Time Evolution of Expectation Values and Ehrenfest's Theorem: Difference between revisions

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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
{{Quantum Mechanics A}}


<math>\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</math>
==Time Evolution of Expecation Values==


Note that
Having described in a previous section how the state vector of a system evolves in time, we may now derive a formula for the time evolution of the expectation value of an operator.  Given an operator <math>\hat{O}(t),</math> we know that its expectation value is given by <math>\langle\hat{O}(t)\rangle=\langle\Psi(t)|\hat{O}(t)|\Psi(t)\rangle.</math>  If we take the time derivative of this expectation value, we get
                                   


<math>\frac{\partial }{\partial t\ }\psi =-\frac{\hbar}{2im}\triangledown ^{2}\Psi +\frac{V}{i\hbar}\psi</math>
<math>\frac{d\langle\hat{O}(t)\rangle}{dt}=\left [\frac{d}{dt}\langle\Psi(t)|\right ]\hat{O}(t)|\Psi(t)\rangle+\langle\Psi(t)|\frac{d\hat{O}(t)}{dt}|\Psi(t)\rangle+\langle\Psi(t)|\hat{O}(t)\left [\frac{d}{dt}|\Psi(t)\rangle\right ]</math>


and
<math>=\left [\frac{d}{dt}\langle\Psi(t)|\right ]\hat{O}(t)|\Psi(t)\rangle+\langle\Psi(t)|\hat{O}(t)\left [\frac{d}{dt}|\Psi(t)\rangle\right ]+\left\langle\frac{d\hat{O}(t)}{dt}\right\rangle.</math>


<math>\frac{\partial }{\partial t\ }\psi\ast  =\frac{\hbar}{2im}\triangledown ^{2}\Psi\ast  -\frac{V}{i\hbar}\psi\ast</math> 
We now use the [[The Schrödinger Equation in Dirac Notation|Schrödinger equation]] and its dual to write this as


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:
<math>\frac{d\langle\hat{O}(t)\rangle}{dt}=\frac{i}{\hbar}\langle\Psi(t)|\hat{H}(t)\hat{O}(t)|\Psi(t)\rangle-\frac{i}{\hbar}\langle\Psi(t)|\hat{O}(t)\hat{H}(t)|\Psi(t)\rangle+\left\langle\frac{d\hat{O}(t)}{dt}\right\rangle</math>


<math>\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</math>
<math>=\frac{i}{\hbar}\langle[\hat{H}(t),\hat{O}(t)]\rangle+\left\langle\frac{d\hat{O}(t)}{dt}\right\rangle.</math>
 
<math>\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</math>
 
We shall now use the identity
<math> \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 )</math>
 
Consider two scalar function <math>x\psi</math> and <math>x\psi</math> that are continuous and differentiable in some volume V
bounded bounded by a surface S. Applying the divergence theorem to the vector field <math>x\psi\psi  \ast </math>(the
left hand side of the identity) we obtain
 
<math>\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</math>
 
Therefore the second integral of first equation can be written as
<math>\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</math>
 
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
 
<math>\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</math>
 
We again used the fact that the surface integral again vanishes, to obtain
 
<math>\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</math>
Thus,
 
<math>\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</math>
<math>=\frac{1}{m}\left \langle p_{x} \right \rangle</math>
 
Since <math>\left \langle x \right \rangle</math> is seen always to be real number from the inherent structure of its definition. The
above equation shows quite incidentally that <math>\left \langle px \right \rangle</math> is real.
 
In similar fashion we can calculate the time rate of change of a component of the momentum of the particle as
 
<math>\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</math>
 
The integral containing laplacins can be transofrmed into a surface integral by Green’s theorem and write
 
<math>\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}
</math>
 
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
 
<math>\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}</math>
 
Therefore,
 
<math>\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</math>
 
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,
 
<math>\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</math>
 
<math>=\int_{V}\psi \ast \triangledown ^{2}\frac{\partial \psi }{\partial x}d^{3}r</math>
 
Thus the Laplacian term vanishes resulting in
 
<math>\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</math>
 
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
 
<math>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</math>
 
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
 
<math>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</math>


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


==Generalized Heisenberg uncertainty relation==
==Ehrenfest's Theorem==
 
If two opperators <math>\hat{A},\hat{B}</math> are Hermitian and
 
:<math>[\hat{A},\hat{B}]=i\hat{C}\;</math><br/>
then
<math>\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</math>
 
Proof:
 
First recall <math>\Delta \hat{O} = \hat{O} - \langle \hat{O} \rangle </math> and note that <math>\Delta \hat{O} </math> is Hermitian if <math>\hat{O} </math> is.
 
Let <math>\alpha</math> be a real scalar and define <math> f(\alpha)</math> as such:
 
:<math> f(\alpha) = |( \alpha \Delta \hat{A} - i\Delta \hat{B})|\psi \rangle|^2 </math>.
 
So <math> f(\alpha) </math> is the norm squared of some arbitrary state vector after operating <math>(\alpha \Delta \hat{A} - i\Delta \hat{B})</math> on it. Hence by the positive semidefinite property of the norm:
:<math> f(\alpha) \geq 0 </math>
 
Proceeding to calculate this norm squared:
 
:<math>
\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}</math>
 
Notice that <math>f(\alpha)</math> is a real valued ( expectation values of Hermitian operators are always real ) quadratic in real  <math>\alpha</math> which is always greater than or equal to zero. This implies that there are no real solutions for <math>\alpha</math> or there is exactly 1. This can be seen by attempting to solve for <math>\alpha</math> by using the "quadratic formula" :
 
:<math>
\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}
</math>
 
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 <math>\alpha</math>) or negative ( yielding 0 real solutions <math>\alpha</math>). Stated more succinctly:
 
:<math>
(\langle  \hat{C} \rangle)^2 -4 \langle  \left(\Delta \hat{A}\right)^2  \rangle  \langle \left(\Delta \hat{B}\right)^2  \rangle \leq 0
</math>
 
<math>\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 </math>
 
which immediately implies what was to be proved.We can try this relation for <math>\hat{A}=\hat{x}</math> and <math>\hat{B}=\hat{p}</math>.
 
So we have <math>[\hat{A},\hat{B}]=ih</math> and <math>\hat{C}=h</math>.
 
Then <math>\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 </math> 
 
<math>\therefore \Delta x\Delta p\geq \frac{h}{2}</math>
 
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 <math> \Delta t\Delta E\geq \hbar </math>
]]
 
== 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 <math>\hat{A}</math> and <math>\hat{B}</math> commute, then there exists a common eigenbasis of these two observables.
 
This theorem is generalized immediately to the case of several observables <math>\hat{A}</math>,<math>\hat{B}</math>,<math>\hat{C}</math> which all commute.
 
                                       
'''Proof.'''
 
Let <math>\left \{| \alpha ,r_{\alpha } \rangle \right \}</math> be the eigenvectors of <math>\hat{A}</math>, where the index <math>r_{\alpha }</math> means that an eigenvector associated with an eigenvalue <math>a_{\alpha }</math> belongs to an eigensubspace
of dimension <math>d_{\alpha }\geq 1</math>,
 
<math>\hat{A|\alpha ,r_{\alpha }}\rangle =a_{\alpha }|\alpha ,r_{\alpha }\rangle</math>  ,  <math>r=1,2,....d_{\alpha }</math>
 
                       
By assumption, we have <math>\left [ \hat{A},\hat{B} \right ]=0</math>, that is,
 
<math>\hat{A}\hat{B}|\alpha ,r_{\alpha }\rangle=\hat{B}\hat{A}|\alpha ,r_{\alpha }\rangle=a_{\alpha }|\alpha ,r_{\alpha }\rangle</math>  , <math>r=1,2,....d_{\alpha }</math>
                                                 
Therefore, the vector <math>\hat{B}|\alpha ,r_{\alpha }\rangle</math> is an eigenvector of A with the eigenvalues <math>a_{\alpha }</math> .
It therefore belongs to the corresponding eigensubspace. We call this vector
<math>|\alpha ,\beta ,k_{\alpha \beta }\rangle</math> ; the index <math>k_{\alpha \beta }</math> means that again this vector may be nonunique.
Therefore, this vector is a linear combination of the vectors <math>\left \{ |\alpha ,r_{\alpha }\rangle \right \}</math>, that is,
<math>
\hat{B}|\alpha ,r_{\alpha }\rangle=\sum_{r_{\alpha }}b_{r_{\alpha }}|\alpha ,r_{\alpha }\rangle</math>
 
                                                                   
which can be diagonalized with no difficulty. In other words, if <math>\hat{A}</math> and <math>\hat{B}</math> 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 <math>\hat{A}</math> and <math>\hat{B}</math> have in common the basis <math>\left \{ |\psi _{n} \rangle\right \}</math> with eigenvalues <math>a_{n}</math> and <math>b_{n}</math> :
 
<math>\hat{A}|\psi _{n}\rangle=a_{n}|\psi \rangle</math>  and    <math>\hat{B}|\psi _{n}\rangle=b_{n}|\psi \rangle</math>
 
                                     
If we apply <math>\hat{B}</math> to the first expression and <math>\hat{A}</math> to the second, and subtract, we obtain
 
<math>\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</math>
 
Because <math>\left \{ \psi _{n} \right \}</math> is a Hilbert basis, we therefore have <math>\left \{ |\psi _{n} \rangle\right \}</math>
 


<math>\left [ \hat{A} ,\hat{B}\right ] |\psi \rangle=0</math> , whatever <math>|\psi  \rangle</math>
We now use the above result to prove Ehrenfest's Theorem, which states that the expecation values of the position and momentum operators obey the same equations that the corresponding classical quantities obey.  Thus, one may consider this theorem to be a manifestation of the [[The Correspondence Principle|correspondence principle]], because the expectation values, in the classical limit, become the values of the corresponding classical quantities.


which means <math>\left [ \hat{A},\hat{B} \right ]=0</math>
Consider the Hamiltonian,


=== Example ===
<math>\hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+V(\hat{\mathbf{r}}).</math>


Consider, for instance, an isotropic two-dimensional harmonic oscillator. The eigenvalue problem of the Hamiltonian is a priori a difficult problem
We are now interested in determining how the expecation values of the position <math>\hat{\mathbf{r}}</math> and momentum <math>\hat{\mathbf{p}}</math> operators evolve in time. Using the formula that we just derived, and noting that neither operator depends explicitly on time, we obtain
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:


<math>\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}}</math>
<math>\frac{d\langle\hat{\mathbf{r}}\rangle}{dt}=\frac{i}{\hbar}\langle[\hat{H},\hat{\mathbf{r}}]\rangle</math> and <math>\frac{d\langle\hat{\mathbf{p}}\rangle}{dt}=\frac{i}{\hbar}\langle[\hat{H},\hat{\mathbf{p}}]\rangle.</math>
                                           
The two operators <math>\hat{H_{x}}</math> and <math>\hat{H_{y}}</math> , which are both operators in one variable and which act on different variables commute obviously. One can solve the eigenvalue problems of <math>\hat{H_{x}}</math> and <math>\hat{H_{y}}</math> separately


<math>\hat{H}\phi _{n}\left ( x \right )=E_{n}\phi _{n}\left ( x \right )</math> <math>\hat{H}\phi _{n}\left ( y \right )=E_{n}\phi _{n}\left ( y \right )</math>
Using the fact that <math>[\hat{p}_i^2,\hat{x}_j]=-2i\hbar p_i\delta_{ij}</math> and <math>[p_i,f(\hat{\mathbf{r}})]=-i\hbar\frac{\partial f(\hat{\mathbf{r}})}{\partial\hat{x}_i},</math> we find that
                                                           
                                                   
The eigenvalues of <math>\hat{H}</math> are the sums of eigenvalues of <math>\hat{H_{x}}</math> and <math>\hat{H_{y}}</math> with eigenfunctions that are the products of corresponding eigenfunctions:


<math>E_{n}=E_{n_{1}}+E_{n_{2}}\left ( n_{1} +n_{2}+1\right )\left ( \hbar\omega  \right )</math> <math>\psi _{n}\left ( x,y \right )    =\phi  _{n_{1}}\left ( x \right )\phi  _{n_{2}}\left ( y \right )</math>
<math>\frac{d\langle\hat{\mathbf{r}}\rangle}{dt}=\frac{\langle\hat{\mathbf{p}}\rangle}{m}</math> and <math>\frac{d\langle\hat{\mathbf{p}}\rangle}{dt}=-\langle\nabla V(\hat{\mathbf{r}})\rangle.</math>


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
These two equations closely resemble equations familar from classical mechanics - the first resembles the statement that momentum is equal to mass times velocity, while the latter looks like Newton's second law.
corresponding eigenfunctions.

Latest revision as of 12:32, 7 March 2014

Quantum Mechanics A
SchrodEq.png
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian , it describes how a state evolves in time.
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
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
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
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
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

Time Evolution of Expecation Values

Having described in a previous section how the state vector of a system evolves in time, we may now derive a formula for the time evolution of the expectation value of an operator. Given an operator we know that its expectation value is given by If we take the time derivative of this expectation value, we get

We now use the Schrödinger equation and its dual to write this as

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

Ehrenfest's Theorem

We now use the above result to prove Ehrenfest's Theorem, which states that the expecation values of the position and momentum operators obey the same equations that the corresponding classical quantities obey. Thus, one may consider this theorem to be a manifestation of the correspondence principle, because the expectation values, in the classical limit, become the values of the corresponding classical quantities.

Consider the Hamiltonian,

We are now interested in determining how the expecation values of the position and momentum operators evolve in time. Using the formula that we just derived, and noting that neither operator depends explicitly on time, we obtain

and

Using the fact that and we find that

and

These two equations closely resemble equations familar from classical mechanics - the first resembles the statement that momentum is equal to mass times velocity, while the latter looks like Newton's second law.