# A couple basic quantum mechanics questions

1. May 18, 2010

### cotyledon

Hi, I have a couple hopefully quick questions about quantum mechanics:

For a momentum eigenstate how does the wave function

$$e^{\frac{2 \pi}{h}ixp_{x}}$$

not violate the uncertainty principle? (since both p and x are present in the function)

Also, when you rewrite the above as

$$e^{ikx}$$

how is that a momentum eigenstate when it has position in it?

Finally, what mathematically is a position eigenstate?

2. May 18, 2010

### phyzguy

I think the point you are missing is that in your momentum eigenstate, px is a constant, but x is a variable. When you take your momentum eigenstate and calculate the probability of finding the particle at a position x, this is just the wave function times its complex conjugate. But this is just a constant everywhere, independent of x. So a particle in a momentum eigenstate has a definite value of momentum, but is equally likely to be found anywhere in space. In other words the uncertainty in momentum is zero, but the uncertainty in position is infinite. Similarly, a position eigenstate is a delta function at position x. It has a definite position, but its momentum is equally likely to have any value. So its position uncertainty is zero, but its momentum uncertainty is infinite. Of course, these two cases are just limiting cases - any real wave function would be a superposition of multiple position or momentum eigenstates, so that it has a finite position uncertainty and a finite momentum uncertainty. Look at Gaussian wave packets, and take the state of definite position to be the limit of an infinitely sharp Gaussian, and the momentum eigenstate to be the limit of an infinitely broad Gaussian.

3. May 18, 2010

### NanakiXIII

The statement "$f$ is an eigenfunction (eigenstate, usually, in QM) of an operator $A$" means

$$Af = \lambda f$$

where $\lambda$ is some constant number. Being a momentum eigenstate means being an eigenfunction of the momentum operator. The momentum operator (in one dimension) is

$$p = -i \hbar \frac{d}{d x}$$.

Applying this to the state you mention, you get

$$p e^{ikx} = -i \hbar \frac{d}{d x} e^{ikx} = \hbar k e^{ikx}$$.

So indeed it is a momentum eigenstate.

This state does not violate the uncertainty principle. This has nothing to do with what kind of expression appears in your function. Though the wavefunction contains $x$, the position of the particle this wavefunction describes is actually completely uncertain ($\Delta x = \infty$).