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 understand the form of operators in use for quantum mechanics such as the momentum operator: $$\hat{\text{P}}=-ih\frac{d}{dx}$$ My question is in what ways can I use it and what am I getting back? For example: if I simply apply th momentum operator to the wave function $$-ih\frac{d}{dx}\Psi$$ Will I get an Equation that will provide the momentum for a given position? Or is that a useless mathematical thing I just did?

If I use it to get an "expected value" by $$\langle \Psi | \hat{\text{ P }} | \Psi \rangle =\int_{-\infty}^\infty \Psi^* \hat{\text{ P } }\Psi$$ am I getting a number representing the probable momentum of that area integrated for? Or the percentage of total momentum there? Essentially is it a probability (if so of what kind?) or a value for the momentum?

I'm trying to understand these basic things because it has always remained unclear. I'm using it to find the momentum of and electron INSIDE the potential energy barrier as it is tunneling, i.e. the electrons between a surface and a Scanning Tunneling Microscope.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

The first thing you did is useless--- multiplying by the momentum doesn't do very much. But if you multiply by functions of the momentum, you can do things like project out the part of the state with a certain momentum.

The momentum operator is most important because if you find its eigenvectors and eignevalues, these are the states of definite momentum.

The expected value is the average of many measurements of the momentum--- it is the average value of momentum measurements. It is given by the expression you wrote down, but only when you integrate over all space. You can't restrict the range of integration to find the momentum in a limited region.

share|improve this answer

Starting from the intuitive meaning of wavefunctions as position amplitudes, the interpretation of the momentum operator is given by the Fourier transform. The eigenfunctions of the momentum operator are the pure (plane wave) frequency components of wavefunctions in coordinate representation.

In order to get any kind of localised position dependency, a wavefunction must be a superposition of momentum eigenfunctions having several frequencies. Actually, the pure momentum states are never observed (see Heisenberg uncertainty), and would also not be allowed mathematically, being non-integrable. The other extreme is the position operator, multiplication by x. Its eigenfunction, the three-dimensional delta, must also not occur on its own.

So in short, forming the expectation value using wavefunctions you always get a momentum (and position) distribution.

share|improve this answer

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.