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.tryingtolearn1 said: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 ..##
tryingtolearn1 said:I know that there is a boundary condition at ##x=0## where the wave function becomes zero
Is this the correct potential?tryingtolearn1 said:$$
V(x)=\left\{
\begin{array}{ll}
\frac{1}{2}m\omega^2x^2, \ x > 0\\
0, \ x < 0
\end{array}
\right.
$$
Opps the correct potential is:vela said:Is this the correct potential?
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 Schrodinger's equation in a quadratic potential. Hint: The wave function is expressed in terms of Hermite polynomials.tryingtolearn1 said:Opps the correct potential is:
$$
V(x)=\left\{
\begin{array}{ll}
\frac{1}{2}m\omega^2x^2, \ x < 0\\
\infty , \ x > 0
\end{array}
\right.
$$
Does it matter?Fred Wright said: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 Schrodinger's equation in a quadratic potential. Hint: The wave function is expressed in terms of Hermite polynomials.
Fred Wright said: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 Schrodinger's equation in a quadratic potential. Hint: The wave function is expressed in terms of Hermite polynomials.
You're right. My bad.etotheipi said: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}}##.
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.tryingtolearn1 said:Oops, the correct potential is:
$$
V(x)=\begin{cases}
\frac{1}{2}m\omega^2x^2, & x < 0 \\
\infty, & x > 0
\end{cases}
$$
Oops, I didn't see it that way. Post deleted.vela said:Well, sure, but the point was to have the OP figure that out for himself/herself.
At x=0, the potential energy is at its minimum, meaning that the particle is in a stable equilibrium position. In quantum mechanics, particles are described by wave functions, and at x=0, the wave function must be symmetric in order to satisfy the boundary conditions. This requirement results in only odd energy levels being allowed at x=0.
The symmetry of the wave function plays a crucial role in determining the allowed energy levels at x=0. In quantum mechanics, particles are described by wave functions that must satisfy certain boundary conditions. At x=0, the wave function must be symmetric, which leads to only odd energy levels being allowed.
In quantum mechanics, there are certain energy levels that are not allowed for a particle in a given system. These are known as forbidden energy levels. At x=0, the potential energy is at its minimum, and the wave function must be symmetric, leading to only odd energy levels being allowed. Any even energy levels would be considered forbidden at x=0.
Quantization is a fundamental concept in quantum mechanics that describes the discrete nature of energy levels in a system. At x=0, the potential energy is at its minimum, and the wave function must be symmetric, leading to only odd energy levels being allowed. This discrete nature of energy levels is a result of quantization.
There are certain cases where even energy levels may be allowed at x=0. This can occur if there is an external potential present that breaks the symmetry of the wave function. In these cases, both odd and even energy levels may be allowed at x=0.