Unitary equivalence of QM (not for energy?)

In summary: 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.
  • #1
center o bass
560
2
Since all the observables in QM is on the form

[tex]\langle \alpha |A| \beta \rangle[/tex]

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

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

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

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

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

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

[tex] \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.[/tex]

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?
 
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  • #2
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).
 
  • #3
center o bass said:
However by using the Schrödinger equation one can show that the Hamiltonian transforms like

[tex] H = H' = UHU^{-1} + i\hbar \frac{dU}{dt} U^{-1}[/tex]
Please show this.
 
  • #4
Dickfore said:
Please show this.

[tex] 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 [/tex]

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.
 
  • #5
IsometricPion said:
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?
 
  • #6
center o bass said:
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.
center o bass said:
where I've used the Schrödinger equation in the second equality
This is not quite correct see Wikipedia-Heisenberg picture.
 
  • #7
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 [itex]\vert \Psi(t) \rangle[/itex] is isolated (and its time-evolution is given by Schroedinger's equation through the Hamiltonian H), then the system described by [itex]\vert \psi'(t) \rangle \equiv U(t) \, \vert \psi(t) \rangle[/itex] is, in general, open, and its energy is not conserved (and it's evolution is described by another time-dependent Hamiltonian [itex]H' = U \, H \, U^{\dagger} + i \, \hbar \, \dot{U} \, U^{\dagger}[/itex]).
 
  • #8
wait, I think you might be mixing up some time dependence and the derivatives... maybe
consider the following:
[tex]
|\psi(t)\rangle = U(t,t_0)|\psi(t_0)\rangle
[/tex]
Now if we re-do this derivation watching the [itex]t[/itex]s carefully, we get a different result
[tex]
\begin{align}
i\hbar \frac{d}{dt}|\psi (t)\rangle &= 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 \\
&= i\hbar \dot{U}U^{-1}|\psi (t)\rangle = H|\psi (t)\rangle
\end{align}
[/tex]
because the time derivative doesn't hit both the [itex]U[/itex] and the state. then
[tex]
i\hbar \frac{d U}{dt} = HU
[/tex]
which is normal and I imagine [itex]UHU^{-1}[/itex] still holds... but I might be mixed up.
 
  • #9
jfy4 said:
wait, I think you might be mixing up some time dependence and the derivatives... maybe
consider the following:
[tex]
|\psi(t)\rangle = U(t,t_0)|\psi(t_0)\rangle
[/tex]
Now if we re-do this derivation watching the [itex]t[/itex]s carefully, we get a different result
[tex]
\begin{align}
i\hbar \frac{d}{dt}|\psi (t)\rangle &= 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 \\
&= i\hbar \dot{U}U^{-1}|\psi (t)\rangle = H|\psi (t)\rangle
\end{align}
[/tex]
because the time derivative doesn't hit both the [itex]U[/itex] and the state. then
[tex]
i\hbar \frac{d U}{dt} = HU
[/tex]
which is normal and I imagine [itex]UHU^{-1}[/itex] 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.
 
  • #10
IsometricPion said:
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
[tex]i\hbar \dot{|\psi(t)\rangle }= H|\psi(t)\rangle.[/tex]

I have to agree with your explanation though. Thanks :)
 
  • #11
Another way to derive what you did is:[tex][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[/tex]. However, this derivation is only correct for U that do not explicitly depend on time. Per the wikipedia page, in general [itex][H,A]=-i\hbar\frac{dA}{dt}+i\hbar\frac{\partial{A}}{ \partial{t}}[/itex]. The correspoding result being: [tex]UHU^{-1}=i\hbar{}(\frac{dU}{dt}-\frac{\partial{U}}{ \partial{t}})U^{-1}+H[/tex]. When U is the time-evolution operator it so happens that [itex]\frac{dU}{dt}=\frac{\partial{U}}{ \partial{t}}[/itex], 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.
 

1. What is unitary equivalence in quantum mechanics?

Unitary equivalence in quantum mechanics refers to the concept of two different quantum systems having the same mathematical structure and therefore being equivalent. This means that they have the same physical properties and can be described by the same mathematical formalism.

2. How is unitary equivalence different from energy equivalence in quantum mechanics?

Unitary equivalence is a broader concept that encompasses energy equivalence. While energy equivalence only focuses on the energy states of a system, unitary equivalence takes into account all physical properties of a system, including its energy states, momentum, and spin.

3. Why is unitary equivalence important in quantum mechanics?

Unitary equivalence is important in quantum mechanics because it allows us to compare and understand different quantum systems. It also helps us to identify patterns and similarities between seemingly different systems, leading to new insights and discoveries in quantum physics.

4. How is unitary equivalence determined in quantum mechanics?

Unitary equivalence is determined by comparing the Hamiltonian operators of two different quantum systems. If the Hamiltonians are found to be unitarily equivalent, then the two systems are considered to be equivalent.

5. Can two quantum systems be unitarily equivalent but have different energies?

Yes, two quantum systems can be unitarily equivalent but have different energies. This is because unitary equivalence takes into account all physical properties of a system, not just its energy states. Therefore, even if the energy states are different, if the overall mathematical structure is the same, the systems are considered to be unitarily equivalent.

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