Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am taking my second course in QM, and my head is starting to spin as it probably should. But I would very much like to clear up my head about a few details regarding the wave function. As I know it is impossible to predict where particles are and one can only give a probability of where it should be.

The simplest case is a "frictionless" particle "bouncing" back and forth inside a infinite square well. Eg a particle in the following potential

$$ V(x) = \left( \begin{array}{cc} 0 \ , & \text{for} \ 0 \leq x \leq a \\ \infty \ , & \text{elsewhere} \end{array} \right) $$ Which gives rise to the following normalized solution $$ \psi_n(x) = \sqrt{\frac{2}{a}} \sin\left( \frac{\pi n}{a}x \right) $$ My problem is what the nodes in the square function represents. If I draw $|\Psi_2(x,0)|^2=|\psi_2(x)|^2$ I obtain a graph similar to the one below.

enter image description here

What is the physical explenation that finding the particle around a small region around $a/2$ is close to zero? Or why is it so much less likely to find it near $a/2$ than $a/4$? Eg why is $$ P(a/2 -\varepsilon \leq X \leq a/2+\varepsilon) = \int_{a/2-\varepsilon}^{a/2+\varepsilon} \left| \psi_2(x) \right|^2 \,\mathrm{d}x \sim 0 $$ for small $\varepsilon$

share|improve this question
2  
An analogy is manifest with the modes of a vibrating string with fixed extremities. You have different harmonics, and for each harmonic, you have nodes. –  Trimok Aug 25 '13 at 14:28
    
I like this question and would like to see a good answer. In principle the node is "caused" by the assumptions of QM and these in turn are justified by a large body of evidence. This is counter-intuitive when coming from a macroscopic perspective but inherent in QM. –  Alexander Aug 25 '13 at 22:35
add comment

2 Answers

Recall that the functions $\psi_n$ are energy eigenstates; these states are very special. A generic quantum state of the system is simply some continuous, square-integrable function $\psi$ on $[0,a]$ that vanishes at the endpoints of the interval; within these requirements, it can have any shape. Moreover, any such function can be written as a linear combination of the energy eigenfunctions $\psi_n$.

The precise functional form of the wavefunction is determined by the way in which the state of the system is prepared. In particular, a generic state of the system need not have any nodes.

share|improve this answer
add comment

Physically, what you have is a standing wave. Even though it's this bizarre seeming probability wave, it still shares properties with waves on a string or in the water. It must always be zero at the boundaries, so at x=0 and x=a it is pinned at zero. In the middle you have a wave function. It will rarely be found at a/2 for n=2 because of the physical restraints you have placed on it. For n=1, it will almost always be found at a/2, and for n=3 it will almost never be found at a/3.

Physically, an infinite square well is not really possible. We can approximate it, but it is not physically possible. The properties of the waves you are studying give rise to many of the physical properties we see in the quantum world, such as quantized energy levels in atoms. Don't get too caught up in the probability distribution piece, because ladder operators will be much more of a mind trip than that. :-)

share|improve this answer
    
Reading about the ladder operators in Griffiths now... I think I will need more Vodka if this continues, thanks for the rope analogy it was just the piece missing. I have taken a bit classical mechanics beforehand. –  N3buchadnezzar Aug 25 '13 at 19:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.