The Heisenberg Picture: Equations of Motion for Operators: Difference between revisions
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It is important to stress that the above oscillatory solution is for the position and momentum ''operators''. | It is important to stress that the above oscillatory solution is for the position and momentum ''operators''. | ||
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>. | 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>. | ||
==Motion in electromagnetic field== | |||
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 as follows: | |||
:<math> H=\frac{1}{2m}\left(\mathbf{p}-\frac{e}{c}\bold A(\bold r,t)\right)^2+e\phi(\bold r,t)</math> | |||
where <math> \bold{A(\bold r,t)} \!</math> is the vector potential and <math>{\phi(\bold r,t)}\!</math> is the Coulomb potential of the electromagnetic field. In a problem, if there is a momentum operator <math>\bold p\!</math>, it must be replaced by | |||
:<math>\left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)</math> | |||
if a particle is under the influence of an electromagnetic field. | |||
Let's find out the [[Heisenberg and interaction picture: Equations of motion for operators#The Heisenberg Equation of Motion|Heisenberg equations of motion]] for the position and velocity operators. | |||
For position operator<math>\bold r\!</math>, we have: | |||
:<math> | |||
\begin{align} | |||
\frac{d\bold r}{dt} &= \frac{1}{i\hbar} \left[\bold r,H \right] \\ | |||
&= \frac{1}{i\hbar} \left[ \bold r, \frac{1}{2m} \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)^2 + e\phi(\bold r,t)\right] \\ | |||
&= \frac{1}{2im\hbar} \left[\bold r, \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)^2\right] \\ | |||
&= \frac{1}{2im\hbar} \left[\bold r, \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)\right]\left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) + \frac{1}{2im\hbar} \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) \left[\bold r, \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)\right] \\ | |||
&= \frac{1}{2im\hbar} \left[\bold r, \bold p\right] \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) + \frac{1}{2im\hbar} \left(\bold p - \frac{e}{c}\bold A(\bold r,t)\right) \left[\bold r, \bold p\right] \\ | |||
&= \frac{1}{2im\hbar}i\hbar \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) + \frac{1}{2im\hbar} \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)i\hbar \\ | |||
&= \frac{1}{m}\left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right), | |||
\end{align} | |||
</math> | |||
where (<math>\bold r \!</math> does not depend on <math>t \!</math> explicitly) | |||
is the equation of motion for the position operator <math>\bold r</math>. | |||
This equation also defines the velocity operator <math>\bold v</math>: | |||
:<math>\bold v= \frac {1}{m}\left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)</math> | |||
The Hamiltonian can be rewritten as: | |||
:<math>H=\frac {m}{2}\bold v \cdot \bold v+e\phi</math> | |||
Therefore, the Heisenberg equation of motion for the velocity operator is: | |||
:<math> | |||
\begin{align} | |||
\frac{d\bold v}{dt} &=\frac {1}{i\hbar}\left[\bold v,H\right]+\frac{\partial \bold v}{\partial t} \\ | |||
&= \frac {1}{i\hbar}\left[\bold v,\frac{m}{2}\bold v \cdot \bold v\right]+\frac {1}{i\hbar}\left[\bold v,e\phi\right]-\frac{e}{mc} \frac{\partial \bold A}{\partial t} | |||
\end{align} | |||
</math> | |||
(Note that <math>\bold p\!</math> does not depend on <math>t\!</math> expicitly) | |||
Let's use the following commutator identity: | |||
:<math>\left[\bold v,\bold v \cdot \bold v\right]=\bold v \times \left(\bold v \times \bold v\right)-\left(\bold v \times \bold v\right) \times \bold v </math> | |||
Substituting, we get: | |||
:<math> | |||
\frac{d\bold v}{dt} = \frac{1}{i\hbar} \frac{m}{2} | |||
\left(\bold v \times (\bold v \times \bold v) - (\bold v \times \bold v) \times \bold v \right) + \frac{1}{i\hbar} e[\bold v,\phi] - \frac{e}{mc} \frac{\partial \bold A}{\partial t}</math> | |||
Now let's evaluate <math>\bold v \times \bold v \!</math> and <math>[\bold v,\phi] \!</math>: | |||
:<math> | |||
\begin{align} | |||
(\bold v \times \bold v)_i &= \epsilon_{ijk} v_j v_k \\ | |||
&= \epsilon_{ijk}\frac{1}{m} \left(p_j-\frac{e}{c}A_j(\bold r,t)\right) | |||
\frac{1}{m}\left(p_k-\frac{e}{c}A_k(\bold r,t)\right) \\ | |||
&= -\frac{e}{m^2c} \epsilon_{ijk}\left(p_j A_k(\bold r,t) + | |||
A_j(\bold r,t)p_k\right) \\ | |||
&= -\frac{e}{m^2c}\epsilon_{ijk}p_jA_k(\bold r,t) - \frac{e}{m^2c} | |||
\epsilon_{ijk} A_j(\bold r,t) p_k \\ | |||
&= -\frac{e}{m^2c}\epsilon_{ijk} p_j A_k(\bold r,t)-\frac{e}{m^2c} | |||
\epsilon_{ikj} A_k(\bold r,t) p_j \mbox{(Switching indices in the second terms)} \\ | |||
&= -\frac{e}{m^2c}\epsilon_{ijk} p_j A_k(\bold r,t) + \frac{e}{m^2c} | |||
\epsilon_{ijk} A_k(\bold r,t) p_j \\ | |||
&= -\frac{e}{m^2c}\epsilon_{ijk}\left[p_j,A_k(\bold r,t)\right] \\ | |||
&= -\frac{e}{m^2c}\epsilon_{ijk}\frac{\hbar}{i} \nabla_j A_k(\bold r,t) \\ | |||
&= i\hbar\frac{e}{m^2c}\left(\nabla \times \bold A\right)_i | |||
\end{align} | |||
</math> | |||
:<math> | |||
\rightarrow \left[\bold v \times \bold v\right]=i\hbar\frac{e}{m^2c}\left(\nabla \times \bold A\right) = i\hbar\frac{e}{m^2c}\bold B | |||
</math> | |||
and | |||
:<math> | |||
\begin{align} | |||
\left[\bold v,\phi\right] &= \frac{1}{m} \left[\bold p-\frac{e}{c}\bold A(\bold r, t),\phi(\bold r,t)\right] \\ | |||
&= \frac{1}{m} \left[\bold p,\phi(\bold r,t) \right] \\ | |||
&= \frac{1}{m} \frac{\hbar}{i}\nabla\phi | |||
\end{align} | |||
</math> | |||
Substituting and rearranging, we get: | |||
:<math> | |||
m\frac{d\bold v}{dt} = \frac{e}{2c} | |||
\left(\bold v \times \bold B-\bold B \times \bold v \right) + e\bold E | |||
</math> | |||
where | |||
:<math> | |||
\bold E = -\nabla \phi - \frac{1}{c} \frac{\partial \bold A}{\partial t} | |||
</math> | |||
Above is the quantum mechanical version of the equation for the acceleration of the particle in terms of the Lorentz force. | |||
These results can also be deduced in Hamiltonian dynamics due to the similarity between the Hamiltonian dynamics and quantum mechanics. |
Revision as of 15:27, 12 August 2013
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 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:
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} U(t,t_0) = H U(t,t_0) }
The solution to this differential equation depends on the form of .
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 U} , 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.
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 U|\alpha(0)\rangle= |\alpha(t)\rangle. }
Therefore, if we know the initial state of a system, we can obtain the expectation value of an operator, 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 A \rangle(t) = \langle \alpha(t)|A|\alpha(t)\rangle= \langle \alpha(0)|U^{\dagger}AU|\alpha(0)\rangle.}
We can make a redefinition by claiming 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 A_H(t) = U^{\dagger}AU}
and taking as our state kets the time independent, initial valued state ket Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\alpha(0)\rangle} .
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).
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:
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(0)\rangle }
as the state at t = 0.
At another time t, this becomes
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^{-iHt/\hbar} |\Psi(0)\rangle } ,
which in this form solves the Schrödinger equation
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|\Psi(t)\rangle}{\partial t}=H|\Psi(t)\rangle } .
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:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(t) = e^{iHt/\hbar}Ae^{-iHt\hbar} } ,
which means the expectation value 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 \langle A \rangle_t = \langle\Psi(0)| A(t)| \Psi(0)\rangle } .
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:
- 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} 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{-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}\left(\left[A,H\right]\right)_H+\left(\frac{\partial A}{\partial t}\right)_H. \end{align} }
The last equation is known as the Ehrenfest's Theorem.
For example, if we have a hamiltonian of the form,
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{p^2}{2m}+V(r),}
then we can find the Heisenberg equations of motion for p and r.
The position operator in 3D 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 i\hbar\frac{d \hat{r}(t)}{dt} = \left[ \hat{r}(t),H\right] }
Since the Hamiltonian as given above has a V(r,t) term and a momentum term the two commuators are 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 \left[ \hat{r}(t),V(r,t)\right] = 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 \left[ \hat{r}(t), \frac{p(t)^{2}}{2m}\right] = \frac{i\hbar p(t)}{m} }
this yields 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{r}(t)_{H}}{dt} = \frac{ \hat{p}(t)_{H}}{m} } .
To find the equations of motion for the momentum you need to evaluate 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{p}(t), V(r(t))\right] }
,
which equals, 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{p}, V(r)\right] = -i\hbar \nabla V(r) } .
This yields 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}=-\nabla V(\hat{r}_H).}
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.
In particular, if we apply these equations to the 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}}\!}
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}}
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}. }
we can solve the above equations of motion and find
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)}
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 oscillatory 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)\!} correspond to 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\!} .
Motion in electromagnetic field
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 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 H=\frac{1}{2m}\left(\mathbf{p}-\frac{e}{c}\bold A(\bold r,t)\right)^2+e\phi(\bold 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 \bold{A(\bold 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(\bold r,t)}\!}
is the Coulomb potential of the electromagnetic field. In a problem, if there is a momentum 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 \bold p\!}
, it must be replaced 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 \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)}
if a particle is under the influence of an electromagnetic field.
Let's find out the Heisenberg equations of motion for the position and velocity operators. For position operatorFailed 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 \bold r\!} , we have:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{d\bold r}{dt} &= \frac{1}{i\hbar} \left[\bold r,H \right] \\ &= \frac{1}{i\hbar} \left[ \bold r, \frac{1}{2m} \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)^2 + e\phi(\bold r,t)\right] \\ &= \frac{1}{2im\hbar} \left[\bold r, \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)^2\right] \\ &= \frac{1}{2im\hbar} \left[\bold r, \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)\right]\left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) + \frac{1}{2im\hbar} \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) \left[\bold r, \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)\right] \\ &= \frac{1}{2im\hbar} \left[\bold r, \bold p\right] \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) + \frac{1}{2im\hbar} \left(\bold p - \frac{e}{c}\bold A(\bold r,t)\right) \left[\bold r, \bold p\right] \\ &= \frac{1}{2im\hbar}i\hbar \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) + \frac{1}{2im\hbar} \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)i\hbar \\ &= \frac{1}{m}\left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right), \end{align} }
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 \bold r \!} does not depend on 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 \!} explicitly) is the equation of motion for 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 \bold r} . This equation also 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 \bold 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 \bold v= \frac {1}{m}\left(\bold p-\frac{e}{c}\bold A(\bold 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=\frac {m}{2}\bold v \cdot \bold v+e\phi}
Therefore, the Heisenberg equation of motion for the velocity 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 \begin{align} \frac{d\bold v}{dt} &=\frac {1}{i\hbar}\left[\bold v,H\right]+\frac{\partial \bold v}{\partial t} \\ &= \frac {1}{i\hbar}\left[\bold v,\frac{m}{2}\bold v \cdot \bold v\right]+\frac {1}{i\hbar}\left[\bold v,e\phi\right]-\frac{e}{mc} \frac{\partial \bold A}{\partial t} \end{align} }
(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 \bold p\!} does not depend on 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\!} expicitly)
Let's use the following commutator identity:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[\bold v,\bold v \cdot \bold v\right]=\bold v \times \left(\bold v \times \bold v\right)-\left(\bold v \times \bold v\right) \times \bold v }
Substituting, we get:
Now let's evaluate and :
and
Substituting and rearranging, we get:
where
Above is the quantum mechanical version of the equation for the acceleration of the particle in terms of the Lorentz force.
These results can also be deduced in Hamiltonian dynamics due to the similarity between the Hamiltonian dynamics and quantum mechanics.