2
$\begingroup$
  1. What does a "STATE" exactly mean in quantum mechanics?

  2. What is the equivalence of "STATE" in classical mechanics?

  3. If we have a wave function $\Psi$ , its absolute square $|\Psi|^2$ is the probability density of finding the particle somewhere in the space, and I know it can be written as $\langle \Psi |\Psi \rangle$. But what is the physical meaning of $\langle \Phi | \Psi \rangle$, where $\Phi$ and $\Psi$ are two different wave functions. Why we need to take inner product of two different wave functions.

$\endgroup$

closed as too broad by BMS, Brian Moths, Kyle Kanos, Robin Ekman, Kyle Oman Jul 14 '14 at 16:04

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ I suggest you consult the first few chapters of Feynman's excellent Lectures on Physics, Volume III. The book is available online for free and offers a good introduction to the topic, answering the questions you have. $\endgroup$ – Danu Jul 14 '14 at 14:49
  • $\begingroup$ I suggest to collect a little bit of reputation, and then try to make these questions step-by-step. $\endgroup$ – peterh Jul 14 '14 at 16:49
1
$\begingroup$

I will give you an example from classical wave theory. Take Melde's experiment with a rope where you can have different modes in this rope. Those modes are discrete, like say a photon in a box. And you have a dispersion relation that relates the wave number $k$ to the frequency $\omega$.

Therefore, if we try to make an analogy with a photon in a box, the displacement of the rope (position of the rope $y(x)$) is the analogue to the wave function of the photon in the box. And a state for the photon means a certain eigenvalue. Transpose to the rope, it's mean you're are in a eigenmode of the rope.

Now, let's say we have two states $\Psi(x)$ and $\Phi(x)$. Those are two function. You want to see if they are the same or different, how different etc... So we actually consider the space of the wave function as a space with an hermitian product and we compare the wave function with this tool.

$\endgroup$
  • $\begingroup$ I made an edit to fix the orthographic problems of your answer. However, I didn't understand it myself completely. What do you mean by "Transpose to the rope"? $\endgroup$ – Jonas Jul 14 '14 at 15:11
1
$\begingroup$

The wavefunction, $\psi$, is the most complete possible description of a particle (or collection of particles). From the wavefunction, one can calculate a probability distribution for the the outcome of any measurement. Remember, quantum mechanics is probabilistic - one cannot make exact predictions for all measurable quantities, even if you know the wavefunction.

There is nothing equivalent to the wavefunction in classical mechanics. The observable quantities in classical mechanics, for example position and momentum, do have equivalents in quantum mechanics - they are promoted to operators in quantum mechanics.

The physical meaning of $|\langle \phi |\psi \rangle|^2$ is that it is the probability that $\psi$ is found to be in the state $\phi$, upon measurement. Suppose that $\phi$ is an eigenstate of the Hamiltonian with energy $E$. $|\langle \phi |\psi \rangle|^2$ is the probability of measuring the energy of the particle described by $\psi$ and finding $E$. See the Born rule.

The physical meaning of $\langle \phi |\psi \rangle$ is less obvious - it is an amplitude, which is something like the square root of a probability. Remember, in quantum mechanics, we add amplitudes ($\sim$ square roots of probabilities) then square. This is obviously different to ordinary probability, in which one would add probabilities.

$\endgroup$
1
$\begingroup$

1) It's basically just a description of what a particle's doing. This involves its energy level, the probability of where to "find it", its spin, etc.

2) I'd say it's somewhat analogous to describing a ball rolling down a hill in terms of its energy, where you describe its potential energy, kinetic energy, and rotational energy. But there's no "equivalence" in classical mechanics per se.

3) A state works similar to a vector, and $\langle \psi | \phi \rangle$ is like the projection of $\psi$ onto $\phi$. If the states are identical, you get 1, and if they're orthogonal you get 0, just like if vectors A and B are in the same direction and length $\frac{\sqrt{2}}{2}$ you get 1, and if they're at right angles you get 0.

$\endgroup$

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