What is the condition for one-dimensional motion in an infinite potential well?

[matematikuvol]

In one dimensional problem of infinite square potential well wave function is $\phi_n(x)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}$ and energy is $E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}$. Questions: What condition implies that motion is one dimensional. Did wave function describes motion of particle? When we say particle did we mean electron? What kind of particle mass are good enough to be treated in this model? And also why we don't draw $\phi_n(x)$.

 

[Naty1]

one distance variable x, implies one dimension.

In quantum mechanics, the analogue of Newton's law is Schrödinger's equation for a quantum system, usually atoms, molecules, and subatomic particles; free, bound, or localized. It is not a simple algebraic equation, but (in general) a linear partial differential equation. The differential equation describes the wavefunction of the system, also called the quantum state or state vector. In the standard interpretation of quantum mechanics, the wavefunction is the most complete description that can be given to a physical system. Solutions to Schrödinger's equation describe not only molecular, atomic, and subatomic systems, but also macroscopic systems, possibly even the whole universe.[1].....

In quantum mechanics, particles do not have exactly determined properties, and when they are measured, the result is randomly drawn from a probability distribution. The Schrödinger equation predicts what the probability distributions are, but fundamentally cannot predict the exact result of each measurement.
The Heisenberg uncertainty principle is a famous example of the uncertainty in quantum mechanics.

http://en.wikipedia.org/wiki/Schrödinger_equation

be sure to see the dynamic illustration near the beginning of the article here...http://en.wikipedia.org/wiki/Infinite_square_well

 

[naturale]

In one dimensional problem of infinite square potential well wave function is $\phi_n(x)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}$ and energy is $E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}$. Questions: What condition implies that motion is one dimensional. Did wave function describes motion of particle? When we say particle did we mean electron? What kind of particle mass are good enough to be treated in this model? And also why we don't draw $\phi_n(x)$.

The picture you must draw is a string (one dimension) vibrating between two fixed point (the walls of the potential) - distance L. The harmonics are the quantum excitations of the same particle or elementary system - p_n = n h / \lambda . This gives you the quantized momentum spectrum, that is the harmonic discrete spectrum a guitar string. The energy spectrum follows from the non-relativistic dispersion relation E = p^2 / 2 m.

 

[matematikuvol]

I didn't get the answer that I'm looking. My questions:
Did wave function describes motion of particle?

What kind of particle mass are good enough to be treated in this model?

And

And also why we don't draw $\phi_{-n}(x)$?

 

[geoduck]

I didn't get the answer that I'm looking. My questions:
Did wave function describes motion of particle?

For motion you need to solve the Schrodinger equation. The wavefunctions you are talking about are solutions to the energy eigenvalue equation, but it is the Schrodinger equation that gives you the evolution of state in time.

What kind of particle mass are good enough to be treated in this model?

Anything I guess. You have an expression involving m, and you can plug in whatever you want for m. If m is large then the energy levels are closely spaced which is good as that is common experience.

And also why we don't draw $\phi_{-n}(x)$?

\phi_{-n}(x) is the same state as \phi_{n}(x) as they only differ by a phase.

 

[Naty1]

I didn't get the answer that I'm looking. My questions:

What about post #2 did you not understand?

Did you read the links I posted...they will help a lot.

" Interpretations of quantum mechanics address questions such as what the relation is between the wavefunction, the underlying reality, and the results of experimental measurements."

There are different interpretations about exactly what the wavefunctions means...there are different ways to think about it.

 

[ZealScience]

Usually in one dimensional problem you can constrain a particle between two infinitely large parallel plates, that makes the particle only subject to the boundary conditions in one dimension.

Also, I don't think that particles are described by equations of motion in QM. In QM we describe particles in Hilbert space, in classical mechanics we use phase space which describe the equations of motion of a particle.

Basically any particles follows QM, but when n is large, by Correspondence Principle, classical mechanics must be a good approximation of QM

 

[matematikuvol]

One more question. By solving Sroedinger eq we get
$\varphi_n(x)=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}$
$E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}$
But for example solution of Sroedinger eq is also
$C_1sin\frac {\pi x}{a}+C_2\sin\frac{2\pi x}{a}$
in which state is particle?
What is the energy level? How could I know in which state is particle?

 

[jtbell]

But for example solution of Sroedinger eq is also
$C_1sin\frac {\pi x}{a}+C_2\sin\frac{2\pi x}{a}$
in which state is particle?
What is the energy level? How could I know in which state is particle?

The particle is in a superposition of the n=1 and n=2 states. When something happens that in effect measures the particle's energy, it will be either E₁ or E₂, chosen randomly according to probabilities that depend on C₁ and C₂.

Before the measurement happens, the answer to the question, "which of the two energy states is the particle really in?" depends on which interpretation of QM you subscribe to.

 

[matematikuvol]

Well from
\int^{a}_0 (C_1\sin \frac {\pi x}{a}+C_2 \sin \frac{2\pi x}{a})^2=1
I get
$C_1^2+C_2^2=\frac{2}{a}$
what next?

 

[jtbell]

It's better to start out by normalizing each individual function:

$$\psi(x) = C_1 \psi_1(x) + C_2 \psi_2(x)\\
\psi(x) = C_1 \sqrt{\frac{2}{a}} \sin {\left(\frac{\pi x}{a}\right)}
+ C_2 \sqrt{\frac{2}{a}} \sin {\left(\frac{2 \pi x}{a}\right)}$$

Then when you normalize $\psi$ as a whole by doing an integral like yours, you end up with

$$C_1^2 + C_2^2 = 1$$

We interpret $C_1^2$ as the probability that the particle will be measured to have energy E₁, and $C_2^2$ as the probability that the particle will be measured to have energy E₂. The two probabilities add to 1 because they are the only two possibilities for this particle as far as energy is concerned.

 

[matematikuvol]

Ok but from $C_1^2+C_2^2=1$ and some other physical behavior could I say something about $C_1^2$ and $C_2^2$?

 

[jtbell]

Obviously you need additional information about your specific situation in order to find C₁ and C₂. It's impossible to say more unless you have a specific situation in mind.

 

[matematikuvol]

I understand that. But could you give me example of some specific situation.