Time-dependent unitary transformations of the Hamiltonian

[Muninn]

TL;DR Summary

Under time-dependent unitary transformations, the Hamiltonian transforms in a different way than other operators, and I'm trying to understand why.

Hi!

I recently came across a quantum mechanics problem involving a change of basis to a rotating basis. As part of the solution, I wanted to transform the Hamiltonian operator into the rotating basis. Since the new basis is rotating, the basis change operator is time-dependent. This led to a different transformation law for the Hamiltonian than what I expected, and I'm trying to understand how one could expect this.

The transformation law for operators which I normally use is of course
$$A' = UAU^\dagger$$.
The proof that I remember for this is to introduce an eigenbasis $|a\rangle$ for $A$ so that $A|a\rangle = a|a\rangle$, and then asking that, if $|a'\rangle=U|a\rangle$, $A'|a'\rangle = a|a'\rangle$. In that case:
$$U^\dagger A'U|a\rangle = U^\dagger A'|a'\rangle = aU^\dagger |a'\rangle = a|a\rangle = A|a\rangle \Rightarrow A' = UAU^\dagger$$.

However, if one assumes that $U$ is time-dependent and inserts $|\psi'\rangle = U|\psi\rangle$ into the Schrödinger equation for $|\psi'\rangle$, one finds ($\hbar=1$):
$$i\frac{d}{dt}|\psi'\rangle = iU\frac{d}{dt}|\psi\rangle + i(\frac{d}{dt}U)|\psi\rangle = H'|\psi'\rangle=H'U|\psi\rangle$$
which after rearrangement is
$$i\frac{d}{dt}|\psi\rangle = U^\dagger(H'U - i(\frac{d}{dt}U))|\psi\rangle$$
which prompts the identification $H = U^\dagger H'U - iU^\dagger\frac{d}{dt}U$, i.e. the transformation law
$$H' = UHU^\dagger + i(\frac{d}{dt}U)U^\dagger$$
which reduces to the "normal" law for time-independent transformations but is otherwise different.

I guess one has to accept this result, but my question is: Where does the proof of the "normal" law fail for the case of the Hamiltonian? I don't see the hole which let's the Hamiltonian be exempt from the first proof. Any clarification of this would be of much help!

Thanks!

 

[PeroK]

I suspect that if you make $U(t)$ a function of time, then you are changing the dynamics of the system, so the form of your Hamiltonian will change.

The operator $\hat{p}$ is the canonical momentum (the generator of translations). But, you also have the kinematic momentum $\hat{\pi}$.

If you have an arbitrary time-dependent change of basis, then the Hamiltonian may need to be expressed in terms of the kinematic momentum, which may no longer coincide with the canonical momentum.

For example, if you take a free particle and rotate the coordinate system, then you still have a free particle. But, if you take a rotating coordinate system, then it no longer has the dynamics of a free particle.

In short, your assumption that the form of the Hamiltonian remains the same may be the issue.

PS in the classical case, you would add a fictitious force, which would translate to adding a potential to the Hamiltonian in the QM case. That would change the form of $H$. I think that is the issue.

 

[PeroK]

Here's another, perhaps simpler, explanation. If you look at the derivation of the Schrodinger equation (e.g. from Sakurai), $H$ is defined in terms of the time-evolution operator. Different systems have different time-evolution, hence different Hamiltonians.

By making your change of basis time-dependent, you change the time evolution of your system, hence you change $H$ in a more fundamental way - i.e. you need to consider this as a change of time-evolution, not a change of basis.

 

[Muninn]

This makes a lot of sense. Just as you're saying, in the classical case we would expect to generate centrifugal terms in the Hamiltonian as we switch to the rotating coordinate system. Then we should expect this to happen also in quantum mechanics.

Since the form of the Hamiltonian changes, clearly the assumption that there is an operator $H'$ with the same eigenvalue spectrum in the transformed system breaks in some way. When I find time I'll try investigating this a bit more carefully to see if I can find out exactly how. But I suspect one can see it directly from
$$H' = UHU^\dagger + i(\frac{d}{dt}U)U^\dagger\ \text{;}$$
when the last term is nonzero the eigenvalues of $H'$ will at a minimum be shifted w.r.t. those of $H$ which breaks the demand $A'|a'\rangle = a|a'\rangle$.

Thanks for the help, PeroK!

By the way, I'm not too familiar with your notion of the kinematical momentum $\hat{\pi}$. What I am familiar with is the separation between the canonical and the "physical" momentum for motion in external magnetic fields. The textbook I've spent the most time with (Ballentine) introduces a velocity operator $\hat{v}$ to simplify results in this case, making them similar to classical expressions. Is the kinematical momentum something similar, i.e. $\hat{\pi} = m\hat{v}$? Do you have a literature reference where I can read more about it?

 

[PeroK]

This makes a lot of sense. Just as you're saying, in the classical case we would expect to generate centrifugal terms in the Hamiltonian as we switch to the rotating coordinate system. Then we should expect this to happen also in quantum mechanics.

Since the form of the Hamiltonian changes, clearly the assumption that there is an operator $H'$ with the same eigenvalue spectrum in the transformed system breaks in some way. When I find time I'll try investigating this a bit more carefully to see if I can find out exactly how. But I suspect one can see it directly from
$$H' = UHU^\dagger + i(\frac{d}{dt}U)U^\dagger\ \text{;}$$
when the last term is nonzero the eigenvalues of $H'$ will at a minimum be shifted w.r.t. those of $H$ which breaks the demand $A'|a'\rangle = a|a'\rangle$.

Thanks for the help, PeroK!

By the way, I'm not too familiar with your notion of the kinematical momentum $\hat{\pi}$. What I am familiar with is the separation between the canonical and the "physical" momentum for motion in external magnetic fields. The textbook I've spent the most time with (Ballentine) introduces a velocity operator $\hat{v}$ to simplify results in this case, making them similar to classical expressions. Is the kinematical momentum something similar, i.e. $\hat{\pi} = m\hat{v}$? Do you have a literature reference where I can read more about it?

The kinematic momentum was a red herring. One thing you have to be careful of in QM is when you talk about an "operator", how is that operator defined.

I'm not saying anything different in this post, it's just a bit more detail.

In this case, H was already defined relative to the time evolution of the system. If you do something that doesn't change the time-evolution, then H will not change (as an operator). If you simply change the basis, then H does not change (as an operator) but it's representation will change. The usual change of basis formula.

Introducing a time-dependent change of basis, however, changes the dynamics of the system and (crucially) changes H (as an operator).

The equation you got is correct. I've just checked by going back to the derivation of $H$ in terms of the time-evolution operator.

What your first equation says is that, under a change of basis, $H'_0 = UH_0U^\dagger$, where $H_0$ is the Hamiltionan associated with the system.

If you have a time-dependent change to the system, then you need to find a new Hamiltonian $H_1$, say, that is associated with the time-evolution of your new system. In this case you really have:

$H_1 = H_0 + iU^\dagger \frac{\partial U}{\partial t}$

This is your new Hamiltonian defined by an operator equation. Then, represented in your new basis, you have:

$H'_1 = UH_0U^\dagger + i\frac{\partial U}{\partial t}U^\dagger = H'_0 + i\frac{\partial U}{\partial t}U^\dagger$

 

[PeroK]

Just to pull all this together. The Hamiltonian is defined as the generator of time translations. This results in the Schrodinger (operator) equation:
$$i \hbar \frac{\partial T}{\partial t} = HT$$
Where $T(t, t_0)$ is the time-evolution operator, which maps the state at time $t_0$ to the state at time $t$.

If we imagine that $U(t)$ represents a rotation of the system, say. Then we have a new time evolution: $T_1 = UT$ and new Hamiltonian, $H_1$ that satisfy:
$$i \hbar \frac{\partial T_1}{\partial t} = H_1T_1$$
$$i \hbar (\frac{\partial U}{\partial t}T + U\frac{\partial T}{\partial t}) = H_1UT$$
$$i \hbar \frac{\partial U}{\partial t}T + UHT = H_1UT$$
$$H_1 =H + i \hbar \frac{\partial U}{\partial t}U^{\dagger}$$
(As $H$ commutes with $U$.)

That tells you how the Hamiltonian has changed. You can then apply the change of basis to this. Although, you don't have to as this gives you the new Schrodinger equation (for wave functions) directly.

 

[vanhees71]

I think the question has been correctly answered, but maybe it helps to remember that also in classical mechanics the Hamiltonian changes under time-dependent canonical transformations, and that's the analogue to the unitary transformation considered here:

If you change from one set $(q,p)$ of phase-space coordinates to another one, $(Q,P)$, you can do this with a generating function like $g(q,Q,t)$. Then
$$P=\partial_q g, \quad p=-\partial_Q g, \quad H'=H+\partial_t g.$$
Then the Hamilton equations for $(q,p)$ using $H=H(q,p,t)$ are equivalent to the ones for $(Q,P)$ using $H'=H'(Q,P,t)$.

The same is the case in QM. The entire formalism is invariant under arbitrary unitary transformations (which can also dependent on time, which is a parameter in QM).

The Hamiltonian is special in the sense that it governs the (physical) time evolution of the system, and that's why it has to be transformed differently than the other operators.

It's most easily seen in the Schrödinger picture of time evolution. I consider a spin less particle with $\hat{\vec{x}}$ and $\hat{\vec{p}}$ the fundamental operators from which the observable algebra is built via the use of the Heisenberg algebra (saying basically that momentum is the generator for spatial translations from Noether's theorem applied to classical Newtonian mechanics):
$$[\hat{x}_j,\hat{x}_k]=[\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar.$$

Then you have the Hamiltonian $\hat{H}(\hat{\vec{x}},\hat{\vec{p}},t)$, and in the Schrödinger picture by definition the $\hat{\vec{x}}$ and $\hat{\vec{p}}$ are time-independent,
$$\mathrm{d}_t \hat{\vec{x}}=0, \quad \mathrm{d}_t \hat{\vec{p}}=0,$$
and the state kets evolve with the full Hamiltonian, i.e.,
$$\mathrm{i} \hbar \partial_t |\psi(t) \rangle=\hat{H}(\hat{\vec{x}},\hat{\vec{p}},t) |\psi(t) \rangle.$$
That's all you need to define non-relativistic QM of a single particle.

Now the entire formalism is invariant under an arbitrary unitary time-dependent transformation, except that you change the picture of time evolution, i.e., you are no longer in the Schrödinger picture since the fundamental operators $\hat{\vec{x}}$ and $\hat{\vec{p}}$ are no longer time-independent, because you have
$$\hat{\vec{x}}'=\hat{U} \hat{\vec{x}} \hat{U}^{\dagger}, \quad \hat{\vec{p}}'=\hat{U} \hat{\vec{p}} \hat{U}^{\dagger},$$
and thus
$$\mathrm{d}_t \hat{\vec{x}}' = \mathrm{d}_t \hat{U} \hat{\vec{x}} \hat{U}^{\dagger} + \hat{U} \hat{\vec{x}} \mathrm{d}_t \hat{U}^{\dagger}=(\mathrm{d}_t \hat{U}) \hat{U}^{\dagger} \hat{\vec{x}}' + \hat{\vec{x}}' \hat{U} \mathrm{d}_t \hat{U}^{\dagger}. \qquad (*)$$
Now because of $\hat{U} \hat{U}^{\dagger} =1$ you have
$$(\mathrm{d}_t \hat{U}) \hat{U}^{\dagger}=-\hat{U} \mathrm{d}_t \hat{U}^{\dagger}.$$
Now we define
$$(\mathrm{d}_t \hat{U}) \hat{U}^{\dagger}=-\frac{1}{\mathrm{i} \hbar} \hat{H}_0.$$
Then we have because of the previous equation $\hat{H}_0^{\dagger}=\hat{H}_0$ and from (*)
$$\mathrm{d}_t \hat{\vec{x}}'=\frac{1}{\mathrm{i} \hbar} [\hat{\vec{x}}',\hat{H}_0].$$
The analogous equation holds for $\hat{\vec{p}}'$.

At the same time the state kets have also to be transformed,
$$|\psi'(t) \rangle=\hat{U} |\psi(t) \rangle.$$
Taking now the time derivative, you get
$$\mathrm{i} \hbar \mathrm{d}_t |\psi'(t) \rangle = \mathrm{i} \hbar (\mathrm{d}_t \hat{U}) |\psi(t) \rangle + \hat{U} \mathrm{i} \hbar \mathrm{d}_t |\psi(t) \rangle = \mathrm{i} \hbar (\mathrm{d}_t \hat{U}) \hat{U}^{\dagger}|\psi'(t) \rangle + \hat{U} \hat{H} \hat{U}^{\dagger} |\psi'(t) \rangle= (\hat{U} \hat{H} \hat{U}^{\dagger}-\hat{H}_0) |\psi'(t) \rangle.$$
Now you call
$$\hat{H}_1=\hat{U} \hat{H} \hat{U}^{\dagger} - \hat{H}_0 = \hat{H}'-\hat{H}_0,$$
where
$$\hat{H}'=\hat{U} \hat{H} \hat{U}^{\dagger} = \hat{U} \hat{H}(\hat{\vec{x}},\hat{\vec{p}},t) \hat{U}^{\dagger} = \hat{H}(\hat{\vec{x}'},\hat{\vec{p}}',t).$$
So you have now split the time evolution into a time evolution of the observable operators $\hat{\vec{x}}$ and $\hat{\vec{p}}$, defined by a self-adjoint operator $\hat{H}_0$ through the above equation of motion and the time evolution of the state ket with the self-adjoint operator $\hat{H}_1$.

 

[Muninn]

Thank you for your excellent and very clear answers, PeroK and vanhees71. This has been an enlightening discussion. Sorry for replying this late.

The kinematic momentum was a red herring. [...]

Indeed, but it's still a concept I would like to be more familiar with. Do you know of any textbooks which discuss it? Thanks.

 

[PeroK]

Thank you for your excellent and very clear answers, PeroK and vanhees71. This has been an enlightening discussion. Sorry for replying this late.Indeed, but it's still a concept I would like to be more familiar with. Do you know of any textbooks which discuss it? Thanks.

Sakurai (Modern QM) covers it in his analysis of Electromagnetism. The Hamiltonian (in cgs units) is:
$$H = \frac{1}{2m} (\vec p - \frac{e \vec{A}}{c})^2 + e\phi$$
Here $\vec p$ is the Canonical Momentum (generator of spatial translations), while:
$$\vec \pi = \vec p - \frac{e \vec{A}}{c}$$
Is the Kinematic Momentum, where we also have:
$$\vec \pi = m\frac{d \vec x}{dt}$$
That's the only place in his book he uses it though!

 

[Muninn]

Sakurai (Modern QM) covers it in his analysis of Electromagnetism. The Hamiltonian (in cgs units) is:
$$H = \frac{1}{2m} (\vec p - \frac{e \vec{A}}{c})^2 + e\phi$$
Here $\vec p$ is the Canonical Momentum (generator of spatial translations), while:
$$\vec \pi = \vec p - \frac{e \vec{A}}{c}$$
Is the Kinematic Momentum, where we also have:
$$\vec \pi = m\frac{d \vec x}{dt}$$
That's the only place in his book he uses it though!

Ok! That is identical to Ballentine's treatment, only that he works with the velocity operator
$$\vec{v} = \frac{d\vec{x}}{dt}$$
instead of introducing the kinematic momentum.

Thanks a lot!