# Operator Algebra III – Time Evolution and the Hamiltonian

At the end of my last post I promised we would spend some time thinking about the transformations that are brought about by Hermitian operators.  This may sound horribly abstract, so let me remind you of a very concrete claim that I have made a few times in the past: that the Hamiltonian generates time translations.  The goal of this post is to pick apart this statement and understand it on a fundamental level. I believe it is well within our reach now that we have the commutator in our pocket.  Note that the logic presented here generalizes, and I will speak to this generality later, but for now the specific case of the Hamiltonian provides a useful illustration of how to go about studying transformations induced by quantum mechanical operators.

As I said just a moment ago, we know an important fact about the Hamiltonian – i.e. that it generates time translations – and I have, up until now, parroted this fact without providing any intuition for why it is true.  Off the top of my head I can think of two ways to demonstrate the veracity of my claim.  The first is totally superficial, but worth seeing if you haven’t already.  Starting from the time-dependent Schrodinger equation, this is an almost trivial endeavor:

$i\hbar \frac{\partial }{{\partial t}}\left| {\psi \left( t \right)} \right\rangle = \hat H\left| {\psi \left( t \right)} \right\rangle \Leftrightarrow \frac{\partial }{{\partial t}}\left| {\psi \left( t \right)} \right\rangle = \frac{{\hat H}}{{i\hbar }}\left| {\psi \left( t \right)} \right\rangle \Leftrightarrow \left| {\psi \left( t \right)} \right\rangle = e^{\hat Ht/i\hbar } \left| {\psi \left( 0 \right)} \right\rangle$

(Edit: note that we have taken the Hamiltonian to be time-independent, and will continue to do so for the rest of this post.  Thank you Alec for making me realize I should have pointed this out explicitly!)

While formally correct, this statement is totally opaque.  It doesn’t admit any physical insight because (in my mind) the time-dependent Schrodinger equation is subordinate to the fact that the Hamiltonian generates time translations.  In short, we are letting the tail wag the dog, and we should instead try to understand how the Hamiltonian generates time translations, and then use this intuition to “derive” the TD Schrodinger equation.

I think a brief jaunt through world of classical mechanics will help us achieve this goal. Suppose we have an uncharged, non-relativistic particle moving about in regular 3-d space through a potential that varies with position.  The particle’s trajectory in position space is dictated by its kinetic energy (a function of momentum), while the particle’s trajectory in momentum space is dictated by the potential (a function of position).  In other words, as we march forward in time by an infinitesimal step $\varepsilon$, the particle’s position in phase space (which is, speaking loosely, the direct sum of position and momentum space) evolves according to

$\begin{array}{l} \vec x\left( {t + \varepsilon } \right) = \vec x\left( t \right) + \varepsilon \frac{{d\vec x\left( t \right)}}{{dt}} = \vec x\left( t \right) + \frac{\varepsilon }{m}\vec p\left( t \right) = \vec x\left( t \right) + \varepsilon \nabla _p T\left( {\vec p\left( t \right)} \right) \\ \vec p\left( {t + \varepsilon } \right) = \vec p\left( t \right) - \varepsilon \nabla _x V\left( {\vec x\left( t \right)} \right) \\ \end{array}$

Where $T\left( {\vec p\left( t \right)} \right) = \vec p^2 \left( t \right)/2m$ and $V\left( {\vec x\left( t \right)} \right)$ represent the potential and kinetic energy (respectively) of the particle; and ${\nabla _p }$ and ${\nabla _x }$ operate on the position and momentum components of vector describing the particle’s location in phase space. These observations can be written suggestively as

$\begin{array}{l} \frac{d}{{dt}}\left( {\vec x\left( t \right);\vec p\left( t \right)} \right) = \left( {\nabla _p - \nabla _x } \right)\left( {T\left( {\vec x\left( t \right);\vec p\left( t \right)} \right) + V\left( {\vec x\left( t \right);\vec p\left( t \right)} \right)} \right) = ... \\ ... = \left( {\nabla _p - \nabla _x } \right)H\left( {\vec x\left( t \right);\vec p\left( t \right)} \right) \\ \end{array}$

Here, we see that the time evolution of the system is given by the symplectic gradient $\left( {\nabla _p - \nabla _x } \right)$ of the Hamiltonian.  This is the mechanism through which the Hamiltonian generates time translation in classical mechanics. We don’t have to work too hard to extend this result to the quantum Hamiltonian. At this point, the commutator becomes quite useful. Recall that the commutator can be thought of as a directional derivative with respect to conjugation

$\left[ {A,B} \right] = \mathop {\lim }\limits_{t \to 0} \frac{{e^{ - itA} Be^{itA} - B}}{{it}}$

for Hermitian operators A and B. With this in mind, let’s consider the transformation that the Hamiltonian induces on ${\hat x}$ and ${\hat p}$.

$\left[ {\hat H,\hat x} \right] = \left[ {\frac{{\hat p^2 }}{{2m}},\hat x} \right] = \frac{1}{{2m}}\left( {\hat p\left[ {\hat p,\hat x} \right] + \left[ {\hat p,\hat x} \right]\hat p} \right) = \frac{\hbar }{{im}}\hat p = \frac{\hbar }{i}\frac{{\partial \hat T\left( {\hat p} \right)}}{{\partial \hat p}}$

$\begin{array}{l} \left[ {\hat H,\hat p} \right] = \left[ {\hat V\left( {\hat x} \right),\hat p} \right] = \left[ {\sum\limits_n {a_n \hat x^n } ,\hat p} \right] = \sum\limits_n {a_n \left[ {\hat x^n ,\hat p} \right]} = ... \\ ... = \sum\limits_n {na_n \hat x^{n - 1} \left[ {\hat x,\hat p} \right]} = - \frac{\hbar }{i}\frac{{\partial \hat V\left( {\hat x} \right)}}{{\partial \hat x}} \\ \end{array}$

These results look familiar! It is not a coincidence that we can make the identifications

$\begin{array}{l} \frac{{d\vec x}}{{dt}} = \nabla _p T\left( {\vec p} \right) \Leftrightarrow i/\hbar \left[ {\hat H,\hat x} \right] = \frac{{\partial \hat T\left( {\hat p} \right)}}{{\partial \hat p}} \\ \frac{{d\vec p}}{{dt}} = - \nabla _x V\left( {\vec x} \right) \Leftrightarrow i/\hbar \left[ {\hat H,\hat p} \right] = - \frac{{\partial \hat V\left( {\hat x} \right)}}{{\partial \hat x}} \\ \end{array}$

Next, we take advantage of the fact that expectation values necessarily behave classically.  It therefore makes sense to write $\frac{{d\left\langle {\hat x} \right\rangle }}{{dt}} = \frac{i}{\hbar}\left\langle {\left[ {\hat H,\hat x} \right]} \right\rangle$ and $\frac{{d\left\langle {\hat p} \right\rangle }}{{dt}} = \frac{i}{\hbar}\left\langle {\left[ {\hat H,\hat p} \right]} \right\rangle$.  I will leave out the details, but these two results imply that $\frac{{d\left\langle {\hat \Omega } \right\rangle }}{{dt}} = \frac{i}{\hbar}\left\langle {\left[ {\hat H,\hat \Omega } \right]} \right\rangle$ for an arbitrary operator ${\hat \Omega }$, since ${\hat \Omega }$ can be written in terms containing powers of ${\hat x}$ and ${\hat p}$.

This result – Ehrenfest’s theorem – is one of the most important in quantum mechanics, and we will use it to our advantage. Let’s examine the implications of this theorem, working from the definition of the commutator (vide supra).

$\begin{array}{l} \frac{i}{\hbar }\left\langle {\left[ {\hat H,\hat \Omega } \right]} \right\rangle = i\left\langle {\left[ {\hat H/\hbar ,\hat \Omega } \right]} \right\rangle = \left\langle {\mathop {\lim }\limits_{t \to 0} \left( {e^{ - it\hat H/\hbar } \hat \Omega e^{it\hat H/\hbar } - \hat \Omega } \right)/t} \right\rangle = ... \\ ... = \mathop {\lim }\limits_{t \to 0} \left( {\left\langle {\psi \left( 0 \right)} \right|e^{ - it\hat H/\hbar } \hat \Omega e^{it\hat H/\hbar } \left| {\psi \left( 0 \right)} \right\rangle - \left\langle {\psi \left( 0 \right)} \right|\hat \Omega \left| {\psi \left( 0 \right)} \right\rangle } \right)/t \\ \end{array}$

For this result to be physically meaningful, it must be true that

$\left\langle {\psi \left( 0 \right)} \right|e^{ - it\hat H/\hbar } \hat \Omega e^{it\hat H/\hbar } \left| {\psi \left( 0 \right)} \right\rangle = \left\langle {\psi \left( t \right)} \right|\hat \Omega \left| {\psi \left( t \right)} \right\rangle$

It follows that the Hamiltonian generates the dynamics of our system, and we get the time-dependent Schrodinger equation for free:

$e^{it\hat H/\hbar } \left| {\psi \left( 0 \right)} \right\rangle = \left| {\psi \left( t \right)} \right\rangle \Leftrightarrow i\hbar \frac{\partial }{{\partial t}}\left| {\psi \left( t \right)} \right\rangle = \hat H\left| {\psi \left( t \right)} \right\rangle$

Hopefully this post has provided some insight into why the Hamiltonian generates time translations (in both classical and quantum mechanics).  In the next post, I will generalize the logic presented here to arbitrary operators, and wrap up this subject with a discussion of Noether’s theorem, symmetry, and conservation laws.  Thanks for reading!

-B

Filed under Group Theory in QM

### 4 responses to “Operator Algebra III – Time Evolution and the Hamiltonian”

1. Love the concept behind your blog! I’m a puny organic chemist who recoils from mathematics like a … I can’t think of a non-sexual metaphor. In any case, I like that you try to refer things back to physical interpretation, as I guess that’s ultimately the point.

Added you to our ‘blogroll’, will be following keenly!

2. Will

I didn’t check all your steps but the basic idea seems right. There’s so much to the classical-quantum correspondence that is generally unappreciated. I think you could very realistically just say (though perhaps you don’t like the handwaving) that since the classical Hamilton equations obviously link to H to time evolution, so to it must be in quantum mechanics.

3. Alec

Just a minor correction — the first equation is only true for a time-independent hamiltonian. Otherwise, really great article.

• Thanks for the comment Alec – you are absolutely right. There are a couple of equations in which I implicitly take the Hamiltonian to be time independent (e.g. Ehrenfest), and I should have mentioned this explicitly. I apologize if this caused any confusion!