The free particle solution in stationary state (with definite energy) to the Schrödinger equation is

$$\psi(x,t) =Ae^{i(kx-\omega t)} + Be^{-i(kx+\omega t)}$$

Since the energy is definite, and hence the momentum is definite, the uncertainty in position must be infinite. How is this reflected by the probability distribution function:

$$\Psi = |\psi(x,t)\psi^*(x,t)| $$

The book that I am using just look at the first term of the solution, and derive that the probability distribution function is $A^2$. However, I do not understand why we can do that?

Does it imply that if wave function is made up of n terms such that each individual term has a constant probability distribution function, the whole wave function also has a constant probability distribution function? If so, how can I prove it?

I know my question might be very vague but that is precisely the problem I am facing now, I don't even know how to ask about the things that I don't understand.

  • $\begingroup$ By "the energy is definite," do you mean it has no uncertainty? The eigenvalues of the Hamiltonian are expectation values, not an exact prediction of the energy. $\endgroup$
    – JamalS
    Commented Mar 28, 2014 at 11:25
  • $\begingroup$ Means U = 0 I am not sure about Hamiltonian because this is my first quantum mechanics course. $\endgroup$ Commented Mar 28, 2014 at 11:32

3 Answers 3


For any function of $x$ and $t$ that depends on the combination $x\pm vt$ (for constant $v$ represents a wave with a fixed shape that travels in the $\mp x$ direction with speed $v$. That is to say, $$ x\pm vt={\rm constant} $$ In your wave function, $$ \psi(x)=Ae^{i(kx-\omega t)}+Be^{-i(kx+\omega t)}\tag{0}, $$ the first term represents a right-moving wave while the second term represents a left-moving wave. Since they differ only by the sign of $k$, we can simplify the wavefunction to $$ \psi(x)=Ae^{i(kx-\omega t)}\tag{1} $$ where $k>0$ means the right-moving wave (the first term of Eq (0) ) and $k<0$ means a left-moving wave (the second term of Eq (0)).

Using Equation (1) and multiplying by its complex conjugate, we get $$ \psi^*\psi(x)=\left(Ae^{-i(kx-\omega t)}\right)\left(Ae^{i(kx-\omega t)}\right)=A^2 $$


The terms in $A$ and $B$ represent waves traveling in opposite directions. Their values are set by the initial conditions of the problem. Some problems will involve waves traveling to the right, others to the left, others a combination of the two. Evidently your book chose to focus on a wave traveling to the right for simplicity.

Note that this does imply that the wave function can be constructed as a sum of terms. Calculating probabilities densities for each term is usually not what you want to do. And if you do that for some reason, you do not add the results together. In your case, you want the probability density for the wave function as a whole.

Now, if you complete the calculation of $\psi \psi^*$ for the entire wave function, you will find that the result is just another real number. So the logic still stands, but the probability distribution function is something other than $A^2$. I'll leave it to you to calculate what that value is.

  • $\begingroup$ I get A^2 + B^2 + ABe^(2i(kx-wt)) + ABe^(-2i(kx+wt)). How is this a real number? $\endgroup$ Commented Mar 28, 2014 at 15:10
  • $\begingroup$ Ah.. I probably should have said "real function". But you didn't get the algebra quite right in the second two terms. Certainly $\psi \psi^*$ should be real, shouldn't it? $\endgroup$
    – garyp
    Commented Mar 28, 2014 at 18:29

A free particle moves in infinite region ,no matter at which instant,potencial of it converted into kinetic entirely because of its infinite region.and then Schrödinger equation becomes in the form of momentum.which is used to detect the features of free particle.since the region is infinite,so we can easily say that the particle has same probability through the space,,,,without any instant b/c we can not detect the exact position of a particle in the wave function,,as well as de Broglie wavelength goes to infinity....

  • $\begingroup$ Are u convenienced........ $\endgroup$ Commented Oct 30, 2017 at 5:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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