My question concerns the following theorem (not to be confused with quantum teleportation protocols): what is the exact mathematical content of this theorem? After all any (pure) quantum state (say in $\mathbb{C}^2$) is determined by a unit vector (up to phase factor) so it is clear that any finite amount of classical bits (zeros and ones) is insufficient to determine it completely. Stated in this way, mathematically this is trivial: more problematic is the infinite case (an infinite sequence of bits is enough to determine two complex numbers). Of course I understand that the interest in this theorem has to do with applications and so on but nevertheless I'm curious about purely mathematical formulation (and proof as well!) of no teleportation theorem.

  • $\begingroup$ I'd even put the apparent triviality somewhere else: sure you cannot describe an arbitrary quantum state by a finite number of bits. But it's even stronger than that. If I give you a quantum state, how much classical information can you get about it? Basically the best you can do is a projective measurement that returns 1 bit of information. $\endgroup$
    – DaftWullie
    Commented Jul 12, 2023 at 13:56
  • $\begingroup$ In some hand wavy sense, if you had an (continuum-many) infinite number of copies of your state, you could perform all the possible measurements an infinite number of times, deduce the probabilities from the law of large numbers. But this is very hand-wavy :p $\endgroup$
    – Plop
    Commented Jul 14, 2023 at 10:17
  • $\begingroup$ I believe the no-cloning theorem essentially says that if we have a wave function $|\Psi\rangle = a|0\rangle + b|1\rangle$, then there is just no way we can find out what exactly is $a$ and $b$. So, the proof by counter example would be to find and show an algorithm that can coax $a$ and $b$ out of this wavefunction... $\endgroup$
    – James
    Commented Jul 14, 2023 at 20:13

1 Answer 1


Disclaimer: I am just a mathematician that likes to try to understand the math behind quantum things. I am very bad at physics.

As far as I understand the link you provide, I think a mathematical formulation of the theorem is as follows.

Let $\mathcal{H}$ be a (finite-dimensional) Hilbert space and $O$ be an observable on $\mathcal{H}$. Let $f :\mathbb{R} \rightarrow \mathfrak{S}(\mathcal{H})$ (where $\mathfrak{S}(\mathcal{H})$ designates the set of states on $\mathcal{H})$. Let $\rho$ be a state. Let us measure $O$ on it. We get a result $x$, and we consider the state $\rho' := f(x)$.

A map $f$ is a teleportation map if for all state $\rho$, then the probability that $\rho' = \rho$ is $1$ (that is, outcomes $x$ such that $f(x) \neq \rho$ appear with $0$ probability).

The theorem asserts that there is no teleportation map.

One could generalize: taking observables $O_1,\cdots, O_n$, a map $f: \mathbb{R}^n \rightarrow \mathfrak{S}(\mathcal{H})$ would be called a teleportation if, whenever $x_1,\cdots, x_n$ appear as the result of measuring $O_1 \otimes Id \otimes \cdots \otimes Id ,\cdots, Id \otimes \cdots \otimes Id \otimes O_n$ on the state $\rho \otimes \cdots \otimes \rho$, then $f(x_1,\cdots,x_n) = \rho$. Then such a generalized teleportation map does not exist.

A word about the proof: there is an infinite number of states, but there is only a finite number of outcomes of an observable, so most of states won't be attained by any $f$.

A recap on probabilities of outcomes of measurements of observables.

If $\rho$ is a state, i.e. a trace one selfadjoint operator with nonnegative eigenvalues, and if $O$ is an observable, i.e. a selfadjoint operator, then for all eigenvalue $x$ of $O$, if $P_x$ denotes the orthogonal projection on the eigenspace associated to $x$, then we say that « the probability of finding outcome $x$ after performing the measurement associated to $O$ on a system whose state is represented by $\rho$ is $Tr(\rho P_x)$ ». Now, of course, making sense of this sentence is an open problem in the foundations of quantum mechanics, and it isn’t our concern here. With this, you should be able to explicitly write down the probability I am talking about. If you don’t, I will try to add details.

PS: I don’t really know how to state the theorem for an infinite-dimensional Hilbert space.

  • $\begingroup$ Thank you for your answer! Still I have some problems: 1. When you are saying that we are measuring the observable $O$ in the state $\rho$ what exactly do you mean? Are you taking $Tr(\rho O)$? If yes, then it is not true that there are only finitely many outcomes. Maybe you would prefer to take eigenvalue of $O$-but in this case I don't know how this can be formalized as a map (since there is no mapping corresponding to the measurment outcome)... $\endgroup$
    – truebaran
    Commented Jul 13, 2023 at 17:26
  • $\begingroup$ 2. Moreover I don't understand to which probability are you referring to: if measurment outcome $x$ is meant to be $Tr(\rho O)$ then probably you are interested in the measure on $\mathfrak{S}({\mathcal{H}})$ but what probability measure you have in mind then? If $x$ is obtained by just taking some eigenvalue of an observable $O$ then I don't know where our state $\rho$ enters. The third option, that your measure is defined on $\mathbb{R}$ also seems problematic since after all our $x$ is somehow defined with the help of $\rho$ and $O$ so this probability... $\endgroup$
    – truebaran
    Commented Jul 13, 2023 at 17:37
  • $\begingroup$ ... measure should come as a transport of a probability measure on the set of states-and again we need to take some probability measure on $\mathfrak{S}(\mathcal{H})$. 3. There is some indication on wikipedia that this theorem is also valid in the infinite dimensional case. I suppose that the crucial thing for a proof is just cardinality comparison so one have to assume that the set of possible outcomes of $O$ should be countable-so in other words, we exclude cases when $O$ is allowed to have continuous spectrum $\endgroup$
    – truebaran
    Commented Jul 13, 2023 at 17:42
  • $\begingroup$ I would be very grateful if you could clarify these issues! Thank you! $\endgroup$
    – truebaran
    Commented Jul 13, 2023 at 17:43
  • 1
    $\begingroup$ Thank you, now everything seems to work just fine! $\endgroup$
    – truebaran
    Commented Jul 15, 2023 at 8:29

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