The equivalence between Heisenberg and Schroedinger pictures

Question

In quantum mechanics, the two pictures of Schroedinger and Heisenberg are taken as equivalent, where in the former wavefunctions are time variants and operators are not, and in the latter it is the other way around. I think it is important to understand equivalences in physics in general, but this one I have never grasped.

• I wonder, based on what criteria we see them as equivalent?

• Is there a simplified way of mathematically showing (or at least hinting at) how they are describing the same thing?

Answer 1

I will try to make it as simple and intuitive as possible. In the Schrödinger picture, the expectation value of a given operator $\hat{\xi}$ (which itself is frozen in time) is defined as follows (with $\psi(t)$ the wavefunction of our system at time $t$):

$$\langle \hat{\xi} (t) \rangle = \langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle$$

Which is just the average value of the observable corresponding to $\hat{\xi}$ if a measurement is made at time $t.$ Now exactly because the expectation value creates a direct link between what we predict with our theory in QM with what we observe experimentally, then logically however one goes about defining quantum mechanics, we should obtain the same values for $\langle \hat{\xi} (t) \rangle$ to ensure that we're going to predict the correct experimentally expected values (and hence be able to claim then that the two pictures are equivalent).

To show this equivalence, we first use an important property of the unitary time evolution operator, namely

$$\psi(t_1) = \hat{U}(t_1,t_0) \psi(t_0)$$

i.e. we propagate our wavefunction in time be acting $\hat{U}$ on it. With this, we can now redefine the wavefunction at time $t$ as its value at time $t=0$ upon which we act $\hat{U}(t,0).$ So we rewrite (by a simple substitution) our original expression for $\langle \hat{\xi} (t) \rangle$ as:

$$ \langle \psi (t) \lvert \hat{\xi} \rvert \psi(t) \rangle = \langle \psi (0) \lvert \hat{U}^{\dagger}(t,0) \hat{\xi} \hat{U}(t,0)\rvert \psi(0) \rangle $$ From the above you can already see the freedom of choice, i.e. to decide to act the time operators either on the wavefunctions or on the operator, by choosing the latter we get:

$$ \langle \psi (0) \lvert \left(\hat{U}^{\dagger}(t,0) \hat{\xi} \hat{U}(t,0)\right)\rvert \psi(0) \rangle = \langle \psi (0) \lvert \hat{\xi}(t) \rvert \psi(0) \rangle $$ Hence we have successfully shown that the time dependence can also be implemented in the operators, instead of wavefunctions while obtaining the same expectation values for our chosen observable, so let's call $\psi(0) = \psi_h$ with $h$ for Heisenberg, and similarly $\hat{\xi}(t) = \hat{\xi}_h(t).$ With this notation then you can easily relate the operator in the Schrödinger picture with that of the Heisenberg picture by:

$$\hat{\xi}_h(t)=\hat{U}^{\dagger}(t,0) \hat{\xi}_{\rm Schrödinger} \hat{U}(t,0)$$ Finally, from here you can straightforwardly obtain the expression of Heisenberg's equation of motion (although you didn't ask for it, but we've come all this way, may as well show it...):

Take the time derivative of $\hat{\xi}_h(t)$ (using the last equation derived) and by using the relation $d\hat{U}/dt=-\frac{i}{\hbar}\hat{H}\hat{U}$ (and also that $[\hat{H},\hat{U}]=0$):

$$ \begin{align*} \frac{d\hat{\xi}_h (t)}{dt} &= \frac{d\hat{U}^{\dagger}}{dt} \hat{\xi} \hat{U} + \hat{U}^{\dagger} \hat{\xi}\frac{d\hat{U}}{dt} \\ &= \frac{-1}{i\hbar}(\hat{U}^{\dagger}\hat{H}\hat{\xi}\hat{U}-\hat{U}^{\dagger}\hat{\xi}\hat{H}\hat{U})\\ &=\frac{1}{i\hbar}[\hat{\xi}_h (t),\hat{H}]. \end{align*} $$

Answer 2

On their non-equivalence. Yes, this is largely a folklore result. There are many ways for them to be non-equivalent. A few examples

https://arxiv.org/abs/1404.6775

https://www.sciencedirect.com/science/article/abs/pii/S0375960102015086

https://arxiv.org/abs/0706.3867

More generally, in curved spacetimes, the Heisenberg Picture treats all coordinates on equal ground, while the Schrödinger Picture has, as a precondition, that there be a universal time variable with respect to which states evolve. There may not be or (if there is) it may lead to a fundamental inconsistency. Indeed, the Problem of Time is, itself, is that very inconsistency: a proof by contradiction that the condition is false and that, hence, no Schrödinger Picture exists at all. So, they are inequivalent in that setting.

The formal equivalence of the two pictures also neglects half the foundation of quantum theory itself. There are not one but two von Neumann postulates to consider: the Evolution Postulate (states evolve in accordance with the Schrödinger equation) and the Projection Postulate (a state upon measurement coughs up an eigenvalue and collapses to an eigenstate, in accordance with Born's Rule). It seems everybody keeps forgetting about that other postulate.

The equivalence of the picture only applies to the first postulate. The Heisenberg Picture version of Evolution, of course, being the Heisenberg Equations. There is no equivalence between the two pictures for the second postulate -- because there is no Heisenberg Picture version of the Born Rule at all! If you try to formulate one, you will see revealed an interesting new infrastructure, that is not present in the Schrödinger Picture, but which is required to properly handle multiple applications of the Born Rule in the Heisenberg Picture. Contained in it is a distinguished "now" and a sense of time flowing with respect to it. But the "flow" is not within the Evolution postulate; rather it is stemming from the Projection postulate!

The question of what the Born Rule is and how it is to be handled, interpreted, explained or explained away is the crux of what's called the Measurement Problem. The different answers to this question then produce the different Interpretations of quantum theory (Bohm, Many Minds, Many Worlds, Consistent Histories, Physical Collapse, each of which may be threaded by the analyses provided of Decoherence).

Here, too, there is a gap. The same question asked of the Born Rule is now passed down to each of their putative replacements: what's the Heisenberg Picture version? And is there even one at all? For instance: Many Worlds and Bohm.

Answer 3

Preliminaries Recall that a representation of an algebra on a Hilbert space is a map from the algebra to the bounded operators on a certain Hilbert space. Also recall the Heisenberg canonical commutation relations $$[q_i,p_k]=i\delta_{ik}I$$ A representation of such relations is a set of operators on some Hilbert space that satisfy the same commutation relations. A typical example is the Schrödinger representation, which in one dimension is realised by the multiplication operator $q=M_s$ and the differentiation operator $p=-i\frac{\text d}{\text ds}$ on the Hilbert space $L^2(\mathbb R)$ with Lebesgue measure.

The equivalence of the different pictures is a consequence of von Neumann's uniqueness theorem, which states that every irreducible representation of the Heisenberg's uncertainty relations is unitarily equivalent to the Schrödinger representation. So if you start with Heisenberg's matrix mechanics, i.e. you assume that you have a representation by (infinite) matrices of the canonical commutation relations, then there exists a unitary that "translates" the action of these matrices over some Hilbert space into the action of the Schrödinger operators $q$ and $p=-i\hbar\nabla_q$ on the Hilbert space $L^2(\mathbb R^n)$ with Lebesgue measure. The place where these two different pictures meet is the Dirac notation involving bras and kets. One way of actually proving this is a la Dirac-Dixmier, which involves studying the spectrum of the quantum harmonic oscillator and proving that the Hamiltonian is essentially self-adjoint as a consequence of Nelson's criterion.

The rough idea behind the result is that uniqueness of the Schrödinger's representation follows roughly from the fact that the Weyl algebra stemming from the Weyl's operator is isomorphic to the C*-algebra of compact operators on a infinite-dimensional separable Hilbert space, which is known to only have one class of unitary equivalence of irreducible representations. This can be constructed using the magic positive type function on the Heisenberg group $$\phi(z,t)=e^{-\frac{\Vert z\Vert^2}4+it},\qquad(z,t)\in\mathbb C^n\ltimes\mathbb R.$$

Answer 4

I believe what physicists mean is that the two theories specify the same experimental state space, which is a consequence of the Stone-von Neumann Theorem.

One can ask the same question about Lagrangian vs. Hamiltonian mechanics: Why do physicists say the these theories are equivalent when there are many ways in which they are not? I believe because they do end up assigning the same states to any physical system to which both of them can be applied.