The Heisenberg Picture: Equations of Motion for Operators: Difference between revisions

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{{Quantum Mechanics A}}
There are several ways to mathematically approach the change of a quantum mechanical system with time. The Schrödinger picture considers wave functions which change with time while the Heisenberg picture places the time dependence on the operators and deals with wave functions that do not change in time.  The interaction picture, also known as the Dirac picture, is somewhere between the other two, placing time dependence on both operators and wave functions. The two "pictures" correspond, theoretically, to the time evolution of two different things.  Mathematically, all methods should produce the same result.  
There are several ways to mathematically approach the change of a quantum mechanical system with time. The Schrödinger picture considers wave functions which change with time while the Heisenberg picture places the time dependence on the operators and deals with wave functions that do not change in time.  The interaction picture, also known as the Dirac picture, is somewhere between the other two, placing time dependence on both operators and wave functions. The two "pictures" correspond, theoretically, to the time evolution of two different things.  Mathematically, all methods should produce the same result.  


== Definition of the Heisenberg Picture ==
== Definition of the Heisenberg Picture ==


The time evolution operator is at the heart of the "evolution" of a state, and is the same in each picture.  The evolution operator is obtained from the time dependent [[Schrödinger equation]] after separating the time and spatial parts of the wave equation:
The time evolution operator, defined as
 
<math>|\Psi(t)\rangle=\hat{U}(t,t_0)|\Psi(t_0)\rangle,</math>
 
is at the heart of the "evolution" of a state, and is the same in each picture.  The evolution operator is obtained from the time dependent [[Schrödinger Equation|Schrödinger equation]] as follows.  We start by rewriting the equation as


<math>
<math>
i\hbar \frac{\partial}{\partial t} U(t,t_0) = H U(t,t_0)
i\hbar\frac{\partial}{\partial t}\hat{U}(t,t_0)|\Psi(t_0)\rangle = \hat{H}(t)\hat{U}(t,t_0)|\Psi(t_0)\rangle.
</math>
</math>
 
The solution to this differential equation depends on the form of .


If we know the time evolution operator, <math>U</math>, and the initial state of a particular system, all that is needed is to apply  to the initial state ket.  We then obtain the ket for some later time.
Since this equation will hold regardless of our choice of <math>|\Psi(t_0)\rangle,</math> we conclude that


<math>
<math>
U|\alpha(0)\rangle= |\alpha(t)\rangle.
i\hbar\frac{\partial}{\partial t}\hat{U}(t,t_0)=\hat{H}(t)\hat{U}(t,t_0).
</math>
</math>


Therefore, if we know the initial state of a system, we can obtain the expectation value of an operator, A, at some later time:
It follows from the fact that the Schrödinger equation preserves the normalization of the state vector for all times that the time evolution operator is unitary:


<math>
<math>\hat{U}^\dagger(t,t_0)\hat{U}(t,t_0)=\hat{I}</math>
\langle A \rangle(t) = \langle \alpha(t)|A|\alpha(t)\rangle= \langle \alpha(0)|U^{\dagger}AU|\alpha(0)\rangle.</math>


We can make a redefinition by claiming that
If we know the time evolution operator, <math>\hat{U},</math> and the initial state of a particular system, all that is needed to determine the state at a later time is to apply the time evolution operator to the initial state vector:


<math>
<math>
A_H(t) = U^{\dagger}AU</math>
\hat{U}(t,t_0)|\Psi(t_0)\rangle=|\Psi(t)\rangle.
</math>


and taking as our state kets the time independent, initial valued state ket <math>|\alpha(0)\rangle</math>.
Therefore, if we know the initial state of a system, we can also obtain the expectation value of an operator <math>\hat{A}</math> at some later time:


This formulation of quantum mechanics is called the Heisenberg picture. In this picture, operators evolve in time and state kets do not.  (Note that the difference between the two pictures only lies in the way we write them down).
<math>
 
\langle\hat{A}\rangle(t) = \langle\Psi(t)|\hat{A}|\Psi(t)\rangle= \langle \Psi(t_0)|\hat{U}^{\dagger}(t,t_0)\hat{A}\hat{U}(t,t_0)|\Psi(t_0)\rangle</math>
In classical physics we obviously have time evolution of a system - our observable (position or angular momentum or whatever variable we are considering) changes in a way dictated by the classical equations of motion.  We do not talk about state kets in classical mechanics.  Therefore, the Heisenberg, where the operator changes in time, is useful because we can see a closer connection to classical physics than with the Schrödinger picture.
 
== Comparing the Heisenberg Picture and the Schrödinger Picture ==
 
As mentioned above, both the Heisenberg representation and the Schrödinger representation give the same results for the time dependent expectation values of operators. 


In the Schrödinger picture, in which we are most accustomed, the states change over time while the operators remain constant. In other words, the time dependence is carried by the state operators. When the Hamiltonian is independent of time, it is possible to write:
We may view the above equation in two ways.  One way is to think of the operator <math>\hat{A}</math> as time-independent and to consider all of the time dependence of its expectation value as coming from the state vector.  This point of view is known as the Schrödinger picture.  Alternatively, we may think of the ''operator'' as evolving in time, while the state vector stays constant. This is known as the Heisenberg picture.  In this picture, the time evolution of the operator <math>\hat{A}</math> is given by


<math>
<math>\hat{A}_H(t)=\hat{U}^{\dagger}(t,t_0)\hat{A}\hat{U}(t,t_0)</math>
|\Psi(0)\rangle
</math>  


as the state at t = 0.   
and our state vector is just the initial state vector <math>|\Psi(t_0)\rangle.</math> Note that the only difference between the two pictures is in the way that we assign the time dependence of the system; i.e., whether we think of the ''state'' as evolving or of the ''operators'' as evolving.


At another time t, this becomes
In the special case that the Hamiltonian is independent of time, we may obtain an explicit expression for the time evolution operator.  If we solve the differential equation that this operator satisfies, we find that


<math>
<math>\hat{U}(t,t_0)=e^{-i\hat{H}(t-t_0)/\hbar}.</math>
|\Psi(t)\rangle = e^{-iHt/\hbar} |\Psi(0)\rangle
</math>,


which in this form solves the Schrödinger equation
Therefore, the time evolution of a state vector is given by


<math>
<math>|\Psi(t)\rangle=e^{-i\hat{H}(t-t_0)/\hbar}|\Psi(t_0)\rangle</math>
i\hbar\frac{\partial|\Psi(t)\rangle}{\partial t}=H|\Psi(t)\rangle
</math>.
Conversely, in the Heisenberg picture to find the time dependent expectation of a given operator, the states remain constant while the operators
change in time.  In this case, you can write the time dependent operator as:


<math>
and that of an operator in the Heisenberg picture is
A(t) = e^{iHt/\hbar}Ae^{-iHt\hbar}
</math>,


which means the expectation value is,
<math>\hat{A}_H(t)=e^{i\hat{H}(t-t_0)/\hbar}\hat{A}e^{-i\hat{H}(t-t_0)/\hbar}.</math>


<math>
The Heisenberg picture is useful because we can see a closer connection to classical physics than with the Schrödinger picture.  In classical physics, we describe the evolution of a system in terms of the time evolution of the observables, such as position or angular momentum, as dictated by the classical equations of motion.  Classical mechanics does not include the concept of state vectors, as quantum mechanics does.
\langle A \rangle_t = \langle\Psi(0)| A(t)| \Psi(0)\rangle
</math>.


== The Heisenberg Equation of Motion ==
== The Heisenberg Equation of Motion ==
In the Heisenberg picture, the quantum mechanical observables change in time as dictated by the Heisenberg equations of motion.  We can study the evolution of a Heisenberg operator by differentiating equation 2 with respect to time:
 
In the Heisenberg picture, the quantum mechanical observables change in time as dictated by the Heisenberg equations of motion.  We can study the evolution of a Heisenberg operator by differentiating it with respect to time:


:<math>
:<math>
\begin{align}
\begin{align}
\frac{d}{dt} A_H &= \frac{\partial U^{\dagger}}{\partial t} A U + U^{\dagger}\frac{\partial A}{\partial t} U + U^{\dagger} A \frac{\partial U}{\partial t} \\
\frac{d}{dt}\hat{A}_H(t) &= \frac{\partial\hat{U}^{\dagger}}{\partial t}\hat{A}\hat{U}+\hat{U}^{\dagger}\frac{\partial\hat{A}}{\partial t}\hat{U}+\hat{U}^{\dagger}\hat{A}\frac{\partial\hat{U}}{\partial t} \\
&= \frac{-1}{i\hbar} U^{\dagger} HUU^{\dagger}A U + \left(\frac{\partial A}{\partial t}\right)_H + U^{\dagger} AUU^{\dagger}HU \frac{1}{i\hbar} \\
&=-\frac{1}{i\hbar}\hat{U}^{\dagger}\hat{H}\hat{U}\hat{U}^{\dagger}\hat{A}\hat{U}+\left(\frac{\partial\hat{A}}{\partial t}\right)_H+\frac{1}{i\hbar}\hat{U}^{\dagger}\hat{A}\hat{U}\hat{U}^{\dagger}\hat{H}\hat{U} \\
&= \frac{1}{i\hbar}\left(\left[A,H\right]\right)_H+\left(\frac{\partial A}{\partial t}\right)_H.
&=-\frac{i}{\hbar}\left(\left[\hat{A},\hat{H}\right]\right)_H+\left(\frac{\partial\hat{A}}{\partial t}\right)_H.
\end{align}
\end{align}
</math>
</math>


The last equation is known as the [[Ehrenfest's Theorem]]<nowiki />.
As an example, let us consider the Hamiltonian,


For example, if we have a hamiltonian of the form,
<math>\hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+V(\hat{\mathbf{r}}).</math>


<math>H=\frac{p^2}{2m}+V(r),</math>
then we can find the Heisenberg equations of motion for <math>\hat{\mathbf{p}}</math> and <math>\hat{\mathbf{r}}.</math>


then we can find the Heisenberg equations of motion for p and r. 
The equation for the position operator is


The position operator in 3D is:
<math>\frac{d\hat{\mathbf{r}}_H(t)}{dt}=-\frac{i}{\hbar}\left (\left [\hat{\mathbf{r}},\hat{H}\right ]\right )_H.</math>   
<math>i\hbar\frac{d \hat{r}(t)}{dt} = \left[ \hat{r}(t),H\right] </math>   


Since the Hamiltonian as given above has a V(r,t) term and a momentum term the two commuators are as follows:
The commuators of the position operator with respect to the kinetic and potential terms are


<math> \left[ \hat{r}(t),V(r,t)\right] = 0 </math>
<math> \frac{1}{2m}\left [\hat{\mathbf{r}},\hat{\mathbf{p}}^2\right ]=\frac{i\hbar}{m}\hat{\mathbf{p}}</math>


<math> \left[ \hat{r}(t), \frac{p(t)^{2}}{2m}\right] = \frac{i\hbar p(t)}{m} </math>
and


this yields
<math>\left [\hat{\mathbf{r}},V(\hat{\mathbf{r}})\right ]=0.</math>
<math> \frac{d \hat{r}(t)_{H}}{dt} = \frac{ \hat{p}(t)_{H}}{m} </math>.


The equation of motion for <math>\hat{\mathbf{r}}</math> is therefore


To find the equations of motion for the momentum you need to evaluate  <math> \left[ \hat{p}(t), V(r(t))\right] </math>,
<math> \frac{d\hat{\mathbf{r}}(t)_{H}}{dt} = \frac{\hat{\mathbf{p}}(t)_{H}}{m}.</math>


which equals, <math> \left[\hat{p}, V(r)\right] = -i\hbar \nabla V(r) </math>.
We may see this as defining a velocity operator,


This yields <math>\frac{d \hat{p}_H}{dt}=-\nabla V(\hat{r}_H).</math>
<math>\hat{\mathbf{v}}=\frac{\hat{\mathbf{p}}}{m}.</math>


These results are, of course, what we would expect if we only knew classical physics, providing another example of how the Heisenberg picture is more transparently similar to classical physics.
We may find the equation for the momentum similarly.  The commutator of the momentum with the kinetic term is obviously zero, while that with the potential term is


In particular, if we apply these equations to the Harmonic oscillator with natural frequency <math>\omega=\sqrt{\frac{k}{m}}\!</math>
<math>\left [\hat{\mathbf{p}},V(\hat{\mathbf{r}})\right ]=-i\hbar\nabla V(\mathbf{r}).</math>
 
The equation of motion for the momentum is thus
 
<math>\frac{d\hat{\mathbf{p}}_H}{dt}=-\nabla V(\hat{\mathbf{r}}_H).</math>
 
These are just the equations satisfied by the corresponding classical quantities, as expected.
 
In particular, if we apply these equations to a harmonic oscillator with natural frequency <math>\omega=\sqrt{\frac{k}{m}},\!</math> we obtain
 
<math>\frac{d\hat{x}_H }{dt}=\frac{\hat{p}_H }{m}</math>
 
and
 
<math>\frac{d\hat{p}_H}{dt}=-k{\hat{x}_H}.</math>
 
Solving the above equations of motion, we obtain
 
<math>\hat{x}_H(t)=\hat{x}_H(0)\cos(\omega t)+\frac{\hat{p}_H(0)}{m\omega}\sin(\omega t)</math>
 
and
 
<math>\hat{p}_H(t)=\hat{p}_H(0)\cos(\omega t)-\hat{x}_H(0)m\omega\sin(\omega t).\!</math>
 
It is important to stress that the above solution is for the position and momentum ''operators''.  Also, note that <math>\hat x_H(0)\!</math> and <math>\hat p_H(0)\!</math> are just the time independent operators <math>\hat{x}</math> and <math>\hat{p}.</math>
 
==An Example: Charged Particle in an Electromagnetic Field==
 
Recall that the Hamiltonian of a particle of charge <math>e\!</math> and mass <math>m\!</math> in an external electromagnetic field, which may be time-dependent, is given by
 
:<math> H=\frac{1}{2m}\left(\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}(\hat{\mathbf{r}},t)\right )^2+e\phi(\hat{\mathbf{r}},t),</math>
 
where <math>\mathbf{A}(\mathbf{r},t)\!</math> is the vector potential and <math>{\phi(\mathbf{r},t)}\!</math> is the Coulomb potential of the electromagnetic field.  We now wish to determine the Heisenberg equations of motion for the position and velocity operators.
 
We first turn our attention to the position operator, <math>\hat{\mathbf{r}}.</math>  We determine the equation of motion as follows:
 
<math>
\begin{align}
\frac{d\hat{\mathbf{r}}}{dt} &= \frac{1}{i\hbar}\left[\hat{\mathbf{r}},\hat{H}\right] \\
&= \frac{1}{i\hbar} \left[\hat{\mathbf{r}},\frac{1}{2m}\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)^2 + e\phi(\hat{\mathbf{r}},t)\right] \\
&= \frac{1}{2im\hbar} \left[\hat{\mathbf{r}},\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)^2\right] \\
&= \frac{1}{2im\hbar} \left[\hat{\mathbf{r}},\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)\right]\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) + \frac{1}{2im\hbar} \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) \left[\hat{\mathbf{r}},\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)\right] \\
&= \frac{1}{2im\hbar} \left[\hat{\mathbf{r}}, \hat{\mathbf{p}}\right] \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) + \frac{1}{2im\hbar} \left(\hat{\mathbf{p}} - \frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) \left[\hat{\mathbf{r}},\hat{\mathbf{p}}\right] \\
&= \frac{1}{2im\hbar}i\hbar \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) + \frac{1}{2im\hbar} \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)i\hbar \\
&= \frac{1}{m}\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right).\end{align}
</math>
 
This equation defines the velocity operator <math>\hat{\mathbf{v}},</math>
                                               
<math>\hat{\mathbf{v}}=\frac{1}{m}\left(\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}(\hat{\mathbf{r}},t)\right ).</math>                                                                             
 
The Hamiltonian can be rewritten as
 
<math>H=\tfrac{1}{2}m\hat{\mathbf{v}}\cdot\hat{\mathbf{v}}+e\phi.</math>
 
We now determine the equation of motion for the velocity operator:
 
<math>
\begin{align}
\frac{d\hat{\mathbf{v}}}{dt} &=\frac{1}{i\hbar}\left[\hat{\mathbf{v}},\hat{H}\right]+\frac{\partial\hat{\mathbf{v}}}{\partial t} \\
&= \frac{1}{i\hbar}\left[\hat{\mathbf{v}},\tfrac{1}{2}m\hat{\mathbf{v}}\cdot\hat{\mathbf{v}}\right]+\frac{1}{i\hbar}\left[\hat{\mathbf{v}},e\phi\right]-\frac{e}{mc}\frac{\partial\mathbf{A}}{\partial t}
\end{align}
</math>
 
Note that <math>\hat{\mathbf{p}}</math> does not depend on <math>t\!</math> expicitly.
 
Using the identity,
 
<math>\left[\hat{\mathbf{v}},\hat{\mathbf{v}}\cdot\hat{\mathbf{v}}\right]=\hat{\mathbf{v}}\times \left(\hat{\mathbf{v}}\times\hat{\mathbf{v}}\right)-\left(\hat{\mathbf{v}}\times\hat{\mathbf{v}}\right) \times\hat{\mathbf{v}}, </math> 
 
we obtain


<math>
<math>
\frac{d\hat x_H }{dt}=\frac{\hat p_H }{m}</math><br/>
\frac{d\hat{\mathbf{v}}}{dt} = \frac{1}{i\hbar} \tfrac{1}{2}m
\left(\hat{\mathbf{v}}\times (\hat{\mathbf{v}}\times\hat{\mathbf{v}})-(\hat{\mathbf{v}}\times\hat{\mathbf{v}})\times\hat{\mathbf{v}}\right)+\frac{1}{i\hbar}e[\hat{\mathbf{v}},\phi] - \frac{e}{mc}\frac{\partial\mathbf{A}}{\partial t}.</math>
 
We now evaluate <math>\hat{\mathbf{v}}\times\hat{\mathbf{v}}</math> and <math>[\hat{\mathbf{v}},\phi].</math> The former is evaluated as follows:


<math>
<math>
\frac{d\hat p_H }{dt}=-k{\hat x_H}.
\begin{align}
(\hat{\mathbf{v}}\times\hat{\mathbf{v}})_i &= \epsilon_{ijk} v_j v_k \\
&= \epsilon_{ijk}\frac{1}{m} \left(p_j-\frac{e}{c}A_j(\hat{\mathbf{r}},t)\right)
\frac{1}{m}\left(p_k-\frac{e}{c}A_k(\hat{\mathbf{r}},t)\right) \\
&= -\frac{e}{m^2c} \epsilon_{ijk}\left(p_j A_k(\hat{\mathbf{r}},t) +
A_j(\hat{\mathbf{r}},t)p_k\right) \\
&= -\frac{e}{m^2c}\epsilon_{ijk}p_jA_k(\hat{\mathbf{r}},t) - \frac{e}{m^2c}
\epsilon_{ijk} A_j(\hat{\mathbf{r}},t) p_k \\
&= -\frac{e}{m^2c}\epsilon_{ijk} p_j A_k(\hat{\mathbf{r}},t)-\frac{e}{m^2c}
\epsilon_{ikj} A_k(\hat{\mathbf{r}},t) p_j \\
&= -\frac{e}{m^2c}\epsilon_{ijk} p_j A_k(\hat{\mathbf{r}},t) + \frac{e}{m^2c}
\epsilon_{ijk} A_k(\hat{\mathbf{r}},t) p_j \\
&= -\frac{e}{m^2c}\epsilon_{ijk}\left[p_j,A_k(\hat{\mathbf{r}},t)\right] \\
&= -\frac{e}{m^2c}\epsilon_{ijk}\frac{\hbar}{i} \nabla_j A_k(\hat{\mathbf{r}},t) \\
&= i\hbar\frac{e}{m^2c}\left(\nabla \times \bold A\right)_i
\end{align}
</math>
</math>


we can solve the above equations of motion and find
Noting that the curl of the vector potential is just the magnetic field,
 
<math>\mathbf{B}=\nabla\times\mathbf{A},</math>
 
this becomes


<math>
<math>
\hat x_H(t)=\hat x_H(0)\cos(\omega t)+\frac{\hat p_H(0)}{m\omega}\sin(\omega t)</math><br/>
\left (\hat{\mathbf{v}}\times\hat{\mathbf{v}}\right )=i\hbar\frac{e}{m^2c}\mathbf{B}.</math>
 
The second commutator, <math>[\hat{\mathbf{v}},\phi],</math> is


<math>
<math>
\hat p_H(t)=\hat p_H(0)\cos(\omega t)-\hat x_H(0)m\omega\sin(\omega t).\!</math>
\begin{align}
\left[\hat{\mathbf{v}},\phi\right] &= \frac{1}{m}\left[\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}(\hat{\mathbf{r}},t),\phi(\hat{\mathbf{r}},t)\right] \\
&= \frac{1}{m} \left[\hat{\mathbf{p}},\phi(\hat{\mathbf{r}},t) \right] \\
&= \frac{1}{m} \frac{\hbar}{i}\nabla\phi
\end{align}
</math>
 
Substituting and rearranging, we get
 
<math>m\frac{d\hat{\mathbf{v}}}{dt}=\frac{e}{2c}\left(\hat{\mathbf{v}}\times\mathbf{B}-\mathbf{B}\times\hat{\mathbf{v}}\right) + e\mathbf{E},</math>
where the electric field <math>\mathbf{E}</math> is given by


It is important to stress that the above oscillatory solution is for the position and momentum ''operators''.
<math>\mathbf{E}=-\nabla\phi - \frac{1}{c}\frac{\partial\mathbf{A}}{\partial t}.</math>                  
Also, note that <math>\hat x_H(0)\!</math> and <math>\hat p_H(0)\!</math> correspond to the time independent operators <math> \hat x\!</math> and <math> \hat p\!</math>.
                                                                     
This is similar to the classical equation for the motion of a particle under the influence of electric and magnetic (Lorentz) forces.  The reason why the Lorentz force term is slightly different from the classical expression, <math>\mathbf{F}=\frac{e}{c}\mathbf{v}\times\mathbf{B},</math> is because the components of the velocity operator do not commute with those of the magnetic field.

Latest revision as of 10:02, 16 August 2013

Quantum Mechanics A
SchrodEq.png
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.
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

There are several ways to mathematically approach the change of a quantum mechanical system with time. The Schrödinger picture considers wave functions which change with time while the Heisenberg picture places the time dependence on the operators and deals with wave functions that do not change in time. The interaction picture, also known as the Dirac picture, is somewhere between the other two, placing time dependence on both operators and wave functions. The two "pictures" correspond, theoretically, to the time evolution of two different things. Mathematically, all methods should produce the same result.

Definition of the Heisenberg Picture

The time evolution operator, defined 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 |\Psi(t)\rangle=\hat{U}(t,t_0)|\Psi(t_0)\rangle,}

is at the heart of the "evolution" of a state, and is the same in each picture. The evolution operator is obtained from the time dependent Schrödinger equation as follows. We start by rewriting the equation 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{\partial}{\partial t}\hat{U}(t,t_0)|\Psi(t_0)\rangle = \hat{H}(t)\hat{U}(t,t_0)|\Psi(t_0)\rangle. }

Since this equation will hold regardless of our choice 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 |\Psi(t_0)\rangle,} we conclude 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 i\hbar\frac{\partial}{\partial t}\hat{U}(t,t_0)=\hat{H}(t)\hat{U}(t,t_0). }

It follows from the fact that the Schrödinger equation preserves the normalization of the state vector for all times that the time evolution operator is unitary:

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{U}^\dagger(t,t_0)\hat{U}(t,t_0)=\hat{I}}

If we know the time evolution operator, 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{U},} and the initial state of a particular system, all that is needed to determine the state at a later time is to apply the time evolution operator to the initial 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 \hat{U}(t,t_0)|\Psi(t_0)\rangle=|\Psi(t)\rangle. }

Therefore, if we know the initial state of a system, we can also obtain the expectation value of an operator 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}} at some later time:

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{A}\rangle(t) = \langle\Psi(t)|\hat{A}|\Psi(t)\rangle= \langle \Psi(t_0)|\hat{U}^{\dagger}(t,t_0)\hat{A}\hat{U}(t,t_0)|\Psi(t_0)\rangle}

We may view the above equation in two ways. One way is to think of the operator 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}} as time-independent and to consider all of the time dependence of its expectation value as coming from the state vector. This point of view is known as the Schrödinger picture. Alternatively, we may think of the operator as evolving in time, while the state vector stays constant. This is known as the Heisenberg picture. In this picture, the time evolution of the operator 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}} 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 \hat{A}_H(t)=\hat{U}^{\dagger}(t,t_0)\hat{A}\hat{U}(t,t_0)}

and our state vector is just the initial 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(t_0)\rangle.} Note that the only difference between the two pictures is in the way that we assign the time dependence of the system; i.e., whether we think of the state as evolving or of the operators as evolving.

In the special case that the Hamiltonian is independent of time, we may obtain an explicit expression for the time evolution operator. If we solve the differential equation that this operator satisfies, we find 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{U}(t,t_0)=e^{-i\hat{H}(t-t_0)/\hbar}.}

Therefore, the time evolution of a state vector 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 |\Psi(t)\rangle=e^{-i\hat{H}(t-t_0)/\hbar}|\Psi(t_0)\rangle}

and that of an operator in the Heisenberg picture 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}_H(t)=e^{i\hat{H}(t-t_0)/\hbar}\hat{A}e^{-i\hat{H}(t-t_0)/\hbar}.}

The Heisenberg picture is useful because we can see a closer connection to classical physics than with the Schrödinger picture. In classical physics, we describe the evolution of a system in terms of the time evolution of the observables, such as position or angular momentum, as dictated by the classical equations of motion. Classical mechanics does not include the concept of state vectors, as quantum mechanics does.

The Heisenberg Equation of Motion

In the Heisenberg picture, the quantum mechanical observables change in time as dictated by the Heisenberg equations of motion. We can study the evolution of a Heisenberg operator by differentiating it with respect to time:

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} \frac{d}{dt}\hat{A}_H(t) &= \frac{\partial\hat{U}^{\dagger}}{\partial t}\hat{A}\hat{U}+\hat{U}^{\dagger}\frac{\partial\hat{A}}{\partial t}\hat{U}+\hat{U}^{\dagger}\hat{A}\frac{\partial\hat{U}}{\partial t} \\ &=-\frac{1}{i\hbar}\hat{U}^{\dagger}\hat{H}\hat{U}\hat{U}^{\dagger}\hat{A}\hat{U}+\left(\frac{\partial\hat{A}}{\partial t}\right)_H+\frac{1}{i\hbar}\hat{U}^{\dagger}\hat{A}\hat{U}\hat{U}^{\dagger}\hat{H}\hat{U} \\ &=-\frac{i}{\hbar}\left(\left[\hat{A},\hat{H}\right]\right)_H+\left(\frac{\partial\hat{A}}{\partial t}\right)_H. \end{align} }

As an example, let us consider the 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 \hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+V(\hat{\mathbf{r}}).}

then we can find the Heisenberg equations of motion 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{\mathbf{p}}} 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{\mathbf{r}}.}

The equation for the position operator 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 \frac{d\hat{\mathbf{r}}_H(t)}{dt}=-\frac{i}{\hbar}\left (\left [\hat{\mathbf{r}},\hat{H}\right ]\right )_H.}

The commuators of the position operator with respect to the kinetic and potential terms are

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}{2m}\left [\hat{\mathbf{r}},\hat{\mathbf{p}}^2\right ]=\frac{i\hbar}{m}\hat{\mathbf{p}}}

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 \left [\hat{\mathbf{r}},V(\hat{\mathbf{r}})\right ]=0.}

The equation of motion 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{\mathbf{r}}} is 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 \frac{d\hat{\mathbf{r}}(t)_{H}}{dt} = \frac{\hat{\mathbf{p}}(t)_{H}}{m}.}

We may see this as defining a velocity operator,

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{\mathbf{v}}=\frac{\hat{\mathbf{p}}}{m}.}

We may find the equation for the momentum similarly. The commutator of the momentum with the kinetic term is obviously zero, while that with the potential term 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 \left [\hat{\mathbf{p}},V(\hat{\mathbf{r}})\right ]=-i\hbar\nabla V(\mathbf{r}).}

The equation of motion for the momentum is 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{d\hat{\mathbf{p}}_H}{dt}=-\nabla V(\hat{\mathbf{r}}_H).}

These are just the equations satisfied by the corresponding classical quantities, as expected.

In particular, if we apply these equations to a harmonic oscillator with natural frequency 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 \omega=\sqrt{\frac{k}{m}},\!} 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 \frac{d\hat{x}_H }{dt}=\frac{\hat{p}_H }{m}}

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{d\hat{p}_H}{dt}=-k{\hat{x}_H}.}

Solving the above equations of motion, 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 \hat{x}_H(t)=\hat{x}_H(0)\cos(\omega t)+\frac{\hat{p}_H(0)}{m\omega}\sin(\omega t)}

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{p}_H(t)=\hat{p}_H(0)\cos(\omega t)-\hat{x}_H(0)m\omega\sin(\omega t).\!}

It is important to stress that the above solution is for the position and momentum operators. Also, 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 \hat x_H(0)\!} 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 p_H(0)\!} are just the time independent 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{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{p}.}

An Example: Charged Particle in an Electromagnetic Field

Recall that the Hamiltonian of a particle of charge 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\!} and mass 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 m\!} in an external electromagnetic field, which may be time-dependent, 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 H=\frac{1}{2m}\left(\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}(\hat{\mathbf{r}},t)\right )^2+e\phi(\hat{\mathbf{r}},t),}

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 \mathbf{A}(\mathbf{r},t)\!} is the vector potential 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 {\phi(\mathbf{r},t)}\!} is the Coulomb potential of the electromagnetic field. We now wish to determine the Heisenberg equations of motion for the position and velocity operators.

We first turn our attention to the position operator, 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{\mathbf{r}}.} We determine the equation of motion 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 \begin{align} \frac{d\hat{\mathbf{r}}}{dt} &= \frac{1}{i\hbar}\left[\hat{\mathbf{r}},\hat{H}\right] \\ &= \frac{1}{i\hbar} \left[\hat{\mathbf{r}},\frac{1}{2m}\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)^2 + e\phi(\hat{\mathbf{r}},t)\right] \\ &= \frac{1}{2im\hbar} \left[\hat{\mathbf{r}},\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)^2\right] \\ &= \frac{1}{2im\hbar} \left[\hat{\mathbf{r}},\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)\right]\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) + \frac{1}{2im\hbar} \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) \left[\hat{\mathbf{r}},\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)\right] \\ &= \frac{1}{2im\hbar} \left[\hat{\mathbf{r}}, \hat{\mathbf{p}}\right] \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) + \frac{1}{2im\hbar} \left(\hat{\mathbf{p}} - \frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) \left[\hat{\mathbf{r}},\hat{\mathbf{p}}\right] \\ &= \frac{1}{2im\hbar}i\hbar \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right) + \frac{1}{2im\hbar} \left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right)i\hbar \\ &= \frac{1}{m}\left(\hat{\mathbf{p}}-\frac{e}{c}\bold A(\hat{\mathbf{r}},t)\right).\end{align} }

This equation defines the velocity operator 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{\mathbf{v}},}

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{\mathbf{v}}=\frac{1}{m}\left(\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}(\hat{\mathbf{r}},t)\right ).}

The Hamiltonian can be rewritten 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 H=\tfrac{1}{2}m\hat{\mathbf{v}}\cdot\hat{\mathbf{v}}+e\phi.}

We now determine the equation of motion for the velocity operator:

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} \frac{d\hat{\mathbf{v}}}{dt} &=\frac{1}{i\hbar}\left[\hat{\mathbf{v}},\hat{H}\right]+\frac{\partial\hat{\mathbf{v}}}{\partial t} \\ &= \frac{1}{i\hbar}\left[\hat{\mathbf{v}},\tfrac{1}{2}m\hat{\mathbf{v}}\cdot\hat{\mathbf{v}}\right]+\frac{1}{i\hbar}\left[\hat{\mathbf{v}},e\phi\right]-\frac{e}{mc}\frac{\partial\mathbf{A}}{\partial t} \end{align} }

Note that does not depend on expicitly.

Using the identity,

we obtain

We now evaluate and The former is evaluated as follows:

Noting that the curl of the vector potential is just the magnetic field,

this becomes

The second commutator, is

Substituting and rearranging, we get

where the electric field is given by

This is similar to the classical equation for the motion of a particle under the influence of electric and magnetic (Lorentz) forces. The reason why the Lorentz force term is slightly different from the classical expression, is because the components of the velocity operator do not commute with those of the magnetic field.

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