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Let $\{p_j\}_j$ be a set of probabilities, $\sum_j p_j = 1$, let $\{h_j\}_j$ be a set of $n \times n$ Hermitian matrices, and define $ad_h(A) $ be the adjoint.

Define the following linear mapping $$ E(A) = \sum_j p_j e^{-i \ ad_{h_j} }(A). $$

Does $E$ have a trivial kernel (and is therefore invertible)?

Intuitively speaking, it seems like $E$ should have a trivial kernel. In particular, since $e^{-i \ ad_{h_j} }(A) $ corresponds to conjugation by a unitary matrix, then it has a trivial kernel. Then a probabilistic implementation of this also seems like it should have trivial kernel. However, I'm not quite sure how to prove this (if it is true).

If it is not true for general $A$, can we make true by restricting it to some subset of $A$. For example I am interested in the case where $A$ represents a quantum state, and hence $tr[A]=1$ and $A^\dagger = A$.

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    $\begingroup$ Thanks for your answer! I'm not sure I understand your counter example. If $A\in \mathbb{R}$, then surely the linear mapping in your example is E(A) = (1/2)*(-1)A(-1) + (1/2)*(1)A(1) = A? But I feel like I've misunderstood you here? $\endgroup$
    – Hans Schmuber
    Commented Aug 25 at 20:59
  • $\begingroup$ From your answer, I can see that if $n=2$, $A=((0,1)(0,0))$ then it is possible to find unitaries with 1 and -1 eigenvalues, then it looks like it has a non-trivial kernel. $\endgroup$
    – Hans Schmuber
    Commented Aug 25 at 21:23
  • $\begingroup$ If I restrict to the invariant subspace of Hermitian operators, then it looks like this has a trivial kernel, right? $\endgroup$
    – Hans Schmuber
    Commented Aug 25 at 21:24
  • $\begingroup$ Is this the same as asking whether the expectation of a random unitary matrix invertible? (Where "random" does not mean uniformly distributed, but just means having some probability distribution.) $\endgroup$
    – Michael Hardy
    Commented Aug 26 at 18:58

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This is not true, even for the case where $A$ is a quantum state. Notation-wise, I will eschew exponentials of $\text{ad}$ and instead write the unitary transformations explicitly. I will also use the notation $\rho$ for a quantum state, instead of $A$.

There is a nice counterexample from physics known as the depolarizing channel. Consider a single-qubit quantum state $\rho$. Apply one of the four Pauli matrices $I$, $\sigma^x$, $\sigma^y$, $\sigma^z$ with equal probability. Note that these are all Hermitian and unitary matrices.

Then $\rho \to \frac{1}{4}(\rho + \sigma^x \rho \sigma^x + \sigma^y \rho \sigma^y + \sigma^z \rho \sigma^z)$

However, as you can check, the output commutes with the Pauli matrices and so must be proportional to the identity (just Schur's lemma).

That is, $\rho \to \frac{1}{2} I$ for all normalized quantum states $\rho$. (More generally, this transformation would send a generic single-qubit operator $A$ to $\frac{\text{Tr}[A]}{2} I$.)

In particular, this is not an invertible transformation, since all quantum states $\rho$ are mapped to the identity.

I have an additional comment. It's clear that the transformation above is not invertible, but the transformation cannot send any quantum states to $0$! By taking the trace of some generic $E(A)$ made by probabilistic linear combinations of conjugating by unitaries, it's easy to see that the trace of the matrix $A$ is preserved, so if $A$ is not traceless, $E(A)$ cannot be the zero matrix.

Please note that the set of operators corresponding to quantum states is only a subset and not a linear subspace of the set of operators. I believe you're making a mistake when identifying lack of a kernel on this subset of operators with invertibility on this subset of operators. From the depolarizing channel example above, it's clear that one can have $E(\rho) \neq 0$ for all quantum states $\rho$, even when $E$ is not invertible.

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  • $\begingroup$ Ah, this answer is very helpful! Thank you! I'm particularly interested in the case where we can write $E_t = \sum_j p_j e^{-it ad_{h_j}}$ as a semi-group $E_t = e^{tG}$ for some generator $G$. I was hoping this could be proved by the existence of the matrix logarithm , but if $E_t$ is not generally invertible it seems this proof method must fail as the matrix log may not exist. $\endgroup$
    – Hans Schmuber
    Commented Aug 25 at 22:12
  • $\begingroup$ @HansSchmuber Glad to hear it! I might be misinterpreting your case of interest; is it whether you can generically find such a $G$? I worry that your $E_t$ cannot generically be represented with such a simple time-dependence. $\endgroup$
    – user196574
    Commented Aug 25 at 23:30
  • $\begingroup$ @HansSchmuber For example, we can consider the case where you have $A \to E_t(A) = \frac{1}{4} \sum_{\alpha = 0}^4 e^{-i \sigma^\alpha t} A e^{i \sigma^\alpha t}$. When $t=0$, this returns $A$. This $E_t$ is not generically the depolarizing channel; however, when $t=\pi/2$, this recovers the fully depolarizing channel in my answer. The fully depolarizing channel causes traceless matrices to vanish, but $e^{tG}$ shouldn't be able to cause any nonzero matrix to vanish at finite $t$ if $G$ is well-defined. $\endgroup$
    – user196574
    Commented Aug 25 at 23:44
  • $\begingroup$ (Small typo, $\alpha$ should run from $0$ to $3$; i.e. $\sigma^0 = I, \sigma^1 = \sigma^x, \sigma^2 = \sigma^y, \sigma^3 = \sigma^z$.) Let me know if the reasoning is clear; the $E_t$ above is smooth in $t$, and comparing its behavior near $t=0$ and $t=\pi/2$ should rule out a representation of the form $E_t = e^{tG}$. $\endgroup$
    – user196574
    Commented Aug 26 at 0:00
  • $\begingroup$ Sorry, I wasn't very clear! I had intended G to be a function of t in general. So where G appears above I really mean G(t). So the most general case I'm interested in would be to write $E_t = e^{tG(t)}$, provided $t$ is sufficiently small but finite. Naturally, the first thing one might try to do is take logs, this seems to run into problems as discussed above. Thank you again for your help! $\endgroup$
    – Hans Schmuber
    Commented Aug 26 at 2:42
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Suppose the values of a discrete random variable are unitary matrices. I'm not sure I've correctly understood the question, but it looks as if maybe it is whether the expected value of such a random variable is necessarily an invertible matrix.

Consider $$\left[ \begin{array}{rr} \cos(2\pi N/n), & -\sin(2\pi N/n) \\ \sin(2\pi N/n), & \cos(2\pi N/n) \end{array} \right]$$ where $N$ is uniformly distributed in the set $\{\,0,\ldots,n-1\,\}.$

The expectation of that is the $2\times2$ zero matrix.

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  • $\begingroup$ I think you might be misunderstanding the question. The question is whether the map $A \to \sum_{i=1} p_i U_i A U_i^\dagger$ is always invertible for unitary $U_i$. (I use $^\dagger$ for Hermitian conjugate and $^*$ for complex conjugation.) As I show in my answer, it is not always invertible. Phrased a different way, the question is not about averages of $U_i$ but rather about averages of $U_i^* \otimes U_i$. I believe your answer is about averages of $U_i$. $\endgroup$
    – user196574
    Commented Aug 27 at 0:00

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