Eigenfunction of multiple operators simultaneously?

[JaredMTg]

Hello,

I am taking an introductory course in quantum mechanics. One thing I am confused about is, Schrödinger's equation seems to be regarded as the "ultimate" formula which determines a particle's possible wavefunctions and energies, given a certain potential (Hamiltonian of psi = Energy eigenvalue * psi).

But then other operators, such as L^2 (angular momentum) have spherical harmonics as their eigenfunctions. If you tried to plug in the wavefunction that is a solution to the infinite-square well problem (i.e. something of the form psi = C sin sqrt(kx)) into the angular momentum operator, it would not be a solution (i.e. eigenfunction) to the L^2 operator that would return a value for L^2 multiplied by the original wavefunction.

So what if I wanted to measure the angular momentum squared of the particle in the infinite well? Would the wavefunction suddenly change to become a spherical harmonic?

In other words, how can a wavefunction be a solution to Schrödinger's equation as well as the many other operators representing real physical quantities (such as momentum) at the same time? If the measurement of the particular quantity (whether it is energy or something else like angular momentum) is what determines the solution, and therefore the wavefunction, then why is Schrödinger's equation considered any more fundamental than finding the solution of the L^2 operator?

Many thanks in advance for explaining this to me.

 

[DrClaude]

You have to consider what is called the Complete Set of Commuting Observables (CSCO). A quantum system is uniquely defined by the smallest set of quantum numbers that come from operators that commute with each other.

In the case of the infinite square well, you will find that $[\hat{H}, \hat{L}^2] \neq 0$, therefore you cannot simultaneously define the energy of the system and its angular momentum. Were you to measure the angular momentum of a system in an eigenstate of the Hamiltonian, after the measurement the system would be in a (linear combination) of spherical harmonics with a given angular momentum quantum number $l$, but it would no longer be in an energy eigenstate.

 

[vanhees71]

Conceptional wise, one should add that you are completely right in saying that the time-independent Schrödinger equation is in principle nothing else than the eigenvalue equation for the Hamiltonian.

The Hamiltonian, however has a very special status in quantum theory, because it is the operator of time evolution, i.e., it contains the entire dynamics of the system, and the time-dependent Schrödinger equation is the prime dynamical equation of quantum theory (in the position representation), i.e., it tells you, how the state of the system, represented by a wave function, which is the state in the position representation, changes with time due to the dynamics of the system, i.e., knowing the wave function at a time $t_0$ and the Hamiltonian via the Schrödinger equation provides you with the wave function at any later time $t>t_0$.

On the other hand, if you have found a complete set of energy eigenstates $u_n(\vec{x})$, fulfilling the time-independent Schrödinger equation,
$$\hat{H} u_n(\vec{x})=E_n u_n(\vec{x}),$$
which indeed is nothing else than the eigenvalue problem for the Hamiltonian, then you can represent any wave function in terms of the energy eigen functions,
$$\psi(t,\vec{x})=\sum_{n} a_n(t) u_{n}(\vec{x}).$$
Now you can write down the time-dependent Schrödinger equation, which determines the wave function's time evolution in terms of the $a_n$:
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}) \; \Rightarrow \; \sum_{n} \hbar \dot{a}_n(t) u_n(\vec{x}) = \sum_n a_n \hat{H} u_n(\vec{x}) = \sum_n a_n E_n(\vec{x}).$$
Since the expansion of the wave function in terms of the energy eigenfunctions is unique, this implies the equation
$$\mathrm{i} \hbar \dot{a}_n = E_n a_n,$$
which is nothing else than the Schrödinger equation for the wave function in energy representation! But this is trivial to solve:
$$a_n(t)=N_n \exp \left (-\frac{\mathrm{i} E_n}{\hbar}(t-t_0) \right).$$
The $N_n$ must be determined from the initial condition, because now we have
$$\psi(t,\vec{x})=\sum_n N_n \exp \left (-\frac{\mathrm{i} E_n}{\hbar}(t-t_0) \right) u_n(\vec{x}).$$
The initial condition is given by the wave function at $t=t_0$, $\psi(t=t_0,\vec{x})=\phi(\vec{x})$. Thus we have
$$\phi(\vec{x})=\sum_n N_n u_n(\vec{x}) \; \Rightarrow \; N_n=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_n^*(x) \phi(\vec{x}).$$
As you see, now you have a complete solution of the initial value problem of the time-dependent Schrödinger equation in terms of the energy eigenstates, and that's why everybody is so eager to solve the time-independent Schrödinger equations.

This does not imply that these energy eigensolutions are the only possible states. They are the stationary states, because for them the whole time evolution is just the phase factor $\exp(-\mathrm{i} E_n t/\hbar)$, i.e., the probability distribution, which is $|u_n(t,\vec{x})|^2=|u_n(\vec{x})|^2$, is time-independent.

 

[PeroK]

Perhaps something from the perspective of Linear Algebra will help:

Suppose you have a function $\Psi$ (could be any function). And suppose you have a Hermitian Operator $Q$ (could be any Hermitian Operator). Now, the eigenfunctions of $Q$ form a complete, orthonormal set: a "basis" for your Hilbert Space. Therefore, you can express $\Psi$ as a linear combination of eigenfunctions of $Q$.

If, say, $\Psi$ is a solution to the Schrödinger equation (in particular an eigenfunction of the Hamiltonian operator), then that $\Psi$ can be expressed as a linear combination of the eigenfunctions of any other Hermitian Operator ($L^2$, for example). Those functions will generally not be eigenfunctions of the Hamiltonian, hence not solutions to the Schrödinger equation themselves. But, they still form a "basis" for your Hilbert Space. Hence, any solution to the Schrödinger equation can be expressed as a linear combination of those functions.

 

[JaredMTg]

These are all very helpful and great responses. Thank you very much.