Time Evolution of Expectation Values and Ehrenfest's Theorem

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.

Generalized Heisenberg uncertainty relation

If two opperators 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}} are Hermitian 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{A},\hat{B}]=i\hat{C}\;}

then 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}{4}\langle \hat{C}\rangle^2\leq\langle \left(\Delta \hat{A}\right)^2\rangle\langle \left(\Delta \hat{B}\right)^2\rangle}

Proof:

First recall 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 \Delta \hat{O} = \hat{O} - \langle \hat{O} \rangle } and 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 \Delta \hat{O} } is Hermitian 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{O} } is.

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 \alpha} be a real scalar and define 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 f(\alpha)} as such:

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 f(\alpha) = |( \alpha \Delta \hat{A} - i\Delta \hat{B})|\psi \rangle|^2 } .

So 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 f(\alpha) } is the norm squared of some arbitrary state vector after operating 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 \Delta \hat{A} - i\Delta \hat{B})} on it. Hence by the positive semidefinite property of the norm:

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 f(\alpha) \geq 0 }

Proceeding to calculate this norm squared:

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 \begin{align} f(\alpha)&=\langle \psi |( \alpha \Delta \hat{A} - i\Delta \hat{B})^\dagger( \alpha \Delta \hat{A} - i\Delta \hat{B})|\psi \rangle\\ &=\langle \psi |( \alpha \Delta \hat{A} + i\Delta \hat{B})( \alpha \Delta \hat{A} - i\Delta \hat{B})|\psi \rangle\\ &=\langle \psi | \alpha^2 \left(\Delta \hat{A}\right)^2 -i\alpha [\Delta \hat{A},\Delta \hat{B}] + \left(\Delta \hat{B}\right)^2 |\psi \rangle\\ &=\langle \psi |\alpha^2 \left(\Delta \hat{A}\right)^2 |\psi \rangle + \langle \psi |\alpha \hat{C} |\psi \rangle + \langle \psi |\left(\Delta \hat{B}\right)^2 |\psi \rangle\\ &=\alpha^2\langle \left(\Delta \hat{A}\right)^2 \rangle + \alpha \langle \hat{C} \rangle + \langle \left(\Delta \hat{B}\right)^2 \rangle\\ \end{align}}

Notice 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 f(\alpha)} is a real valued ( expectation values of Hermitian operators are always real ) quadratic in 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 \alpha} which is always greater than or equal to zero. This implies that there are no real solutions for 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} or there is exactly 1. This can be seen by attempting to solve for 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} by using the "quadratic formula" :

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=\frac{-\langle \hat{C} \rangle \pm \sqrt{ (\langle \hat{C} \rangle)^2 -4 \langle (\Delta \hat{A})^2 \rangle \langle (\Delta \hat{B})^2 \rangle }}{2\langle (\Delta \hat{A})^2 \rangle} }

Now we exploit our insight about the number of real roots and see that the discriminant, the term under the square root, must be either 0 ( yielding 1 real solution for 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} ) or negative ( yielding 0 real solutions 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} ). Stated more succinctly:

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 (\langle \hat{C} \rangle)^2 -4 \langle \left(\Delta \hat{A}\right)^2 \rangle \langle \left(\Delta \hat{B}\right)^2 \rangle \leq 0 }

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{(\left \langle \hat{C} \right \rangle )^{2}}{4} \leq \left \langle {(\Delta \hat{A})^{2}} \right \rangle \left \langle {(\Delta \hat{B})^{2}} \right \rangle }

which immediately implies what was to be proved.We can try this relation for 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{x}} 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}=\hat{p}} .

So we have $[{\hat {A}},{\hat {B}}]=ih$ and ${\hat {C}}=h$.

Then ${\frac {(\left\langle {\hat {h}}\right\rangle )^{2}}{4}}\leq \left\langle {(\Delta {\hat {x}})^{2}}\right\rangle \left\langle {(\Delta {\hat {p}})^{2}}\right\rangle$

$\therefore \Delta x\Delta p\geq {\frac {h}{2}}$

which is Heisenberg Uncertainty Principle

Question What about the energy-time uncertainty relation?

Answer: We should note that, time is not an operator in quantum mechanics which forbid us to use the commutation relation to get the uncertainty relation. The energy-time uncertainty relation tells us that for short time interval we get broadening in the energy spectrum which we observe. In other words to get a precise energy value we need to wait for a long time. The short/long time intervals are defined by $\Delta t\Delta E\geq \hbar$

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 ${\hat {A}}$ and ${\hat {B}}$ commute, then there exists a common eigenbasis of these two observables.

This theorem is generalized immediately to the case of several observables ${\hat {A}}$,${\hat {B}}$,${\hat {C}}$ which all commute.

Proof.

Let $\left\{|\alpha ,r_{\alpha }\rangle \right\}$ be the eigenvectors of ${\hat {A}}$, where the index $r_{\alpha }$ means that an eigenvector associated with an eigenvalue $a_{\alpha }$ belongs to an eigensubspace of dimension $d_{\alpha }\geq 1$,

${\hat {A|\alpha ,r_{\alpha }}}\rangle =a_{\alpha }|\alpha ,r_{\alpha }\rangle$ , $r=1,2,....d_{\alpha }$

By assumption, we have $\left[{\hat {A}},{\hat {B}}\right]=0$, that is,

${\hat {A}}{\hat {B}}|\alpha ,r_{\alpha }\rangle ={\hat {B}}{\hat {A}}|\alpha ,r_{\alpha }\rangle =a_{\alpha }|\alpha ,r_{\alpha }\rangle$ , $r=1,2,....d_{\alpha }$

Therefore, the vector ${\hat {B}}|\alpha ,r_{\alpha }\rangle$ 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 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, 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 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}} have in common the basis 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 \{ |\psi _{n} \rangle\right \}} with 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_{n}} 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 b_{n}}  :

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}|\psi _{n}\rangle=a_{n}|\psi \rangle} 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}|\psi _{n}\rangle=b_{n}|\psi \rangle}

If we apply 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}} to the first expression 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{A}} to the second, and subtract, 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 \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 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 \{ \psi _{n} \right \}} is a Hilbert basis, we therefore 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 \{ |\psi _{n} \rangle\right \}}

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 ] |\psi \rangle=0} , whatever 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}

which means 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}

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:

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

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{H}\phi _{n}\left ( x \right )=E_{n}\phi _{n}\left ( x \right )}  ; 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{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.