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I'm confused about how energy and time are linked. On the one hand, the Hamiltonian seems to describe the time evolution of the system because in the time dependent Schrodinger equation,

$$ \hat H \left| \psi (t) \right\rangle = i \hbar {\partial\over\partial t} \left| \psi (t) \right\rangle $$

But then it also seems to tell us the energy in the time-independent schrodinger equation:

$$ E\Psi=\hat H \Psi $$

My guess is that this has something to do with the Heisenberg's relationship between energy and time:

$$ \Delta E \Delta t\geq \frac{\hbar}{2} $$

so the relationship seems to be similar to that between position and momentum, and I believe the commutator of position and momentum is $i\hbar$ which crops up in the time-independent Schrodinger equation.

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  • $\begingroup$ I feel like this question's title could be more specific to what's actually being asked. $\endgroup$ – DanielSank Sep 16 '14 at 16:14
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You are right on track.

A great way to think about the Hamiltonian is that it's the thing that "causes translation in time". This just jargon for the fact, which you already observed, that the Hamiltonian tells you how the system moves forward in time, as encapsulated by the Schrodinger equation

$$i\hbar d_t |\Psi\rangle = H |\Psi \rangle . $$

My guess is that this has something to do with the Heisenberg's relationship between energy and time: $$ΔEΔt≥ℏ2$$ so the relationship seems to be similar to that between position and momentum, and I believe the commutator of position and momentum is iℏ which crops up in the time-independent Schrodinger equation.

Indeed. A good way to understand the link is to realize that, similarly to how the Hamiltonian generates translations in time, the momentum operator generates translations in position! Let's express this mathematically. Denote as $|x\rangle$ a quantum state which is an eigenstate of the position operator, i.e. $\hat{x}|x\rangle = x|x\rangle$. Let $\Delta x$ be some amount of position displacement (not an operator). It turns out that $$e^{-i\hat{p} \Delta x/\hbar} |x\rangle = |x+\Delta x\rangle. \qquad (1)$$ In English, this says that the operator $\exp[-i\hat{p}\Delta x / \hbar]$ translates the state $|x\rangle$ by an amount $\Delta x$. This is completely analogous to the fact that the operator $\exp[ -i H t/\hbar]$ translates a state forward in time by an amount $t$. In fact, if we take the derivative of Eq. (1) with respect to $\Delta x$, we get

$$ \begin{align} (-i\hat{p}/\hbar)e^{-i \hat{p}\Delta x / \hbar}|x\rangle &= d_{\Delta x}|x+\Delta x \rangle \\ -i\hat{p}/\hbar |x + \Delta x\rangle &= d_{\Delta x}|x + \Delta x\rangle \\ \text{or} \qquad \hat{p}|\Psi\rangle &= i\hbar d_{\Delta x} |\Psi\rangle \qquad (2) \end{align} $$

Let's compare Eq. (2) with the Schrodinger equation:

$$ \begin{align} \text{Schrodinger}\qquad i\hbar d_t |\Psi\rangle &= \hat{H}|\Psi\rangle \\ \text{Eq. (2)}\qquad i\hbar d_{\Delta x}|\Psi\rangle &= \hat{p}|\Psi\rangle \end{align} $$ As you can see, if the momentum operator is thought of as making translations in position, the Hamiltonian is entirely similarly though of as making translates in time. In an entirely similar manner, the position operator generates translations in momentum.

The truly general statement here is that if you have two operators $\hat{\alpha}$ and $\hat{\beta}$ with a commutator $[\hat{\alpha},\hat{\beta}]=i\gamma$ where $\gamma$ is any real number ($\gamma=\hbar$ for the case of $\hat{x}$ and $\hat{p}$), then $\alpha$ and $\beta$ generate translations of one another [3]. Note that having an $\hbar$ show up in Schrodinger's equation is a convention. You could work with the operator $H/\hbar$, which has units of frequency, instead and never see $\hbar$'s anywhere in your equations of motion. I much prefer this, actually. This is what some people erroneously call "setting $\hbar$ to one".

The analogy between position/momentum and energy/time breaks down at some point though, because there isn't a time operator.

For more information on translation operators, take a look at the wikipedia article on translation operators for a mathematically generic description, and also this Physics SE post.

[3]: This is actually not even really special to quantum mechanics. If you study differential geometry you find relations like this which hold whenever two things have a relation that's kind of similar to a commutator.

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  • $\begingroup$ Can you elaborate more on your footnote [3] in more mathematical terms? I guess you are referring to the Lie Derivative but I am not sure really :) $\endgroup$ – Gonenc May 22 '17 at 18:28

protected by ACuriousMind May 22 '17 at 17:29

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