Time Evolution of Expectation Values and Ehrenfest's Theorem

Quantum Mechanics A

Schrödinger Equation The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.

Physical Basis of Quantum Mechanics

Basic Concepts and Theory of Motion UV Catastrophe (Black-Body Radiation) Photoelectric Effect Stability of Matter Double Slit Experiment Stern-Gerlach Experiment The Principle of Complementarity The Correspondence Principle The Philosophy of Quantum Theory

Schrödinger Equation

Brief Derivation of Schrödinger Equation Relation Between the Wave Function and Probability Density Stationary States Heisenberg Uncertainty Principle Some Consequences of the Uncertainty Principle

Operators, Eigenfunctions, and Symmetry

Linear Vector Spaces and Operators Commutation Relations and Simultaneous Eigenvalues The Schrödinger Equation in Dirac Notation Transformations of Operators and Symmetry Time Evolution of Expectation Values and Ehrenfest's Theorem

Motion in One Dimension

One-Dimensional Bound States Oscillation Theorem The Dirac Delta Function Potential Scattering States, Transmission and Reflection Motion in a Periodic Potential Summary of One-Dimensional Systems

Discrete Eigenvalues and Bound States

Harmonic Oscillator Spectrum and Eigenstates Analytical Method for Solving the Simple Harmonic Oscillator Coherent States Charged Particles in an Electromagnetic Field WKB Approximation

Time Evolution and the Pictures of Quantum Mechanics

The Heisenberg Picture: Equations of Motion for Operators The Interaction Picture The Virial Theorem

Angular Momentum

Commutation Relations Angular Momentum as a Generator of Rotations in 3D Spherical Coordinates Eigenvalue Quantization Orbital Angular Momentum Eigenfunctions

Central Forces

General Formalism Free Particle in Spherical Coordinates Spherical Well Isotropic Harmonic Oscillator Hydrogen Atom WKB in Spherical Coordinates

The Path Integral Formulation of Quantum Mechanics

Feynman Path Integrals The Free-Particle Propagator Propagator for the Harmonic Oscillator

Continuous Eigenvalues and Collision Theory

Differential Cross Section and the Green's Function Formulation of Scattering Central Potential Scattering and Phase Shifts Coulomb Potential Scattering

It is reasonable to expect the motion of a wave packet to agree with the motion of the corresponding classical particle whenever the potential energy changes by a negligible amount over the dimensions of the packet. If we mean by the position and momentum vectors of the packet the weighted averages or expectation values of these quantities, we can show that the classical and quantum mechanics always agree. A component of the velocity of the packet will be the time rate of change of the expectation value of that component of the position; since < x > depends only on the time and the x in the integrand

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}< x > = \frac{\mathrm{d} }{\mathrm{d} t}\int \psi \ast \left ( r \right )x\psi \left ( r \right )d^{3}r=\int \psi \ast\left ( r \right ) x\frac{\mathrm{d} }{\mathrm{d} x}\psi \left ( r \right )d^{3}r+\int \frac{\mathrm{d} }{\mathrm{d} t}\psi \ast \left ( r \right )x\psi \left ( r \right )d^{3}r}

Note that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial }{\partial t\ }\psi =-\frac{\hbar}{2im}\triangledown ^{2}\Psi +\frac{V}{i\hbar}\psi}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial }{\partial t\ }\psi\ast =\frac{\hbar}{2im}\triangledown ^{2}\Psi\ast -\frac{V}{i\hbar}\psi\ast}

This may be simplified by substituting for the time derivatives of the wave function and its complex conjugate and canceling the term involving potential V where we continue to assume that V is real:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}\left \langle x \right \rangle=\int \psi \ast \left ( r \right )x\left ( \left ( \frac{-\hbar}{2im} \right )\triangledown ^{2}\psi +\frac{V}{i\hbar}\psi \right )d^{3}r+\int \left ( \left ( \frac{\hbar}{2im}\right )\triangledown ^{2 }\psi -\frac{V}{i\hbar}\psi\ast \right )x\psi d^{3}r}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{i\hbar}{2m}\int \left \{ \psi \ast \left ( r \right )x\left ( \triangledown ^{2} \psi \right ) -\left ( \triangledown ^{2}\psi \ast \right )x\psi \right \}d^{3}r}

We shall now use the identity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{\triangledown }.\left ( x\psi \vec{\triangledown }\psi \ast \right )=\vec{\triangledown }\left ( x\psi \right ).\vec{\triangledown }\psi \ast +x\psi \left ( \triangledown ^{2} \psi \ast \right )}

Consider two scalar function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\psi} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\psi} that are continuous and differentiable in some volume V bounded bounded by a surface S. Applying the divergence theorem to the vector field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\psi\psi \ast } (the left hand side of the identity) we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{S}x\psi \vec{\triangledown }\psi \ast.d\vec{S} =\int_{V}\left \{ \left ( x\psi \right )\triangledown ^{2}\psi \ast +\left ( \vec{\triangledown }x\psi \right ).\left ( \bar{\triangledown }\psi \ast \right ) \right \}d^{3}r}

Therefore the second integral of first equation can be written as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{V}\left ( \triangledown ^{2}\psi \ast \right )x\psi d^{3}r=-\int \left ( \vec{\triangledown }\psi \ast \right ).\left ( \vec{\triangledown } x\psi \right )d^{3}r+\int_{S}\left ( x\psi \vec{\triangledown }\psi \ast \right ).dS}

where the second integral of the normal component of xψ ψ ∗ over the infinite bounding surface A is zero because a wave packet ψ vanishes at great distances and hence

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{V}\left ( \triangledown ^{2}\psi \ast \right )x\psi d^{3}r=\int \psi \ast \triangledown ^{2}\left ( x\psi \right )d^{3}r}

We again used the fact that the surface integral again vanishes, to obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{V}\left ( \triangledown ^{2}\psi\ast \right )x\psi d^{3}r=\int_{V}\left ( \psi \ast \triangledown ^{2}\left ( x\psi \right ) \right )d^{3}r} Thus,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}\left \langle x \right \rangle=\frac{i\hbar}{2m}\int_{V}\psi \ast \left \{ x\triangledown ^{2}\psi -\triangledown ^{2}\left ( x\psi \right ) \right \}d^{3}r} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{1}{m}\left \langle p_{x} \right \rangle}

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \langle x \right \rangle} is seen always to be real number from the inherent structure of its definition. The above equation shows quite incidentally that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \langle px \right \rangle} is real.

In similar fashion we can calculate the time rate of change of a component of the momentum of the particle as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}\left \langle p_{x} \right \rangle=-\frac{\hbar^{2}}{2m}\int \left \{ \left ( \triangledown ^{2}\psi\ast \right )\frac{\partial \psi }{\partial x}d^{3}r-\psi \ast \triangledown ^{2}\left ( \frac{\partial \psi }{\partial x} \right )+\int V\psi \ast \frac{\partial \psi }{\partial x} d^{3}r-\psi \ast \frac{\partial V\psi }{\partial x} \right \}d^{3}r}

The integral containing laplacins can be transofrmed into a surface integral by Green’s theorem and write

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{V}\left \{ \left ( \triangledown ^{2}\psi \ast \right )\frac{\partial \psi }{\partial x} -\psi \ast \triangledown ^{2}\left ( \frac{\partial \psi }{\partial x} \right )\right \}d^{3}r=\int_{S}\left \{ \frac{\partial \psi }{\partial x} \vec{\triangledown }\psi \ast -\psi \ast \vec{\triangledown }\left ( \frac{\partial \psi }{\partial x} \right )\right \}.d\vec{S} }

It is assumed that the last integral vanishes when taken over a very large surface S. (One can also show that volume integral involving the laplacian vanishes by doing integration by parts twice. For instance, we use the identity

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{\triangledown }.\vec{\triangledown }\psi \ast \left ( \frac{\partial \psi }{\partial x} \right )=\triangledown ^{2}\psi \ast \frac{\partial \psi }{\partial x}+\vec{\triangledown }\psi \ast.\vec{\triangledown }\frac{\partial \psi }{\partial x}}

Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{V}\triangledown ^{2}\psi\ast \frac{\partial \psi }{\partial x}d^{3}r=\int_{V}\vec{\triangledown }.\vec{\triangledown }\psi \ast \left ( \frac{\partial \psi }{\partial x} \right )-\int_{V}\vec{\triangledown }\psi \ast.\vec{\triangledown }\frac{\partial \psi }{\partial x}d^{3}r}

Using the Gauss’s theorem the first on he right hand side can be converted in to a surface integral over the surface that enclose the volume and hence it vanishes. Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{V}\triangledown ^{2}\psi \ast \frac{\partial \psi }{\partial x}d^{3}r=-\int_{V}\vec{\triangledown }\psi .\vec{\triangledown \frac{\partial \psi }{\partial x}}d^{3}r}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\int_{V}\psi \ast \triangledown ^{2}\frac{\partial \psi }{\partial x}d^{3}r}

Thus the Laplacian term vanishes resulting in

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}\left \langle p_{x} \right \rangle=\int \left \{ V\psi \ast \frac{\partial \psi }{\partial x}d^{3}r-\psi \ast \frac{\partial V\psi }{\partial x} \right \}d^{3}r}

We can establish a general formula for the time derivative of the expectation value < F > of any operator F.

The time derivative of the expectation value of any operator F which may be explicitly time dependent, can be written as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\mathrm{d} }{\mathrm{d} t}\left \langle \hat{F} \right \rangle=i\hbar\int \psi \ast \hat{F}\frac{\partial }{\partial t}\psi d^{3}r+i\hbar\int \frac{\partial }{\partial t}\psi \ast \hat{F}\psi d^{3}r+i\hbar\int \psi \ast \frac{\partial }{\partial t}\hat{F}\psi d^{3}r}

In the last step, we have used the Green’s theorem, and vanishing boundary surface terms. The potential energy is, of course, assumed to be real. Compactly, we may write the last result as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\mathrm{d} }{\mathrm{d} t}\left \langle \hat{F} \right \rangle=\left \langle \hat{F}\hat{H}-\hat{H} \hat{F}\right \rangle+i\hbar\left \langle \frac{\partial }{\partial t}\hat{F} \right \rangle}

This formula is of the utmost importance in all facets of quantum mechanics.

Commuting observables

When two observables commute, there is no constraint such as the uncertainty relations. This case is, however, very interesting in practice.

THEOREM

We know that if two matrices commute, one can diagonalize them simultaneously.This remain true in infinite dimensional case. If two observables Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{B}} commute, then there exists a common eigenbasis of these two observables.

This theorem is generalized immediately to the case of several observables Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{B}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{C}} which all commute.

Proof.

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \{| \alpha ,r_{\alpha } \rangle \right \}} be the eigenvectors of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} , where the index Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_{\alpha }} means that an eigenvector associated with an eigenvalue Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{\alpha }} belongs to an eigensubspace of dimension Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d_{\alpha }\geq 1} ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A|\alpha ,r_{\alpha }}\rangle =a_{\alpha }|\alpha ,r_{\alpha }\rangle} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=1,2,....d_{\alpha }}

By assumption, we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left [ \hat{A},\hat{B} \right ]=0} , that is,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}\hat{B}|\alpha ,r_{\alpha }\rangle=\hat{B}\hat{A}|\alpha ,r_{\alpha }\rangle=a_{\alpha }|\alpha ,r_{\alpha }\rangle} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=1,2,....d_{\alpha }}

Therefore, the vector Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{B}|\alpha ,r_{\alpha }\rangle} is an eigenvector of A with the eigenvalues Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{\alpha }} . It therefore belongs to the corresponding eigensubspace. We call this vector Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\alpha ,\beta ,k_{\alpha \beta }\rangle}  ; the index Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{\alpha \beta }} means that again this vector may be nonunique. Therefore, this vector is a linear combination of the vectors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \{ |\alpha ,r_{\alpha }\rangle \right \}} , that is, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{B}|\alpha ,r_{\alpha }\rangle=\sum_{r_{\alpha }}b_{r_{\alpha }}|\alpha ,r_{\alpha }\rangle}

which can be diagonalized with no difficulty. In other words, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} and ${\hat {B}}$ commute, they possess a common eigenbasis.

The reciprocal is simple. The Riesz theorem says that the othonormal eigenvectors of an observable form a Hilbert basis. Suppose ${\hat {A}}$ and ${\hat {B}}$ have in common the basis $\left\{|\psi _{n}\rangle \right\}$ with eigenvalues $a_{n}$ and $b_{n}$ :

${\hat {A}}|\psi _{n}\rangle =a_{n}|\psi \rangle$ and ${\hat {B}}|\psi _{n}\rangle =b_{n}|\psi \rangle$

If we apply ${\hat {B}}$ to the first expression and ${\hat {A}}$ to the second, and subtract, we obtain

$\left({\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}\right)|\psi _{n}\rangle =\left(a_{n}b_{n}-b_{n}a_{n}\right)|\psi _{n}\rangle$

Because $\left\{\psi _{n}\right\}$ is a Hilbert basis, we therefore have $\left\{|\psi _{n}\rangle \right\}$

$\left[{\hat {A}},{\hat {B}}\right]|\psi \rangle =0$ , whatever $|\psi \rangle$

which means $\left[{\hat {A}},{\hat {B}}\right]=0$

Example

Consider, for instance, an isotropic two-dimensional harmonic oscillator. The eigenvalue problem of the Hamiltonian is a priori a difficult problem because it seems to be a partial differential equation in two variables. But the Hamiltonian can be written as the sum of two independent Hamiltonians acting on different variables:

${\hat {H}}={\frac {-\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {1}{2}}m\omega ^{2}x^{2}-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {1}{2}}m\omega ^{2}y^{2}={\hat {H_{x}}}+{\hat {H_{y}}}$

The two operators ${\hat {H_{x}}}$ and ${\hat {H_{y}}}$ , which are both operators in one variable and which act on different variables commute obviously. One can solve the eigenvalue problems of ${\hat {H_{x}}}$ and ${\hat {H_{y}}}$ separately

${\hat {H}}\phi _{n}\left(x\right)=E_{n}\phi _{n}\left(x\right)$  ; ${\hat {H}}\phi _{n}\left(y\right)=E_{n}\phi _{n}\left(y\right)$

The eigenvalues of ${\hat {H}}$ are the sums of eigenvalues of ${\hat {H_{x}}}$ and ${\hat {H_{y}}}$ with eigenfunctions that are the products of corresponding eigenfunctions:

$E_{n}=E_{n_{1}}+E_{n_{2}}\left(n_{1}+n_{2}+1\right)\left(\hbar \omega \right)$  ; $\psi _{n}\left(x,y\right)=\phi _{n_{1}}\left(x\right)\phi _{n_{2}}\left(y\right)$

In other words, a sum of Hamiltonians that commute has for eigenvalues the sum of eigenvalues of each of them, and for eigenfunctions the product of corresponding eigenfunctions.