Please help me clarify my misconception involving generators

[TroyElliott]

I am having a hard time understanding what it means for momentum to be the generator of finite translation. Why would a state |x'> translate to another state |x'+l>, and just stop after a distance l? Wouldn't the state want to keep translating due to the imparted momentum Px, in order to conserve momentum?

This kind of question keeps popping up in my head every time I see a new generator, i.e. Hamiltonian and rotations.

Thank you

 

[Nugatory]

Which textbook or other source are you working from? You'll get better answers if you tell us this.

 

[TroyElliott]

Modern Quantum Mechanics - J.J. Sakurai

 

[A. Neumaier]

Why would a state |x'> translate to another state |x'+l>, and just stop after a distance l?

This happens if you apply $e^{ilp}$ to the state. For each $l$, one has a different translation, and the momentum $p$ generates them all!

 

[TroyElliott]

Thanks A. Neumaier.

I have a slightly different question that gets to the same concern that I have with generators. How is it that when you apply a rotation operator to a |j,m> state, that j (the total angular momentum eigenvalue) remains unchanged? How can we say we are rotating something by some angle but then say we did not impart on it any extra angular momentum?

Mathematically I understand that J^2 commutes with any J_k and thus commutes with any function of J_k, and from here we say that D(R)|j,m> is just an eigenket of J^2. But this doesn't clarify in my mind how is it that we can rotate an object without imparting additional angular momentum to it.

Thanks for any insight!

Maybe I should also ask the following. Are the translations and rotations happening in the physical space that we live in or is it happening in the ket/Hilbert space and thus does not actually correspond to actual rotation or translation in our 3d space?

 

[PeroK]

Thanks A. Neumaier.

I have a slightly different question that gets to the same concern that I have with generators. How is it that when you apply a rotation operator to a |j,m> state, that j (the total angular momentum eigenvalue) remains unchanged? How can we say we are rotating something by some angle but then say we did not impart on it any extra angular momentum?

First, note the difference between "rotated" and "rotating". The first implies a single rotation; the second implies a continuous (in time) rotation. It's the same for a linear translation: it's a single translation of a fixed quantity: $\Delta x$ or $\Delta t$ perhaps. It's not a continuous motion in one direction.

The idea of a rotation in this sense is either:

a) A change of coordinates, where you change to a new coordinate system, rotated with respect to the first. The system is untouched. And neither coordinate system is rotating with respect to the other.

b) A rotation of the system, where you imagine the new system is identical to the one you have, except it is rotated.

It's an axiom of spatial homogeneity and isotropy that neither of these things should change the physical properties of the system.

Sakurai mentions the need to be clear which one you are using.

 

[TroyElliott]

First, note the difference between "rotated" and "rotating". The first implies a single rotation; the second implies a continuous (in time) rotation. It's the same for a linear translation: it's a single translation of a fixed quantity: $\Delta x$ or $\Delta t$ perhaps. It's not a continuous motion in one direction.

The idea of a rotation in this sense is either:

a) A change of coordinates, where you change to a new coordinate system, rotated with respect to the first. The system is untouched. And neither coordinate system is rotating with respect to the other.

b) A rotation of the system, where you imagine the new system is identical to the one you have, except it is rotated.

It's an axiom of spatial homogeneity and isotropy that neither of these things should change the physical properties of the system.

Sakurai mentions the need to be clear which one you are using.

I understand what you said in (a). My confusion comes from what you said in (b). We can only have a new system which is rotated if we rotated it, but to do this we would impart on it some new angular momentum and thus change the total angular momentum of the system. I am trying to reconcile this but am having a difficult time.

Thanks

 

[PeroK]

I understand what you said in (a). My confusion comes from what you said in (b). We can only have a new system which is rotated if we rotated it, but to do this we would impart on it some new angular momentum and thus change the total angular momentum of the system. I am trying to reconcile this but am having a difficult time.

Thanks

That's too practical a view of this. Imagine I create a system: a simple pendulum say. And you create an identical system, but 10m to the right of mine.

Are there any differences in these systems enforced by the fact that one is translated to the right?

If there are external influences, of course. But, theoretically it should make no difference.

You could imagine, instead, picking up my system and moving it 10m to the right without disturbing anything. If we don't disturb anything, then the pendulum should behave the same whether it is in its original position of its new position.

What this definitely doesn't mean is that the practical problem of lifting a pendulum and actually moving it means that space is not homogenous!

If you don't like the idea of actually moving or rotating a system, then imagine simply that the system developed in its rotated state. Two hydrogen atoms, for example, may have different orientations, but they don't have different properties based on some universal notation of what is "the right way up".

 

[TroyElliott]

That's too practical a view of this. Imagine I create a system: a simple pendulum say. And you create an identical system, but 10m to the right of mine.

Are there any differences in these systems enforced by the fact that one is translated to the right?

If there are external influences, of course. But, theoretically it should make no difference.

You could imagine, instead, picking up my system and moving it 10m to the right without disturbing anything. If we don't disturb anything, then the pendulum should behave the same whether it is in its original position of its new position.

What this definitely doesn't mean is that the practical problem of lifting a pendulum and actually moving it means that space is not homogenous!

If you don't like the idea of actually moving or rotating a system, then imagine simply that the system developed in its rotated state. Two hydrogen atoms, for example, may have different orientations, but they don't have different properties based on some universal notation of what is "the right way up".

So basically, when looking at a translation operator, we are just saying that it is something that takes $|x'>$ to $|x'+l>$, but we don't care about how it got the momentum to leave $|x'>$, and we do not care about how it lost the momentum in order to stop at $|x'+l>$?

 

[PeterDonis]

So basically, when looking at a translation operator, we are just saying that it is something that takes $|x'>$ to $|x'+l>$, but we don't care about how it got the momentum to leave $|x'>$, and we do not care about how it lost the momentum in order to stop at $|x'+l>$?

No. The translation operator is best viewed as a change in how we look at the system, not as a change to the system itself. In terms of the cases (a) and (b) that @PeroK described in post #6:

In case (a), we have a system at a particular point in space $x$ (and at a given time $t$, since this state isn't stationary so it will evolve in time to something else, but we can ignore that here). We then choose to re-label all of our points in space by shifting the $x$ coordinates by an amount $l$. That means the state of the system that used to be labeled $|x>$ is now labeled $|x + l>$. But it's still the same state (the same position at the same time).

In case (b), we have a system that is moving, so it is at point $x$ in space at some time, and at point $x + l$ in space at some later time. "Translating" the state $|x>$ to the state $|x + l>$ just means changing which point in time we pick to look at our system; we don't have to do anything to the system itself to "move" it since it's already moving.

Saying that the momentum operator "generates" translations means the same thing mathematically in both cases, but how we interpret the mathematical operation we're doing physically differs. In case (a), we interpret the mathematical operation as a logical implication of changing our coordinates: translating the coordinates by a distance $l$ means we have to re-label states by applying the operation $e^{i l p}$ to each of them. In case (b), we interpret the mathematical operation as describing the system's motion--when it moves through a distance $l$, its state changes as described by the operation $e^{i l p}$.

 

[TroyElliott]

No. The translation operator is best viewed as a change in how we look at the system, not as a change to the system itself. In terms of the cases (a) and (b) that @PeroK described in post #6:

In case (a), we have a system at a particular point in space $x$ (and at a given time $t$, since this state isn't stationary so it will evolve in time to something else, but we can ignore that here). We then choose to re-label all of our points in space by shifting the $x$ coordinates by an amount $l$. That means the state of the system that used to be labeled $|x>$ is now labeled $|x + l>$. But it's still the same state (the same position at the same time).

In case (b), we have a system that is moving, so it is at point $x$ in space at some time, and at point $x + l$ in space at some later time. "Translating" the state $|x>$ to the state $|x + l>$ just means changing which point in time we pick to look at our system; we don't have to do anything to the system itself to "move" it since it's already moving.

Saying that the momentum operator "generates" translations means the same thing mathematically in both cases, but how we interpret the mathematical operation we're doing physically differs. In case (a), we interpret the mathematical operation as a logical implication of changing our coordinates: translating the coordinates by a distance $l$ means we have to re-label states by applying the operation $e^{i l p}$ to each of them. In case (b), we interpret the mathematical operation as describing the system's motion--when it moves through a distance $l$, its state changes as described by the operation $e^{i l p}$.

Thank you very much. That clears up what I was confused about.