# I Understanding the wavefunction for a free particle

1. Sep 6, 2016

### majormuss

Hi everybody,
I was reading about the free particle in a textbook and I got confused by the line:
"If we adopt the convention that k and k are real, then the only oscillating exponentials are the eigenfuntions with positive energy" [Also see the attached picture with the paragraph.]
What is the rationale for that line? Furthermore, does any body have a link where I can read about those exponentials? I couldn't find them online and I am curious to know how and when they are used to solve the Schrodinger Equation!

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2. Sep 6, 2016

### BvU

Insert the $\psi$ in the (time independent) SE and see they satisfy the equation !

3. Sep 6, 2016

### majormuss

I don't doubt that they satisfy the equation. My question is about the line,
If we adopt the convention that k and k are real, then the only oscillating exponentials are the eigenfuntions with positive energy"
Why is there a need for both k and kappa be real? And what are those exponentials e^(+-ikx) and e^(+-kx) called and where can I read more about them?

4. Sep 6, 2016

### Staff: Mentor

Because if they are not real, then it is difficult to separate the two cases. For instance, an imaginary κ would correspond to a real k.

I don't know of any particular name for these exponentials. They come from the mathematics of differential equations.

5. Sep 6, 2016

### PeroK

If you let $k$ be complex, then you get all the solutions with $k^2 = -\frac{2mE}{\hbar^2}$ hence:

$k = \pm i \frac{\sqrt{2mE}}{\hbar}$

And:

$\psi(x) = Ae^{\pm i \frac{\sqrt{2mE}}{\hbar}x}$

So, it amounts to the same thing. Taking $k$ as real initially is just a bit of a shortcut, based to some extent on knowing the solution in advance!

6. Sep 6, 2016

### vanhees71

What's the source? It's very inaccurate.

The rationale is the following: In wave mechanics the vectors of the quantum mechanical Hilbert space are represented by square integrable functions $\psi:\mathbb{R}^3 \rightarrow \mathbb{C}$. The position vector $\vec{x}$ is represented by the multiplication of the wave function with $\vec{x}$, and momentum by $\hat{\vec{p}}=-\hbar \vec{\nabla}$.

These operators are defined on a dense subspace of these Hilbert space of square-integrable functions. Now you can ask for the generalized eigenfunctions of momentum. They obey the equation
$$\hat{\vec{p}} u_{\vec{p}}(\vec{x})=\vec{p} \psi(\vec{x}).$$
The operators have to be understood as self-adjoint operators, and their eigenvalues are thus real. For real $\vec{p}$ the solution of the eigenvalue problem reads
$$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi \hbar)^{3/2}} \exp\left (\frac{\mathrm{i} \vec{x} \cdot \vec{p}}{\hbar} \right).$$
I've normalized this generalized functions "to the $\delta$ distibution", i.e.,
$$\langle u_{\vec{p}'}|u_{\vec{p}} \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}'}^*(\vec{x}) u_{\vec{p}}(\vec{x})=\delta^{(3)}(\vec{p}-\vec{p}').$$
These generalized eigenfunctions do NOT represent states since they are not square integrable, but you can describe all wave functions in terms of generalized momentum eigenstates:
$$\psi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}} (\vec{x}) \tilde{\psi}(\vec{p}),$$
where $\tilde{\psi}(\vec{p})$ is a square integrable function. Then also $\psi(\vec{x})$ is a square integrable function and represents a state of the particle. For the given wave function $\psi(\vec{x})$ you find the $\tilde{\psi}(\vec{x})$ by the inverse transformation (it's a Fourier transformation!):
$$\tilde{\psi}(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u^*(\vec{p})(\vec{x}) \psi(\vec{x}).$$

7. Sep 6, 2016

### BvU

I agree the wording is odd. The time-independent SE leads to a time part of the solution of the time-dependent SE of the form $\exp({-iEt/ \hbar})$. The continuity equation forces $(E-E^*) = 0$ so E is real.

Then: with the most general form, a complex $e^{{\bf \alpha} x}$ where ${\bf \alpha} = \kappa + ik \ \$ ($\kappa$ and $k$ real), you get from the time-independent SE: $\ \ -{\hbar^2\over 2m}{\bf\alpha}^2 = E\$ . Positive, real-valued E requires $\kappa = 0$.

As vanHees points out, these are solutions, eigenfunctions of the Hamiltonian. They are not physically realizable states (not normalizable). But that is (I expect) a subject further on in your curriculum.

8. Sep 7, 2016

### vanhees71

That the plane waves are not representing states, is a very important point. A lot of confusion arises from not making this very clear in the very first encounter of these generalized solutions of the Schrödinger equation!