Suppose I have two quantum channels $\Phi_1:B(\mathcal{H}_1)\rightarrow B(\mathcal{H}_2)$ and $\Phi_2:B(\mathcal{H}_2)\rightarrow B(\mathcal{H}_3)$, and let $\Phi=\Phi_2\circ \Phi_1$.

Stinespring Dilation says there are two auxiliary systems $\mathcal{K}_1$ and $\mathcal{K}_2$ and two isometries $V_1$ and $V_2$ such that $\Phi_i(\rho) = \text{Tr}_{\mathcal{K}_i}(V_i \rho\otimes \vert 0\rangle\langle 0\vert V_i^\dagger)$, but also that there is an auxiliary system $\mathcal{K}$ and isometry $V$ such that $\Phi(\rho)=\text{Tr}_{\mathcal{K}}(V\rho \otimes \vert 0 \rangle\langle 0\vert V^\dagger)$. We could also dilate the last channel by setting $\mathcal{K}=\mathcal{K}_1\otimes \mathcal{K}_2$ and setting $V = (V_2\otimes I_{\mathcal{K}_2})\circ (V_1\otimes I_{\mathcal{K}_1})$ (generally the Stinespring dilation is non-unique).

That's all mathematically accurate. But I can physically interpret the unitaries in each situation as being the evolution of the system under some global, time-independent Hamiltonian. If I'm a very devout acolyte of the church of the larger Hilbert space, I would expect there is some very large Hamiltonian $H$ such that evolving the system plus environment for time $t_1$ under $H$ will give me $\Phi_1$, and evolving the same system plus same environment for time $t_2$ after that will give me $\Phi_2$.

More formally, the physical predictions of the church of the larger Hilbert space say that there should be an environmental system $\mathcal{K}'$ and Hamiltonian $H'$ such that

$$ \Phi_1(\rho) = \text{Tr}_{\mathcal{K}'}(e^{iH't_1/\hbar}\rho\otimes \vert 0\rangle\langle 0\vert e^{-iH't_1/\hbar})$$


$$ \Phi(\rho) = (\Phi_2\circ \Phi_1)(\rho) = \text{Tr}_{\mathcal{K}'}(e^{iH'(t_1+t_2)/\hbar}\rho\otimes \vert 0\rangle\langle 0\vert e^{-iH'(t_1+t_2)/\hbar })$$

But there doesn't seem to be any guarantee that such objects exist just from Stinespring's dilation. That is, we can find some $H$ and $t$ such that $V = e^{iHt/\hbar}$ (where $V$ is an isometry in the Stinespring dilation of $\Phi$), but I don't think there are any guarantees about $e^{iHt_1/\hbar}$.

Is there any way to prove that there is some Stinespring dilation that satisfies my physical intuition, or is there a counterexample? Some pair of channels where no Hamiltonian for the composed channel induces the first channel at some earlier time?

  • $\begingroup$ so you're asking whether there's $H$ such that you get first $\Phi_1$ and then $\Phi_2$ evolving at different times? I don't get your argument for why we should expect this to exist. The Hamiltonian corrisponding to $\Phi$ needs not have anything in common with those of $\Phi_i$. Note that Hamiltonians generating $V_1 V_2$ will be very different than those generating $V_1$ and $V_2$ individually. Also physically speaking doing a time-independent evolution to get $\Phi_1$ and then another for $\Phi_2$ is very different than doing one to get $\Phi_2\circ\Phi_1$ $\endgroup$
    – glS
    Feb 3, 2023 at 22:46
  • $\begingroup$ Right, but since these unitaries and Hamiltonians are non-unique, maybe we can find such an $H$ in the set of all possible $V_1$, $V_2$, and $V$. Physically speaking, I could regard the entire process as some much much larger quantum evolution (e.g., start including the experimenter as part of the environment system) and in this view there is some nearly-universal Hamiltonian that would produce the evolution that causes $\Phi_1$ to happen and then $\Phi_2$. $\endgroup$
    – Sam Jaques
    Feb 4, 2023 at 13:11
  • $\begingroup$ Your intuition seems wrong. Why should that be the case? $\endgroup$ Feb 5, 2023 at 8:48
  • $\begingroup$ The intuition is that quantum mechanics should be capable of describing the universe completely with unitary evolution from time-independent Hamiltonians. And indeed, Adam Zalcman's answer shows that quantum channels and/or time-dependent Hamiltonians can always be approximated by such a theory. $\endgroup$
    – Sam Jaques
    Feb 6, 2023 at 11:57

1 Answer 1


TL;DR: In general, this cannot be done exactly, because of the relationship between the eigenvectors of $A$ and $e^A$.

That said, the Feynman's clock construction gets pretty close to realizing the desired Hamiltonian. In fact, if post-selection is available, the construction provides an exact realization. Otherwise, Feynman's clock with a ballistic program counter allows us to make the probability of successful realization of the desired Hamiltonian as high as we wish. Perhaps this sheds some light on why the physical intuition expressed in the question works in classical mechanics, but fails in quantum mechanics.

Impossibility of exact solution

We construct a counterexample in which the channels $\Phi$, $\Phi_1$ and $\Phi_2$ are unitary. First, we choose two orthonormal bases $|\psi_k\rangle$ and $|\phi_k\rangle$ in $\mathcal{H}$ whose elements are pairwise distinct under the global phase equivalence, i.e. $|\langle\psi_i|\phi_j\rangle|<1$ for all $i,j$. Then we set $$ V_1=\sum_k|\phi_k\rangle\langle\psi_k|,\quad V_2=\sum_k e^{-ia_k}|\psi_k\rangle\langle\phi_k|\tag1 $$ where $a_k\in\mathbb{R}$ are chosen so that $e^{ia_k}$ are distinct. From $V=V_2V_1$, we have $$ V=\sum_k e^{-ia_k}|\psi_k\rangle\langle\psi_k|\tag2 $$ which means that $|\psi_k\rangle$ are the normalized eigenvectors of $V$.

Now, if $|v\rangle$ is eigenvector of operator $A$ associated with eigenvalue $\lambda$ then $|v\rangle$ is an eigenvector of $e^A=\sum_{k=0}^\infty\frac{A^k} {k!}$ associated with eigenvalue $e^\lambda$. In general the converse may not be true: there may exist $|u\rangle$ that is an eigenvector of $e^A$ but not an eigenvector of $A$. This happens when $|u\rangle$ is a linear combination of eigenvectors of $A$ associated with distinct eigenvalues of $A$ that are mapped by the exponential function to the same eigenvalue of $e^A$. In this case, $e^A$ has repeated eigenvalues.

However, $V$ has distinct eigenvalues, so if $H'$ is any Hamiltonian such that $V=e^{-iH't/\hbar}$ then $|v\rangle$ is an eigenvector of $V$ if and only if it is an eigenvector of $H'$. Consequently, all invariant subspaces of $H'$ are one-dimensional and are spanned by the vectors $|\psi_k\rangle$. But, by the choice of $|\phi_k\rangle$, the vectors $|\psi_k\rangle$ are not eigenvectors of $V_1$ and we conclude that $V_1\ne e^{-iH't_1/\hbar}$.

Feynman's clock and post-selection

We are now going to do what classical CPU does: use a program counter to orchestrate orderly execution of a sequence of discrete instructions. The program counter is an extra auxiliary system $\mathcal{C}$ that we adjoin to $\mathcal{H}\otimes\mathcal{K}$ and the Feynman's clock Hamiltonian acts on it using creation and annihilation operators to execute discrete instructions within a continuous time evolution.

We will think of $\mathcal{C}$ as a particle hopping between sites in a one-dimensional lattice (for now we only need three sites of the lattice). Let $c_i^\dagger$ and $c_i$ denote the creation and annihilation operators associated with site $i$ and define interaction strengths $g_1=\hbar/t_1$ and $g_2=\hbar/t_2$. Then, the Feynman's clock Hamiltonian $H'$ acting on $\mathcal{H}\otimes\mathcal{K}\otimes\mathcal{C}$ is $$ H'=g_1V_1c_1^\dagger c_0+g_2V_2c_2^\dagger c_1+\text{h.c.}\tag4 $$ Note that $H'$ conserves the number of particles in the program counter. We assume there is exactly one such particle and that at $t=0$ it is located in site $0$. If at time $t=t_1$ we find the particle in site $1$ then $H'$ will have applied $V_1$ to $\mathcal{H}\otimes\mathcal{K}$ subsystem. Similarly, if at $t=t_1+t_2$ we observe the particle in site $2$ then $H'$ will have applied $V_2V_1$ to $\mathcal{H}\otimes\mathcal{K}$ subsystem. Note that the hermitian conjugate terms in $(4)$ hide the fact that the particle might propagate in both directions. Thus, if at $t=t_1$ or $t=t_1+t_2$ we observe the particle back in site $0$ then the unitaries on $\mathcal{H}\otimes\mathcal{K}$ will have been undone. This is why post-selection is needed: the system has applied $V_2V_1$ to $\mathcal{H}\otimes\mathcal{K}$ only if we observe the program counter particle in site $2$.

Ballistic program counter

We can boost the probability of that happening arbitrarily close to one by turning the program counter into a ballistic particle. To that end, we extend the lattice in both directions$^1$ and create a wave packet propagating from the sites with negative indices in the positive direction. After sufficient time has passed the probability of the program counter particle to be in a site $i$ with $i\geqslant 2$ will be arbitrarily close to one. Thus, the probability that our system will have aplied $V_2V_1$ to $\mathcal{H}\otimes\mathcal{K}$ can be made as close to one as we wish. We cannot achieve deterministic success this way since it would require a wave packet with compact support in both position and momentum space.

I think this construction can be seen as realizing the physical intuition expressed in the question.

$^1$ We also need to add appropriate hopping terms to $H'$.

  • $\begingroup$ Thanks, I stewed on this for a few days and was constructing Fourier series of step functions before realizing I just had a less elegant version of Feynman's Hamiltonian. Two questions: (1) I'm not sure your counterexample works from a quantum channel perspective, since I have freedom to choose the unitaries. That is, could I choose some unitary that would make all the eigenvalues work out? But I think the conclusion still holds and an easier but less rigorous argument is to take a channel induced by evolution of a non-differentiable time-dependent Hamiltonian. $\endgroup$
    – Sam Jaques
    Feb 6, 2023 at 11:52
  • $\begingroup$ (2) I think the Feynman Hamiltonian is $H'=g_1V_1c_1^\dagger c_0 + g_2V_2 c_2^\dagger c_1 + \text{h.c.}$. I suppose to ensure that $t_1$ and $t_2$ match the desired values, the "momentum" should be fairly narrowly concentrated around a particular value, and to ensure the program counter has a definite position, so should position. $\endgroup$
    – Sam Jaques
    Feb 6, 2023 at 11:55
  • 1
    $\begingroup$ Re (1): You're right. I amended the text to point out that the channels constructed in the counterexample are unitary. I think it might be possible to make a more general construction using the fact that the different $V$ that we can use to realize $\Phi$ under Stinespring dilation differ by a unitary on $\mathcal{K}$. $\endgroup$ Feb 7, 2023 at 21:25
  • $\begingroup$ Re (2): $H'=g_1V_1c_1^\dagger c_0+\dots$ may not be hermitian. Yes, in the ballistic case we want to use a wavepacket whose position and momentum are fairly well defined. The closer it is to a classical particle the better since then the amplitude for alternative processes (where $V_2V_1$ hasn't been applied) is vanishingly small. $\endgroup$ Feb 7, 2023 at 21:28
  • 1
    $\begingroup$ $H'=g_1V_1\otimes c_1^\dagger c_0 + \text{h.c.} = g_1V_1 \otimes c_1^\dagger c_0 + g_1V_1^\dagger \otimes c_0^\dagger c_1$ is Hermitian. $\endgroup$
    – Sam Jaques
    Feb 8, 2023 at 10:40

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