Heisenberg/Schrodinger pictures versus passive/active

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In summary: U(t)}.In summary, the conversation discusses the assertion that the Heisenberg picture corresponds to passive transformations while the Schrodinger picture corresponds to active transformations. However, the speaker believes this is not the case and asks for clarification. The conversation then delves into the relationship between the two pictures and how they relate to active and passive views of transformations. It is determined that in the Heisenberg picture, the time evolution operator is associated with the observable instead of the state vector, and the state vector remains unchanged while the basis vectors evolve. This does not align with either the active or passive view of transformations.
  • #1
Geofleur
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In my readings lately, I have come across the assertion that the Heisenberg picture corresponds to passive transformations while the Schrodinger picture corresponds to active transformations; however, such seems not to be the case. I want to know whether my thinking is correct or if I'm confused about something.

Let ##\textbf{R}(\theta)## be a rotation in the plane about angle ##\theta##. Then for the active view, we rotate a vector ##\textbf{v}## thusly $$ \textbf{R}(\theta)[\textbf{v}] = v_i \textbf{R}(\theta)[\textbf{e}_i] = v_i R_{ji} \textbf{e}_j, $$ where I have expanded ##\textbf{R}(\theta)[\textbf{e}_i]## in terms of the basis {##{\textbf{e}_j}##}. For the passive view, on the other hand, we instead transform the basis, rotating each basis vector by ##-\theta##: $$ \bar{\textbf{e}}_i = \textbf{R}(-\theta)[\textbf{e}_i] = \textbf{R}^{-1}(\theta)[\textbf{e}_i] = R_{ji}^T\textbf{e}_j = R_{ij}\textbf{e}_j.$$ Using the orthogonality relation, we can solve for the original basis as ## \textbf{e}_i = R_{ji}\bar{\textbf{e}}_j ##. Then we have $$ \textbf{v} = v_i \textbf{e}_i = v_i R_{ji} \bar{\textbf{e}}_j.$$ So in either the active or passive view, the components of ##\textbf{v}## transform the same way (though it seems strange to de-emphasize that the basis is different between the two cases, which is what happens if we stick to matrices, as many texts do).

Now I will try to relate the Heisenberg and Schrodinger pictures to the above discussion. Let ##\textbf{U}(t)## be a unitary time evolution operator. Then if ##|\psi(0)\rangle## is the state vector at time zero, we can evolve the state as $$|\psi(t)\rangle = \textbf{U}(t)|\psi(0)\rangle.$$ Because ##\textbf{U}(t)|\psi(0)\rangle## is analogous to ##\textbf{R}(\theta)[\textbf{v}]##, evolving the state vector in this way is taking the active view. On the other hand, the expectation value of an observable ##\textbf{A}## is given as $$ \langle \psi | \textbf{A} | \psi \rangle = \langle \psi(0)|{\textbf{U}}^{\dagger}(t) \textbf{A} \textbf{U}(t) | \psi(0)\rangle.$$ In the Heisenberg picture, we associate the time evolution operators with ##\textbf{A}## instead of acting on ##|\psi(0)\rangle## or its corresponding bra, i.e., we define the time dependent operator $$\textbf{A}(t) = \textbf{U}^{\dagger}(t) \textbf{A} \textbf{U}(t),$$ and regard the state vector as unchanging. In this picture, we have done nothing to either the components or the basis vectors of ##|\psi(0)\rangle##. Hence, the Heisenberg picture does not seem to correspond to either the active or the passive view of transformations.

Is all this correct?
 
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  • #2
Geofleur said:
In my readings lately, I have come across the assertion that the Heisenberg picture corresponds to passive transformations while the Schrodinger picture corresponds to active transformations; however, such seems not to be the case. I want to know whether my thinking is correct or if I'm confused about something.

Let ##\textbf{R}(\theta)## be a rotation in the plane about angle ##\theta##. Then for the active view, we rotate a vector ##\textbf{v}## thusly $$ \textbf{R}(\theta)[\textbf{v}] = v_i \textbf{R}(\theta)[\textbf{e}_i] = v_i R_{ji} \textbf{e}_j, $$ where I have expanded ##\textbf{R}(\theta)[\textbf{e}_i]## in terms of the basis {##{\textbf{e}_j}##}. For the passive view, on the other hand, we instead transform the basis, rotating each basis vector by ##-\theta##: $$ \bar{\textbf{e}}_i = \textbf{R}(-\theta)[\textbf{e}_i] = \textbf{R}^{-1}(\theta)[\textbf{e}_i] = R_{ji}^T\textbf{e}_j = R_{ij}\textbf{e}_j.$$ Using the orthogonality relation, we can solve for the original basis as ## \textbf{e}_i = R_{ji}\bar{\textbf{e}}_j ##. Then we have $$ \textbf{v} = v_i \textbf{e}_i = v_i R_{ji} \bar{\textbf{e}}_j.$$ So in either the active or passive view, the components of ##\textbf{v}## transform the same way (though it seems strange to de-emphasize that the basis is different between the two cases, which is what happens if we stick to matrices, as many texts do).

Now I will try to relate the Heisenberg and Schrodinger pictures to the above discussion. Let ##\textbf{U}(t)## be a unitary time evolution operator. Then if ##|\psi(0)\rangle## is the state vector at time zero, we can evolve the state as $$|\psi(t)\rangle = \textbf{U}(t)|\psi(0)\rangle.$$ Because ##\textbf{U}(t)|\psi(0)\rangle## is analogous to ##\textbf{R}(\theta)[\textbf{v}]##, evolving the state vector in this way is taking the active view. On the other hand, the expectation value of an observable ##\textbf{A}## is given as $$ \langle \psi | \textbf{A} | \psi \rangle = \langle \psi(0)|{\textbf{U}}^{\dagger}(t) \textbf{A} \textbf{U}(t) | \psi(0)\rangle.$$ In the Heisenberg picture, we associate the time evolution operators with ##\textbf{A}## instead of acting on ##|\psi(0)\rangle## or its corresponding bra, i.e., we define the time dependent operator $$\textbf{A}(t) = \textbf{U}^{\dagger}(t) \textbf{A} \textbf{U}(t),$$ and regard the state vector as unchanging. In this picture, we have done nothing to either the components or the basis vectors of ##|\psi(0)\rangle##. Hence, the Heisenberg picture does not seem to correspond to either the active or the passive view of transformations.

Is all this correct?

That looks correct to me. There's nothing directly spatial about the Heisenberg and Schroedinger pictures, as it's time evolution. In the special case where the time evolution was a rotation of the system, then there would be a relationship with a change of spatial basis.

PS I guess by "passive" is meant that the state does not evolve in the Heisenberg picture.
 
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  • #3
You can absorb the time evolution into the eigenvectors of [itex]\textbf{A}[/itex]:
[tex]\begin{eqnarray*}
\textbf{A}(t) &=& \textbf{U}^{\dagger}(t) \textbf{A} \textbf{U}(t)\\
&=& \textbf{U}^{\dagger}(t) \left( \sum_i a_i |a_i \rangle \langle a_i| \right) \textbf{U}(t)\\
&=& \sum_i a_i \textbf{U}^{\dagger}(t) |a_i \rangle \langle a_i| \textbf{U}(t)\\
&=& \sum_i a_i |a_i(t) \rangle \langle a_i(t)| \\
\end{eqnarray*}[/tex]
So you can view it like this: In the Schrödinger picture, the state vector evolves in time with [itex]\textbf{U}(t)[/itex] and the eigenvectors of the observables remain fixed. In the Heisenberg picture, the state vector remains fixed and the eigenvectors of the observables evolve in time with [itex]\textbf{U}^{\dagger}(t)[/itex].
 
  • #4
To PeroK: I am thinking of ##\textbf{U}(t)## as a rotation in complex space, since it preserves the inner product of the state vector with itself, just as rotation preserves the inner product of a geometrical vector with itself.

To kith: That makes sense. Writing ##|\psi\rangle## in terms of the eigenstates of ##\textbf{A}## the time evolution equation is ##\textbf{U}(t)|\psi\rangle = b_i \textbf{U}(t) | a_i \rangle##, which is again an active rotation (in complex space). Hence, both the Heisenberg and Schrodinger pictures correspond to active rotations, yes?
 
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  • #5
Geofleur said:
Writing ##|\psi\rangle## in terms of the eigenstates of ##\textbf{A}## the time evolution equation is ##\textbf{U}(t)|\psi\rangle = b_i \textbf{U}(t) | a_i \rangle## [...]
That doesn't seem right to me. (/edit: ##|\psi(t)\rangle = \textbf{U}(t)|\psi_0\rangle## is not valid in the Heisenberg picture. Instead we simply have ##|\psi(t)\rangle = |\psi_0\rangle##.) If you want to write the state vector in terms of eigenstates in the Heisenberg picture, you need to do this: [tex]|\psi_0\rangle = 1|\psi_0\rangle = \sum_i |a_i(t)\rangle \langle a_i(t)| \psi_0\rangle = \sum_i b_i(t) |a_i(t)\rangle = \sum_i b_i(t) \textbf{U}^{\dagger}(t) | a_i \rangle.[/tex] I would agree with the assertion that the Heisenberg picture corresponds to a passive transformation because the state vector remains fixed while the eigenstates evolve in time with the adjoint operator.
 
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  • #6
I think I see - you would do it that way to keep the state vectors from evolving in time.

I'm still having a hard time seeing how the Heisenberg picture corresponds to the passive view. If I keep everything in quantum mechanical terms, the active view would seem to correspond to this operation: $$ \textbf{U}(t) | \psi(0) \rangle = \psi_n U_{mn} |c_m \rangle,$$ where the ##U_{mn}## are the matrix elements of ##\textbf{U}(t)## in some arbitrary (not necessarily energy) basis ##| c_n \rangle##. The passive view, on the other hand, would involve transforming the basis via $$|\bar{c}_m\rangle = U_{nm}^{\dagger}|c_n\rangle,$$ which inverts to give $$|c_n\rangle = U_{mn}|\bar{c}_m\rangle.$$ In terms of these transformed basis vectors, the original state becomes $$|\psi(0)\rangle = \psi_n U_{mn} |\bar{c}_m\rangle.$$ These equations have the same forms as the ones I wrote for spatial rotations, with ##U_{mn} \mapsto R_{ji}##, ## \psi_n \mapsto v_i##, ##|c_m\rangle \mapsto \textbf{e}_j##, and ##|\bar{c}_m\rangle \mapsto \bar{\textbf{e}}_j.##

What I still don't see is how, when I translate the description of passive rotations into quantum mechanical terms, that what I get corresponds to using the Heisenberg picture.
 
  • #7
I have an idea: If the ##|c_n\rangle## depend on time, then we can use ##\textbf{U}^{\dagger}## to "rewind" them to time zero and think of the transformed, time independent basis as the barred one above. That way, we would end up with $$|\psi(t)\rangle = \psi_n U_{mn} |c_m(0)\rangle,$$ which does indeed look Heisenberg-ish. Is that correct?
 
  • #8
One difference between your last equation in post #6 and and my last equation in post #5 is that you have a double sum and I don't. I don't see it right away but if you decompose my [itex]|a_i(t)\rangle[/itex] in terms of the initial basis vectors [itex]|a_j\rangle[/itex] you should get something very similar.
 
  • #9
Geofleur said:
I have an idea: If the ##|c_n\rangle## depend on time [...]
The ##|c_n\rangle## as defined in post #6 shouldn't depend on time because they are the eigenvectors of an operator in the Schrödinger picture.
 
  • #10
Let me see if I understand.

As long as I expand ##| \psi(0) \rangle## in terms of eigenstates for a time-independent operator, the basis vectors in that expansion will be time-independent. However, I could still expand ##|\psi(0)\rangle## in terms of the eigenstates of a time-dependent operator.

You seem to have done just that in post #5, in which case the basis vectors and the components are both time-dependent in a way that cancels out to give a time-independent state. Then, you expressed the basis in terms of that at time zero using the evolution operator, and arrived at an equation similar to mine for passive transformations.

Edit (after much more pondering): All this expansion of ##| \psi (0) \rangle## in terms of time dependent parts seems very strange and inessential to the Heisenberg picture. It seems like I may be trying to take the analogy with rotation too far. With a passive rotation, the basis vectors change and the components of the vector also change, but in such a way that the vector as a whole stays the same. In the Heisenberg picture, does it not make sense to say that nothing at all happens to the state vector, either to its components or its basis? It's the operators that change. So maybe the Heisenberg picture is only passive in the sense that nothing happens to the state vector, while other things (operators) are changing with time.
 
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  • #11
Geofleur said:
Edit (after much more pondering): All this expansion of ##| \psi (0) \rangle## in terms of time dependent parts seems very strange and inessential to the Heisenberg picture.
I agree. When we actually use the Heisenberg picture we are interested in the dynamics of the operators and not in the trivial state vector.

I have also thought a bit more about the whole thing and maybe part of the issue is this: when we use the terms passive and active transformations in linear algebra, we are talking about coordinate spaces, i.e. our vectors are tuples of numbers. The context of QM in which the Heisenberg picture and the Schrödinger picture are discussed is more abstract than this: we talk only about elements of the Hilbert space and don't specify a numerical representation. When this viewpoint is taken talking about the components of a state vector or about its basis just doesn't make sense.
 
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Related to Heisenberg/Schrodinger pictures versus passive/active

1. What is the difference between the Heisenberg and Schrodinger picture in quantum mechanics?

The Heisenberg and Schrodinger pictures are two different ways of representing the time evolution of quantum systems. In the Heisenberg picture, the states of the system remain fixed and the operators evolve with time, while in the Schrodinger picture, it is the states that evolve with time and the operators remain fixed.

2. Which picture is more commonly used in quantum mechanics?

The Schrodinger picture is more commonly used in quantum mechanics because it is simpler to work with mathematically. However, the Heisenberg picture is useful for certain applications in quantum field theory.

3. What is the difference between a passive and active transformation in quantum mechanics?

In quantum mechanics, a passive transformation is a change in the basis or representation of a state or operator, without any change in the physical system itself. An active transformation, on the other hand, involves a physical change in the system, such as a rotation or translation.

4. How are the Schrodinger and Heisenberg pictures related to passive and active transformations?

In the Schrodinger picture, the states and operators are transformed passively, while in the Heisenberg picture, they are transformed actively. This means that the two pictures are related by a unitary transformation, which preserves the inner product of states and the algebra of operators.

5. Which picture is more useful for understanding symmetries in quantum mechanics?

The Heisenberg picture is more useful for understanding symmetries in quantum mechanics, as it allows for a direct treatment of transformations and their generators. This makes it easier to analyze the effects of symmetries on physical systems.

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