Commutation Relations and Simultaneous Eigenvalues

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

Commutator

The commutator of 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{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}\!} is defined as follows:

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}]=\hat{A}\hat{B}-\hat{B}\hat{A}\,\!.}

When 2 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{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 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}]=0.} On the other hand, 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},\hat{B}]\neq 0,} then the operators do not commute, and we can think of the commutator between two operators as a measure of how badly they fail to commute. Note that any operator will commute with an ordinary complex number.

Some Identities:
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}]+[\hat{B},\hat{A}]=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 [\hat{A},\hat{A}]= 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 [\hat{A},\hat{B}+\hat{C}]=[\hat{A},\hat{B}]+[\hat{A},\hat{C}]\!}
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},\hat{C}]=[\hat{A},\hat{C}]+[\hat{B},\hat{C}]\!}
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},\hat{C}]=\hat{A}[\hat{B},\hat{C}]+[\hat{A},\hat{C}]\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{A},\hat{B}\hat{C}]=[\hat{A},\hat{B}]\hat{C}+\hat{B}[\hat{A},\hat{C}]\!}
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},\hat{C}]]+[\hat{C},[\hat{A},\hat{B}]]+[\hat{B},[\hat{C},\hat{A}]]=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 [\hat{A},\hat{B}]=-[\hat{B},\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{A},\hat{B}]^{\dagger} = [\hat{B}^{\dagger},\hat{A}^{\dagger}]}

In addition, if any two operators are Hermitian and their product is also Hermitian, then the operators commute 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 (\hat{A}\hat{B})^{\dagger} = \hat{B}^{\dagger}\hat{A}^{\dagger} = \hat{B}\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{A}\hat{B})^{\dagger} = \hat{A}\hat{B}}

so 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 \hat{A}\hat{B} = \hat{B}\hat{A}.}

One may also prove 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 [\hat{A}^n,\hat{B}]=n\hat{A}^{n-1}[\hat{A},\hat{B}]\!} 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},[\hat{A},\hat{B}]]=0\!} via mathematical induction.

Compatible Observables

An operator which corresponds to some physically measurable property of a system is called an observable. The following is a list of common physical observables and their corresponding operators given in the coordinate representation:
Position: 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 \mathbf{r}\rightarrow\hat{\mathbf{r}} }
Momentum: 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 \mathbf{p}\rightarrow\frac{\hbar}{i}\nabla}
Kinetic energy: 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 T=\frac{p^2}{2m}\rightarrow -\frac{\hbar^2}{2m}\nabla^2}
Potential energy: 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 V(\mathbf{r})\rightarrow V(\hat{\mathbf{r}})}
Total energy: 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 E=T+V\rightarrow -\frac{\hbar^2}{2m}\nabla^2+V(\hat{\mathbf{r}})}

All observables are Hermitian.

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}} are said to be compatible if it is possible to exactly measure both simultaneously; i.e., they possess a common set of eigenstates:

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_{AB}\rangle=a|\Psi_{AB}\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 \hat{B}|\Psi_{AB}\rangle=b|\Psi_{AB}\rangle\!}

It follows from this 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 \hat{A}\hat{B}|\Psi_{AB}\rangle=\hat{A}b|\Psi_{AB}\rangle=b\hat{A}|\Psi_{AB}\rangle=ba|\Psi_{AB}\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}\hat{A}|\Psi_{AB}\rangle=\hat{B}a|\Psi_{AB}\rangle=a\hat{B}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle.\!}

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 \hat{A}\hat{B}-\hat{B}\hat{A}=[\hat{A},\hat{B}]=0.\!}

The same logic works in reverse - if two operators commute, then they have simultaneous eigenkets, and thus they are compatible observables. Let us consider two commuting 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{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}.} Suppose that we know the eigenstates 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};} let us call them 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_{A}\rangle.} We know 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 \hat{A}|\Psi_{A}\rangle=a|\Psi_{A}\rangle.}

We now act on both sides of this equation with 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{B}\hat{A}|\Psi_{A}\rangle=a\hat{B}|\Psi_{A}\rangle,}

or, 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 \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,

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}|\Psi_{A}\rangle=a\hat{B}|\Psi_{A}\rangle.}

We see 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 \hat{B}|\Psi_{A}\rangle} is therefore also an eigenstate 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}} with the same eigenvalue. This means that we have "block diagonalized" 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}} in the sense that it can only mix eigenstates 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}} with other eigenstates with the same eigenvalue. All of the eigenstates 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{B}} are thus also eigenstates of ${\hat {A}},$ meaning that the two observables are compatible.

These observations are especially important when we consider the Hamiltonian of a system. Symmetries of the Hamiltonian are represented by operators that commute with it, and, as we will see later, identification of these symmetries often helps one diagonalize the Hamiltonian and classify its eigenstates.

If $[{\hat {A}},{\hat {B}}]\neq 0$, or, equivalently, if it is not possible to simultaneously diagonalize ${\hat {A}}$ and ${\hat {B}},$ then the two operators are said to be incompatible observables.

Generalized Uncertainty Relation

If two observables are incompatible, then one cannot necessarily simultaneously diagonalize them, so that an eigenstate of one may be a non-trivial linear combination of eigenstates of the other. This fact is embodied in the existence of an uncertainty relation for the two observables, much like that between position and momentum. We will now derive this general uncertainty relation.

For any observable $A$, let us define the uncertainty $\Delta A$ as the standard deviation of said observable from its expectation value:

$(\Delta A)^{2}=\langle (A-\langle A\rangle )\Psi |(A-\langle A\rangle )\Psi \rangle =\langle f|f\rangle$

where $f\equiv (A-\langle A\rangle )\Psi$. Likewise, for any other observable $B$,

$(\Delta B)^{2}=\langle g|g\rangle$ where $g\equiv (B-\langle B\rangle )\Psi$.

Now we invoke the Schwartz inequality. Recall that this is just

$\langle f|f\rangle \langle g|g\rangle \geq |\langle f|g\rangle |^{2}.$

Invoking this expression,

$(\Delta A)^{2}(\Delta B)^{2}\geq |\langle f|g\rangle |^{2}.$

Now, for any complex number $z$,

$|z|^{2}=[{\text{Re}}(z)]^{2}+[{\text{Im}}(z)]^{2}\geq [{\text{Im}}(z)]^{2}=\left({\frac {z-z^{*}}{2i}}\right)^{2}$.

Letting $z=\langle f|g\rangle$,

$(\Delta A)^{2}(\Delta B)^{2}\geq \left({\frac {\langle f|g\rangle -\langle g|f\rangle }{2i}}\right)^{2}$.

The inner products are

$\langle f|g\rangle =\langle (A-\langle A\rangle )\Psi |(B-\langle B\rangle )\Psi \rangle =\langle (A-\langle A\rangle )(B-\langle B\rangle )\rangle$

$=\langle AB-A\langle B\rangle -B\langle A\rangle +\langle A\rangle \langle B\rangle \rangle$

$=\langle AB\rangle -\langle B\rangle \langle A\rangle -\langle A\rangle \langle B\rangle +\langle A\rangle \langle B\rangle =\langle AB\rangle -\langle A\rangle \langle B\rangle$

and

$\langle g|f\rangle =\langle BA\rangle -\langle B\rangle \langle A\rangle .$

Therefore,

$\langle f|g\rangle -\langle g|f\rangle =\langle AB\rangle -\langle BA\rangle =\langle [A,B]\rangle$

and

$(\Delta A)^{2}(\Delta B)^{2}\geq \left({\frac {1}{2i}}\langle [A,B]\rangle \right)^{2},$

or

$\Delta A\,\Delta B\geq \left|{\frac {1}{2i}}\langle [A,B]\rangle \right|.$

This is the general uncertainty relation that we sought.

As a simple demonstration of this relation, suppose that the first observable is position, $A=x$, and the second is momentum, $B=p={\frac {\hbar }{i}}{\frac {d}{dx}}$.

The commutation relation between these two observables is just

$[x,p]=i\hbar ,$

so

$\Delta x\,\Delta p\geq {\frac {\hbar }{2}}.$

We have thus obtained a formal proof of the Heisenberg Uncertainty Principle.

Question: What about the energy-time uncertainty relation?

Answer: We should note that time is not an operator in quantum mechanics, and thus we cannot apply the general uncertainty relation derived above to it. The energy-time uncertainty relation tells us that, for a 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 relation between the energy and time intervals is given by 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 E\,\Delta t\geq \hbar.}

Functions of Operators

It is often useful to consider functions of operators, as the Hamiltonian can be written as such. For example, the exponential function of an operator is represented as a power series,

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 e^\hat{A} = \hat{I} + \frac{\hat{A}}{1!} + \frac{\hat{A}^2}{2!}+ \frac{\hat{A}^3}{3!} + \ldots }

This leads to the identity that for some 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(\lambda) = e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}} , where 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}} are 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 e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}} = \hat{B} + \frac{\lambda}{1!}[\hat{A},\hat{B}] + \frac{\lambda^2}{2!}[\hat{A},[\hat{A},\hat{B}]] + \frac{\lambda^3}{3!}[\hat{A},[\hat{A},[\hat{A},\hat{B}]]] + \ldots }

This follows easily when one considers a Taylor series expansion 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 f(\lambda).\!} We 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{d}{d\lambda}f(\lambda)=e^{\lambda\hat{A}}[\hat{A},\hat{B}]e^{-\lambda\hat{A}}.}

We may then apply this relation recursively to generate the higher-order derivatives required in the Taylor expansion.

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},[\hat{A},\hat{B}]]=\beta\hat{B},} then we can simplify the identity to

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 e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}=\hat{B}\cosh\lambda\sqrt{\beta}+\frac{[\hat{A},\hat{B}]}{\sqrt{\beta}}\sinh\lambda\sqrt{\beta}.}

It is also useful to consider the commutator of an operator function with an operator. Given 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{l}} 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{m}} where

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{l},\hat{m}] = 1 }

it can be shown that, for arbitrary 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(\hat{l})} ,

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(\hat{l}),\hat{m}] = \frac{d}{d\hat{l}}f(\hat{l})}

by use of the power series expansion 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 f(\hat{l})} and the commutator identities in the above section.

Physical Meaning of Various Representations

In a given Hilbert basis, it is obvious that the state 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 |\psi \rangle} is completely determined by the set of its components 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 \{c_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 |\psi\rangle\leftrightarrow\left \{c_{n}=\langle\phi_{n}|\psi\rangle\right \}}

which we can write as a column vector; the corresponding bra is then represented by the conjugate line vector. This representation of the state vector is completely equivalent to the wave 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 \Psi(\mathbf{r},t).}

Therefore, there are not only two, but an infinite number of equivalent representations of the state of the system. What is their physical meaning?

In the basis of the eigenstates of the Hamiltonian, the interpretation of this representation is simple and crystal clear: the coefficients 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 c_{n}\!} are the probability amplitudes for obtaining 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 E_{n}\!} in an energy measurement.

Therefore,

• The position representation, $\Psi (\mathbf {r} ,t),$ is more convenient if we are interested in the properties of the particle in space.

• Its Fourier transform, $\Phi (\mathbf {p} ,t),$ is more convenient if we are interested in its momentum properties.

• The coefficients $c_{n}\!$ in the basis of energy eigenstates are more convenient if we are interested in the energy of the particle.

Owing to Riesz’s theorem, this can be done with any observable, such as the angular momentum, which we examine later on and also has discrete eigenvalues. This can be thought of as a “generalization” of the Fourier transform.

Position and Momentum Operators

An extremely useful example is the commutation relation of the position ${\hat {x}}$ and momentum ${\hat {p}}$ operators. In the position representation, the position operator is ${\hat {x}}\rightarrow x$ and the momentum operator is ${\hat {p}}\rightarrow {\frac {\hbar }{i}}{\frac {\partial }{\partial x}}$. Similarly, in the momentum representation, the momentum operator is ${\hat {p}}\rightarrow p$ and position operator ${\hat {x}}\rightarrow i\hbar {\frac {\partial }{\partial p}}$.

Applying ${\hat {x}}$ and ${\hat {p}}$ to an arbitrary state vector, we can see that

$\left[{\hat {x}},{\hat {p}}\right]=i\hbar .$

The position and momentum operators are thus incompatible. This provides a fundamental contrast to classical mechanics in which $x$ and $p$ obviously commute. Such a commutation relation holds for any coordinate and its canonically conjugate momentum; the commutation relations for all coordinates and their corresponding momenta are known as the canonical commutation relations.

In three dimensions, the canonical commutation relations are

$\left[{\hat {x}}_{i},{\hat {p}}_{j}\right]=i\hbar \delta _{ij},$

$\left[{\hat {x}}_{i},{\hat {x}}_{j}\right]=0,$

and

$\left[{\hat {p}}_{i},{\hat {p}}_{j}\right]=0,$

where the indices stand for the $x,$ $y,$ and $z$ components of the 3-vectors.

It is again interesting to consider functions of these operators. For some function of a coordinate ${\hat {q}},$ $f({\hat {q}}),$ it is easily shown that

$[{\hat {p}},f({\hat {q}})]=-i\hbar {\frac {\partial }{\partial q}}f({\hat {q}}).$

Additionally, the expression $e^{i\lambda {\hat {p}}/\hbar }f({\hat {q}})e^{-i\lambda {\hat {p}}/\hbar }=f({\hat {q}}+\lambda );$ i.e., the operator ${\hat {T}}(\lambda )=e^{i\lambda {\hat {p}}/\hbar }$ is a translation operator that shifts the coordinate ${\hat {q}}$ by an amount $\lambda .\!$

Connection between Classical and Quantum Mechanics

There is a wonderful connection between classical and quantum mechanics. The Hamiltonian is a function that appears in classical mechanics and is "promoted" to an operator in quantum mechanics. Classically, the Hamiltonian is defined as

$H(q,p;t)=\sum _{k}p_{k}{\dot {q}}_{k}-L(q,{\dot {q}};t),$

where $L(q,{\dot {q}};t)$ is the Lagrangian. There are two properties of this quantity.

1. If $L$ does not depend explicitly on time, then $H$ is conserved.
2. Furthermore, if the potential energy and the constraints of the system are time-independent, then $H$ is not only conserved, but $H$ is the energy of the system.

The equations of motion for the system that we may derive from the Hamiltonian are

${\dot {q}}_{k}={\frac {\partial H}{\partial p_{k}}}$

and

${\dot {p}}_{k}=-{\frac {\partial H}{\partial q_{k}}}.$

These equations are called Hamilton's equations of motion. We also define the quantity,

$[A,B]=\sum _{k}\left({\frac {\partial A}{\partial q_{k}}}{\frac {\partial B}{\partial p_{k}}}-{\frac {\partial A}{\partial p_{k}}}{\frac {\partial B}{\partial q_{k}}}\right),$

known as the Poisson bracket. We may observe an interesting connection between this purely classical concept and the commutators that we encounter in quantum mechanics as follows. Let us calculate this quantity with $A$ and $B$ different coordinates and momenta.

$[p_{i},p_{j}]=\sum _{k}\left({\frac {\partial p_{i}}{\partial q_{k}}}{\frac {\partial p_{j}}{\partial p_{k}}}-{\frac {\partial p_{i}}{\partial p_{k}}}{\frac {\partial p_{j}}{\partial q_{k}}}\right)=0$

$[q_{i},q_{j}]=\sum _{k}\left({\frac {\partial q_{i}}{\partial q_{k}}}{\frac {\partial q_{j}}{\partial p_{k}}}-{\frac {\partial q_{i}}{\partial p_{k}}}{\frac {\partial q_{j}}{\partial q_{k}}}\right)=0$

$[p_{i},q_{j}]=\sum _{k}\left({\frac {\partial p_{i}}{\partial q_{k}}}{\frac {\partial q_{j}}{\partial p_{k}}}-{\frac {\partial p_{i}}{\partial p_{k}}}{\frac {\partial q_{j}}{\partial q_{k}}}\right)=\sum _{k}-\delta _{ik}\delta _{jk}=-\delta _{ij}.$

These relations are very similar to the canonical commutation relations discussed earlier. In fact, if we make the replacement,

$\delta _{ij}\rightarrow i\hbar \delta _{ij},$

and replace the Poisson brackets with commutators, then we obtain the canonical commutation relations. This identification is called canonical quantization.

As a final important remark, there is no classical analogue to the Heisenberg uncertainty relation for conjugate classical variables in terms of, say, their Poisson brackets. This is because, in classical mechanics, the objects that we study are point particles (or collections thereof) with well-defined positions and momenta. In quantum mechanics, on the other hand, we study the state of a system as encoded in its wave function. This wave function describes the probability of finding a particle at a given position or with a given momentum; the particles that make up the system do not have definite positions or momenta.

Problems

(1) Let $f(x)\!$ be a differentiable function. Using the fact that $[{\hat {x}},{\hat {p}}_{x}]=i\hbar ,$ prove the following identities:

(a) $[{\hat {x}},{\hat {p}}_{x}^{2}f({\hat {x}})]=2i\hbar {\hat {p}}_{x}f({\hat {x}})$

(b) $[{\hat {x}},{\hat {p}}_{x}f({\hat {x}}){\hat {p}}_{x}]=i\hbar [f({\hat {x}}){\hat {p}}_{x}+{\hat {p}}_{x}f({\hat {x}})]$

(c) $[{\hat {p}}_{x},{\hat {p}}_{x}^{2}f({\hat {x}})]=-i\hbar {\hat {p}}_{x}^{2}{\frac {df({\hat {x}})}{dx}}$

(d) $[{\hat {p}}_{x},{\hat {p}}_{x}f({\hat {x}}){\hat {p}}_{x}]=-i\hbar {\hat {p}}_{x}{\frac {df({\hat {x}})}{dx}}{\hat {p}}_{x}$

Solution

(2) (From Sakurai, Modern Quantum Mechanics, Problem 1.30)

The translation operator for a finite (spatial) displacement $\mathbf {l}$ is given by ${\hat {T}}(\mathbf {l} )=\exp \left(-{\frac {i{\hat {\mathbf {p} }}\cdot \mathbf {l} }{\hbar }}\right),$ where ${\hat {\mathbf {p} }}$ is the momentum operator.

(a) Evaluate $[{\hat {x}}_{i},{\hat {T}}(\mathbf {l} )].$

(b) Using your result from (a), demonstrate how the expectation value $\langle {\hat {\mathbf {x} }}\rangle$ changes under translation.

Solution