Eigenfunctions and wave functions

[semc]

TL;DR Summary

Is my understanding correct : Wave function is a vector and eigen functions are the basis set that span the space.

I saw this statement from the textbook "Quantum physics of atoms, molecules, solids, nuclei, and particles" second edition pg 166. According to the text, is the author saying the solution to the TISE is the eigenfunction and when you multiply the time dependent part, you get the wave function? I always thought that the eigenfunctions are a basis set that forms the wave function

 

[PeroK]

Summary:: Is my understanding correct : Wave function is a vector and eigen functions are the basis set that span the space.

View attachment 272154

I saw this statement from the textbook "Quantum physics of atoms, molecules, solids, nuclei, and particles" second edition pg 166. According to the text, is the author saying the solution to the TISE is the eigenfunction and when you multiply the time dependent part, you get the wave function? I always thought that the eigenfunctions are a basis set that forms the wave function

You've got the right idea, but there is no sharp distinction between eigenfunctions and wavefunctions. Eigenfunctions are simply the special case of wavefunctions corresponding to eigenvalues of some operator.

Also, you can talk about ##\psi(x)## as the (time-independent) wavefunction or ##\Psi(x, t)## as the wavefunction. For example, you often talk about ##\Psi(x, 0)## as the initial wavefunction.

 

[semc]

I see. So in a sense they can be used interchangeably. Thanks!

 

[PeroK]

I see. So in a sense they can be used interchangeably. Thanks!

I wouldn't say that. Take basis vectors and vectors as an example. Basis vectors are just ordinary vectors, but as a set they have a special property (of collectively being a basis). You can't use "basis vector" and "vector" interchangeably.

Similarly, eigenfunctions are wave-functions. You can have a wave-function comprising a single eigenfunction - this is called a stationary state. (Note: I assume from the extract that we are talking about energy eigenfunctions here.) But not all wave-functions are eigenfunctions.

 

[vanhees71]

It's important to distinguish different uses of "wave functions" in this context. It took me a while to understand it, when I learned quantum theory, because it's sometimes not made too clear in textbooks.

In wave mechanics you start with the idea that (for a single non-relativistic particle) there's a wave function describing the (pure) state of the particle, ##\psi(t,\vec{x})##. The physical meaning is that for a particle in this state the probability distribution for its position is given by
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2$$
provided you have properly normalized the wave function,
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x P(t,\vec{x})=1.$$
Further the observables are represented by self-adjoint operators on the corresponding Hilbert space of square-integrable functions, e.g., ##\hat{\vec{x}}=\vec{x}## (i.e., the position operator, applied to the wave function is just the multiplication of the wave function with ##\vec{x})## or ##\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}## (i.e., the momentum operator is given by the gradient of the wave function times ##-\mathrm{i} \hbar##).

The self-adjoint operators have eigenfunctions, sometimes generalized ones if the eigenvalues form a continuous set. Let's take the momentum eigenstates, which must fulfill the equation
$$-\mathrm{i} \hbar \vec{\nabla} u_{\vec{p}}(\vec{x})=\vec{p} u_{\vec{p}}(\vec{x}).$$
The solution is
$$u_{\vec{p}}(\vec{x})=N \exp(\mathrm{i} \vec{p} \cdot \vec{x}/\hbar), \quad \vec{p} \in \mathbb{R}^3.$$
These momentum eigenfunctions are not square integrable but "normalizable to a ##\delta## distribution", i.e., the usual choice of the phase factor is such that
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x u_{\vec{p}}^*(\vec{x}) u_{\vec{p}'}(\vec{x})=\delta^{(3)}(\vec{p}-\vec{p}').$$
From the theory of Fourier integrals one gets ##N=(2 \pi \hbar)^{-3/2}##, i.e., the conveniently normalized momentum eigenfunctions are
$$u_{\vec{p}}(\vec{x})=\frac{1}{\sqrt{(2 \pi \hbar)^3}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}/\hbar).$$
The importance of such eigenfunctions of self-adjoint operators that represent observables is that you get the probabilities (or probability distribution for continuous eigenvalues) by the corresponding projections. E.g., the probability distribution for momentum for a particle prepared in the state represented by the square-integrable wave function ##\psi## is given by
$$\tilde{P}(t,\vec{p})=|\tilde{\psi}(t,\vec{p})|^2$$
where the momentum-space wave function is given by the projection to the momentum eigenfunctions,
$$\tilde{\psi}(t,\vec{p})=\langle u_{\vec{p}}|\psi(t) \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x u_{\vec{p}}^*(\vec{x}) \psi(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 x \frac{1}{\sqrt{(2 \pi \hbar)}^3} \exp(-\mathrm{i} \vec{p} \cdot \vec{x}) \psi(t,\vec{x}).$$
It is also important to know that usually the eigenfunctions of self-adjoint operators are "complete" in the sense that you can write any wave function (to be more careful any wave function in the domain, where the self-adjoint operator is well defined, which build a dense subspace of the Hilbert space of square-integrable functions) as "superposition" of the eigenstates. In the case of operators with continuous eigenvalues the "superposition" is rather an integral than a sum. For the momentum space eigenfunktions you indeed get
$$\int_{\mathbb{R}^3} \mathrm{d}^3 p |u_{\vec{p}} \rangle \langle u_{\vec{p}}|=\hat{1}.$$
This is the "completeness relation" for the momentum eigenstates.

For the wave functions this means that
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 p \langle u_{\vec{p}}|\psi(t) \rangle u_{\vec{p}}(\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{1}{\sqrt{(2 \pi \hbar)^3}} \tilde{\psi}(t,\vec{p}) \exp(\mathrm{i} \vec{p} \cdot \vec{x}/\hbar).$$

 

[semc]

If I understand correctly, you are saying the eigen functions span the entire space which necessary means that the wavefunction can be expressed as combination of eigen functions. Depending on the operators, you can write your wave function in different basis (mometum/space...). It seems similar to my original understanding. In that case, how can the book be right when they say wavefunctions are solution to the TDSE while eigenfunctions are solution to the TISE? Did I misunderstood?

 

[PeroK]

In that case, how can the book be right when they say wavefunctions are solution to the TDSE while eigenfunctions are solution to the TISE? Did I misunderstood?

Strictly speaking we have two different sets of functions: time-independent functions and time-dependent functions. The eigenfunctions of the Hamiltonian span the space of time-independent functions - in principle almost any square-integrable function can be the wavefunction at an instant. So, for any (time-independent) wave-function: $$\psi(x) = \sum c_n \psi_n(x)$$ Then we have the time-dependent wave-functions that are solutions to the SDE: not every function of ##(x,t)## is a solution, as the wave-function must evolve over time according to the SDE for the given potential/Hamiltonian. For a time-independent potential we have: $$\Psi(x, t) = \sum c_n \psi_n(x) \exp(-i\frac{E_n}{\hbar}t)$$
We talk about ##\Psi(x, t)## being the wave-function and also ##\psi(x) = \Psi(x, T)## being the wave-function at some fixed time ##T##.

 

[semc]

I see. Thank you very much for the detailed explanation.