < 0\\
\infty , \ x > 0
\end{array}
\right.
$$

 

[Fred Wright]

Opps the correct potential is:

$$

V(x)=\left\{
\begin{array}{ll}
\frac{1}{2}m\omega^2x^2, \ x < 0\\
\infty , \ x > 0
\end{array}
\right.
$$

I think the dimensions of your potential are incorrect. In SI units energy is $kg(\frac{m}{s})^2$, where $kg$ is kilograms, $m$ is meters and $s$ is seconds, but the units of your potential are $kg(\frac{m}{s})^2(radians)^2$. If you can fix the units issue you face the problem solving Schrödinger's equation in a quadratic potential. Hint: The wave function is expressed in terms of Hermite polynomials.

 

[LCSphysicist]

I think the dimensions of your potential are incorrect. In SI units energy is $kg(\frac{m}{s})^2$, where $kg$ is kilograms, $m$ is meters and $s$ is seconds, but the units of your potential are $kg(\frac{m}{s})^2(radians)^2$. If you can fix the units issue you face the problem solving Schrödinger's equation in a quadratic potential. Hint: The wave function is expressed in terms of Hermite polynomials.

Does it matter?

 

[etotheipi]

I think the dimensions of your potential are incorrect. In SI units energy is $kg(\frac{m}{s})^2$, where $kg$ is kilograms, $m$ is meters and $s$ is seconds, but the units of your potential are $kg(\frac{m}{s})^2(radians)^2$. If you can fix the units issue you face the problem solving Schrödinger's equation in a quadratic potential. Hint: The wave function is expressed in terms of Hermite polynomials.

Angles are dimensionless, and the potential energy of a harmonic oscillator, $V(x) = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2$, has dimensions $[V] = ML^2 T^{-2}$. The units of angular velocity can be taken to be $\mathrm{s^{-1}}$.

 

[Fred Wright]

Angles are dimensionless, and the potential energy of a harmonic oscillator, $V(x) = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2$, has dimensions $[V] = ML^2 T^{-2}$. The units of angular velocity can be taken to be $\mathrm{s^{-1}}$.

You're right. My bad.

 

[vela]

Oops, the correct potential is:
$$
V(x)=\begin{cases}
\frac{1}{2}m\omega^2x^2, & x < 0 \\
\infty, & x > 0
\end{cases}
$$

OK, so what you want to do is follow @LCSphysicist's suggestion in post #2. That should lead you to the answers to your question and @etotheipi's in post #3. If you get stuck, post again and show us your work, and we can go from there.

 

[vela]

Well, sure, but the point was to have the OP figure that out for himself/herself.

 

[kuruman]

Well, sure, but the point was to have the OP figure that out for himself/herself.

Oops, I didn't see it that way. Post deleted.