I have a question about the notion of change of picture in QM.

When I saw it for the first time, it was introduced by the condition that we want to conserve the average value of an observable.

For example from Schrödinger to Heisenberg we can write :

$$ \langle \psi(t) | A_s | \psi(t) \rangle = \langle \psi(t_0) | U^{\dagger}(t,t_0) A_s U(t,t_0) | \psi(t_0) \rangle$$

And we can say that : $ | \psi_H \rangle = | \psi(t_0) \rangle $ and $U^{\dagger}(t,t_0) A_s U(t,t_0)=A_H(t)$ and now our observable change in time and our ket are constants.

But in my opinion this explanation was not really clear because we only asked to keep the average value of observable : what about conservation of the amplitudes of probability to find a given measurement for example.

So I read other introductions to it and the one that I found the best was with saying that a change of picture is just a change of basis in the Hilbert space by a unitary operator.

Indeed, We have : $ | \psi_H \rangle = U^{\dagger}(t,t_0) | \psi_S \rangle$ and $U^{\dagger}(t,t_0) A_s U(t,t_0) = A_H(t)$ with this example.

So a "general" change of picture would be given by a unitary operator $O$. And it would act like a change of coordinates :

$| \psi_O \rangle = O^{\dagger} | \psi_S \rangle$ and $O^{\dagger} A_s O=A_O$

But there are things I don't understand with this vision of things. This would mean that $|\psi_S\rangle$ is not a vector of a Hilbert space but a basis. And $| \psi_O \rangle$ would be another basis of the Hilbert space.

Indeed, the expression of a vector doesn't change : if $u$ is a vector of a vector space, it has different expressions in different basis : $u^i e_i = u'^j e'_j$ but the vector $u$ doesn't change.

Here my $|\psi_S\rangle$ change under my unitary transformation in $|\psi_O\rangle$.

But when we introduce Q.M, we define the ket in schrodinger point of view as a vector that doesn't depend on any choice of basis.

So I would like that you help me to better see things : if $|\psi_S\rangle$ describes my system it should'nt depend on any choice of basis (its components would depend on a choice of basis but not the vector itself). And here when I do the transformation of coordinates, $|\psi_S\rangle$ turns to $|\psi_O\rangle$... so I don't understand.

Where I am wrong ? Maybe if someone could properly help me to set the problem I could better see where I do a mistake.


But in my opinion this explanation was not really clear because we only asked to keep the average value of observable : what about conservation of the amplitudes of probability to find a given measurement for example.

It works too. Given an eigenvalue $a$,










That said, in view of your comment, I think your issue is the frequent confusion between active and passive transforms, and indeed this is very general. Right, let's be general then. I'll consider an operator $A$ and an unitary operator $U$.

Active transform



Then let's apply $U$ to the both of those kets,

$$\begin{aligned} \ket{x'}&=U\ket{x},\\ \ket{y'}&=U\ket{y}. \end{aligned}\tag{Ia}$$

We have therefore




Passive transform

Let's introduce a basis ${\ket{i}}$ and express components

$$\begin{aligned} \ket{y}&=\sum_iy_i\ket{i},\\ \ket{x}&=\sum_ix_i\ket{i}. \end{aligned}$$

Let's do a change of basis now,

$$\ket{i'} = \sum_j U_{ij} \ket{j},$$

and introduce the new coordinates $$\begin{aligned} \ket{y'}&=\sum_iy'_i\ket{i'},\\ \ket{x'}&=\sum_ix'_i\ket{i'}. \end{aligned}$$

Let's consider the (column) vector of coordinates $X,Y,X',Y'$. They are related by

$$\begin{aligned} Y&=U^\dagger X,\\ Y'&=U^\dagger X'. \end{aligned}\tag{Ip}$$

Then $\ket{y}=A\ket{x}$ can be expressed with the matrix of $A$ as

$$y_i = \sum_j A_{ij}x_i,$$

or, using matrix notations,

$$Y = \mathcal{A}X.$$

Then the vector of new coordinates are related by

$$Y' = \mathcal{A}'X'$$


$$\mathcal{A}' = \mathcal{U}^\dagger\mathcal{A}\mathcal{U}.\tag{IIp}$$


Now compare equations (Ia) and (Ip) on one hand and equations (IIa) and (IIp) on the other hand. They look pretty much the same but they are conceptually very different. The confusion arises if we start to blindly identify the operators $A$ and $U$ with their matrices $\mathcal{A}$ and $\mathcal{U}$. This is therefore more frequent a confusion in classical mechanics than in quantum mechanics, as the notation used for the latter protect from that mistake somewhat. But I digress: the point I wanted to get to, is that the change from Schrödinger to Heisenberg picture is an active transformation, not a passive one.

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  • $\begingroup$ Yes I know I can prove things like this but I would like to understand the "change of basis" point of view because it seems to be very general. $\endgroup$ – StarBucK Sep 24 '17 at 9:08
  • $\begingroup$ Thanks a lot for your answer. I think it is more clear but I would like to check something. So in a sense, I can see $| \psi_H \rangle$ not totally as an abstract vector but as a column vector representing the coordinates of $| \psi \rangle$ in a time-dependant basis. So when I write $| \psi_H \rangle = U | \psi_S \rangle$ I can see it as $[Y]=U[X]$ where $[Y]$ and $[X]$ represent a same vector in two different basis (here it is the column vector associated to it) ? Also in the first interpretation we implicitly say that 2 different vectors can represent the same state. $\endgroup$ – StarBucK Oct 5 '17 at 12:11
  • $\begingroup$ But in practice it is not the chosen one right ? Because in QM book states of systems are always represented by unique abstract vectors (if we omit the class of equivalence between proportional vectors). Here it would mean that two different abstract vectors could represent the same QM state which is not how MQ in introduced in general. Just to be sure that I understood well what you said. $\endgroup$ – StarBucK Oct 5 '17 at 12:17
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    $\begingroup$ Yes, I think so. And as you know, you do not just have those two: there is also the interaction picture. Actually, there is an infinite number of such pictures I guess. $\endgroup$ – user154997 Oct 5 '17 at 12:31
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    $\begingroup$ I'll make my conclusion clearer in my answer anyway, so that people don't have to figure it out by reading our discussion! $\endgroup$ – user154997 Oct 5 '17 at 12:35

There are two ways that you could choose to view the different pictures of time evolution in quantum mechanics. These two descriptions are complementary to each other, so they should not be used together.

1) The state of the system could be described in the different pictures by $\newcommand{\ket}[1]{|{#1}\rangle}\ket{\psi_S(t)}, \ket{\psi_I(t)}$, and $\ket{\psi_H}$, respectively. These abstract vectors are NOT equal, and the unitary transformation that you apply to the state in one picture to obtain the state in another picture is not a change of basis of the state, but rather simply a basis-independent transformation of the state. These states all live in the same Hilbert space. Note that we haven't used any choice of basis in this interpretation of the different pictures. In fact, this interpretation is entirely basis-independent. In summary, in this interpretation the difference between the pictures is contained in the abstract state.

2) The state of the system is described by one time-independent abstract vector $\ket{\psi}$ which is independent of the picture used. The pictures then differ in the choice of basis used to describe $\ket{\psi}$. In the Heisenberg picture, a time-independent basis is used. In the Schrodinger and interaction pictures, a time-dependent basis is used that is related to the Heisenberg basis by a time-dependent unitary transformation. Although the state of the system $\ket{\psi}$ is the same across all pictures, it is the basis-dependent representations $\vec{\psi_S}(t), \vec{\psi_I}(t)$, and $\vec{\psi_H}$ that differ. Because the Schrodinger and interaction representations of $\ket{\psi}$ are defined with respect to time-dependent bases, their representations exhibit time-dependence. Thus, in this interpretation, the transformation between the pictures is formally a change of basis. In summary, in this interpretation the difference between the pictures is contained in the basis chosen.

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  • $\begingroup$ Please re-work the maths in this answer to use only standard LaTeX. $\endgroup$ – Emilio Pisanty Sep 24 '17 at 1:14
  • $\begingroup$ I am not sure to understand well your answer. You say that I have to fix $| \psi_S \rangle =| \psi_I \rangle =| \psi_H \rangle $ for all time and then apply the time dependent unitary transformation that would give me abstract vector equals. But there is a unitary transformation between thoose kets so they can't be equal. There is something I misunderstand... $\endgroup$ – StarBucK Sep 24 '17 at 9:10
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    $\begingroup$ @StarBucK. I edited my answer. Hopefully this is more clear now. $\endgroup$ – Ian Sep 25 '17 at 17:47
  • $\begingroup$ @Ian Thanks a lot. I think it is more clear with your answer and the Luc one's. "In practice" we prefer to use the second interpretation right ? Because everywhere on MQ books it is said that the state is represented by a unique abstract vector (if we omit the fact that $\alpha | \psi \rangle $ and $|\psi \rangle$ represent the same state). So the first interpretation of different abstract vectors is not generally the chosen one ?? $\endgroup$ – StarBucK Oct 5 '17 at 12:13
  • $\begingroup$ I would say the first interpretation is more standard. The different pictures are interpreted not as different representations, but rather as different abstract vectors altogether, which are related by time dependent unitary transformations. $\endgroup$ – Ian Oct 5 '17 at 16:11

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