mmgbmm said:
In the context of observables in QM the rate of change of the In the contest of observables in QM the rate of change of the expectation value of an observable A is defined by:
d<A>/dt= d< Y | A Y>/dt = < d Y/dt | A Y > + < Y | dA/dt . Y > + < Y | A dY/dt>
My question is about the second term (< Y | dA/dt . Y >), is it mathematically correct to differentiate an operator!? Can anyone explain to me what's going on? After all d/dt is an operator that works on a function or on another operator as well? I am confused.
Differentiation is expressed using limits.
In what follows, I'll proceed formally with very little regard to rigour. Knowing how much rigour to use is a difficult issue - if high standards of rigour had been insisted on at all times, then the amazing progress that has been made in applying quantum theory to physical systems would not have been made. However, this does not mean that rigour is something that can be completely ignored. While traversing Roger Penrose's Road to Reality, I came across the following relevant passage
"Quantum mechanics is full of irritating issues of this kind. As the state of the art stands, one can either be decidedly sloppy about such mathematical niceties, or else spend the whole time insisting on getting the mathematics right, in which case there is a contrasting danger of getting trapped in a 'rigour mortis'. I am doing my best to steer a middle path, but I am not at all sure what the correct answer is for making progress in the subject!"
Interesting words from someone who, controversial views notwithstanding, has deep understanding of both pure mathematics and theoretical physics.
First, a look at what A(t) means. Let S be the state space for the system. For each t, A(t): \mathbb{R} \rightarrow S [/itex]. <br />
<br />
Following your lead, I'm not going to use bras for elements of S. The inner product is also a mapping, i.e., &lt;|&gt;: S \times S \rightarrow \mathbb{C}. Let \psi be an S-valued function of t, and let \phi (t) = A(t) \psi (t). Thus, for A(t) self-adjoint, &lt;\psi | \phi&gt; is a real-valued function of a real variable (time), and standard intro calculus applies. So, <br />
<br />
<br />
\begin{equation*} <br />
\begin{split} <br />
\frac {d} {dt} &lt;\psi|\phi&gt;(t) &amp;= \lim_{h\rightarrow 0} \frac {&lt;\psi|phi&gt;(t+h) - &lt;\psi|phi&gt;(t)} {h} \\ <br />
&amp;= \lim_{h\rightarrow 0} \frac {[&lt;\psi(t+h)|phi(t+h)&gt; - &lt;\psi(t+h)|phi(t)&gt;] + [&lt;\psi(t+h)|phi(t)&gt; - &lt;\psi(t)|phi(t)&gt;]} {h}. <br />
\end{split} <br />
\end{equation*} <br />
<br />
The 1/h can certainly be "pulled inside" the inner products, but, can the limits also be pulled inside, as you have implicitly done? I think so, but I'm not sure. In intro calculus, f[/itex] continuous guarantees that \lim_{x\rightarrow a} f(g(x)) =f(\lim_{x\rightarrow a} g(x)). Without further ado, pull the limits inside, which results in <br />
<br />
&lt;br /&gt;
\frac {d} {dt} &amp;lt;\psi|\phi&amp;gt; = &amp;lt;\frac {d\psi} {dt}|\phi&amp;gt; + &amp;lt;\psi|\frac {d\phi} {dt}&amp;gt;. <br />
<br />
Now, to your question: What does \frac {d\phi} {dt} = \frac {d} {dt} (A\psi) mean? <br />
<br />
&lt;br /&gt;
\begin{equation*} &lt;br /&gt;
\begin{split} &lt;br /&gt;
\frac {d\phi} {dt}(t) &amp;amp;= \lim_{h\rightarrow 0} \frac {A(t+h)\psi(t+h) - A(t)\psi(t)} {h}\\ &lt;br /&gt;
&amp;amp;= \lim_{h\rightarrow 0} \frac {[A(t+h)\psi(t+h) - A(t+h)\psi(t)] + [A(t+h)\psi(t) - A(t)\psi(t)]} {h}\\ &lt;br /&gt;
&amp;amp;= A(t) \frac {d\psi} {dt} (t) + \frac {dA} {dt} (t) \psi(t). &lt;br /&gt;
\end{split} &lt;br /&gt;
\end{equation*} <br />
<br />
Here, I have written <br />
<br />
\frac {dA} {dt} (t) = \lim_{h\rightarrow 0} \frac {A(t+h) - A(h)} {h}, <br />
<br />
so, I have only succeeded in shifting attention away from your question about differentiation of operators to the question: What does this limit of operators mean? If the operators are bounded, then the limit is with respect to the standard norm on the space of bounded operators. If the operators are unbounded, ... <br />
<br />
Unfortunately, I have had to put all but a handful of my physics and math books into storage, so I can&#039;t say more. Again, I&#039;ve done formal manipulation and not much &quot;real&quot; functional analysis. A rigourous treatment needs to take into account stuff like dextercioby mentioned. I&#039;m not sure you want to see the real thing, but if you do, Reed and Simon is a good place to look. <br />
<br />
Regards, <br />
George
