Unitary equivalence of QM (not for energy?)

[center o bass]

Since all the observables in QM is on the form

\langle \alpha |A| \beta \rangle

where A is an observable, one can transform the observables and states like

A \to A' = UAU^{-1} \ \ \ |\beta \rangle \to |\beta '\rangle = U |\beta \rangle
where U is a unitary transformatioin. These descriptions of the theory is equivalent because

\langle \alpha' |A'| \beta' \rangle = \langle \alpha |U^{-1} U A U U^{-1}| \beta \rangle = \langle \alpha |A| \beta \rangle.

However by using the Schrödinger equation one can show that the Hamiltonian transforms like

H = H' = UHU^{-1} + i\hbar \frac{dU}{dt} U^{-1}

which means that the expectation value of H in the transformed representation is

\langle \psi'| H'|\psi '\rangle = \langle \psi| H \psi \rangle + i\hbar \langle \psi |U^{-1}\frac{dU}{dt} |\psi \rangle \neq \langle \psi| H \psi \rangle.

What is the meaning of this inequivalence? Is not the expectation value of H supposed to be equal in two descriptions which differ by a unitary transformation?

 

[IsometricPion]

I think the assumption is usually that U≠U(t). This seems intuitively reasonable since time-dependent rotations (a subset of U(t)), such as the one required to transform classically from an inertial frame to a rotating frame, do not generally preserve the (classical) energy (they don't even preserve Newton's laws).

 

[Dickfore]

However by using the Schrödinger equation one can show that the Hamiltonian transforms like

H = H' = UHU^{-1} + i\hbar \frac{dU}{dt} U^{-1}

Please show this.

 

[center o bass]

Please show this.

i\hbar \dot{|\psi '(t) \rangle } = i\hbar (\dot{U}(t) |\psi(t) \rangle + U \dot{|\psi(t)\rangle} = (UH + i\hbar \frac{dU}{dt} ) |\psi(t)\rangle = (UHU^{-1} ih \frac{dU}{dt}U^{-1})|\psi'(t)\rangle = H' |\psi'(t) \rangle

where I've used the Schrödinger equation in the second equality and in the third equality I've expressed the non-transformed state in terms of the transformed one.

 

[center o bass]

I think the assumption is usually that U≠U(t). This seems intuitively reasonable since time-dependent rotations (a subset of U(t)), such as the one required to transform classically from an inertial frame to a rotating frame, do not generally preserve the (classical) energy (they don't even preserve Newton's laws).

But then what about the Heisenberg picture? There certainly U=U(t), in fact it's the inverse of the time evolution operator. Is the energy of the system different in the Heisenberg picture?

 

[IsometricPion]

But then what about the Heisenberg picture? There certainly U=U(t), in fact it's the inverse of the time evolution operator. Is the energy of the system different in the Heisenberg picture?

I was thinking of the usual simple cases, such as rotational transformations. You are correct that some time-dependent unitary transformations do not change the expectation of H.

where I've used the Schrödinger equation in the second equality

This is not quite correct see Wikipedia-Heisenberg picture.

 

[Dickfore]

Ok, what you are doing reminds me of the derivation of the Berry phase. Basically, because you make a time-dependent similarity transformation, if the system described by the ket \vert \Psi(t) \rangle is isolated (and its time-evolution is given by Schroedinger's equation through the Hamiltonian H), then the system described by \vert \psi'(t) \rangle \equiv U(t) \, \vert \psi(t) \rangle is, in general, open, and its energy is not conserved (and it's evolution is described by another time-dependent Hamiltonian H' = U \, H \, U^{\dagger} + i \, \hbar \, \dot{U} \, U^{\dagger}).

 

[jfy4]

wait, I think you might be mixing up some time dependence and the derivatives... maybe
consider the following:
<br /> |\psi(t)\rangle = U(t,t_0)|\psi(t_0)\rangle<br />
Now if we re-do this derivation watching the ts carefully, we get a different result
<br /> \begin{align}<br /> i\hbar \frac{d}{dt}|\psi (t)\rangle &amp;= i\hbar \frac{d}{dt}U(t,t_0)|\psi (t_0)\rangle =i\hbar \dot{U}(t,t_0)|\psi (t_0)\rangle = i\hbar \dot{U}(t,t_0)U^{-1}U|\psi (t_0)\rangle \\<br /> &amp;= i\hbar \dot{U}U^{-1}|\psi (t)\rangle = H|\psi (t)\rangle<br /> \end{align}<br />
because the time derivative doesn't hit both the U and the state. then
<br /> i\hbar \frac{d U}{dt} = HU<br />
which is normal and I imagine UHU^{-1} still holds... but I might be mixed up.

 

[center o bass]

wait, I think you might be mixing up some time dependence and the derivatives... maybe
consider the following:
<br /> |\psi(t)\rangle = U(t,t_0)|\psi(t_0)\rangle<br />
Now if we re-do this derivation watching the ts carefully, we get a different result
<br /> \begin{align}<br /> i\hbar \frac{d}{dt}|\psi (t)\rangle &amp;= i\hbar \frac{d}{dt}U(t,t_0)|\psi (t_0)\rangle =i\hbar \dot{U}(t,t_0)|\psi (t_0)\rangle = i\hbar \dot{U}(t,t_0)U^{-1}U|\psi (t_0)\rangle \\<br /> &amp;= i\hbar \dot{U}U^{-1}|\psi (t)\rangle = H|\psi (t)\rangle<br /> \end{align}<br />
because the time derivative doesn't hit both the U and the state. then
<br /> i\hbar \frac{d U}{dt} = HU<br />
which is normal and I imagine UHU^{-1} still holds... but I might be mixed up.

If I understand you correctly it seems like you've assumed that U is the time evolution operator, but I meant a general time dependent, unitary transformation.

 

[center o bass]

I was thinking of the usual simple cases, such as rotational transformations. You are correct that some time-dependent unitary transformations do not change the expectation of H.This is not quite correct see Wikipedia-Heisenberg picture.

I think it is, I did use the fact that for a state in the Schrödinger picture
i\hbar \dot{|\psi(t)\rangle }= H|\psi(t)\rangle.

I have to agree with your explanation though. Thanks :)

 

[IsometricPion]

Another way to derive what you did is:[U,H]=i\hbar{}\frac{dU}{dt}\Rightarrow{}[U,H]U^{-1}=UHU^{-1}-HUU^{-1}=UHU^{-1}-H=i\hbar{}\frac{dU}{dt}U^{-1}\Rightarrow{}UHU^{-1}=i\hbar{}\frac{dU}{dt}U^{-1}+H. However, this derivation is only correct for U that do not explicitly depend on time. Per the wikipedia page, in general [H,A]=-i\hbar\frac{dA}{dt}+i\hbar\frac{\partial{A}}{ \partial{t}}. The correspoding result being: UHU^{-1}=i\hbar{}(\frac{dU}{dt}-\frac{\partial{U}}{ \partial{t}})U^{-1}+H. When U is the time-evolution operator it so happens that \frac{dU}{dt}=\frac{\partial{U}}{ \partial{t}}, so the equation reduces to UHU^-1=H. This is simply a statement that H commutes with the time-evolution operator, something already known to be true.