I am reading this book “Quantum Computing Explained” by David McMahon. I found the following statement on page 74

Note that the order of the tensor product is not relevant, meaning $|\phi\rangle \otimes |\chi\rangle = |\chi\rangle \otimes |\phi\rangle$

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I found it surprising to face this in a quantum computing book, because this is not true. Have any of you read the book or have any idea how it could be true? (Though as I said, I don’t think it is true).

  • $\begingroup$ @MarkS It's not only about the question. It's the error in a popular book that was concerning and I strongly believe this question is serving some value to the community providing information regarding a book which showed this information. $\endgroup$
    – user27286
    Feb 24, 2021 at 23:40
  • $\begingroup$ I myself haven't read the book in question but it has been mentioned a couple of times on this site, see, e.g., here. $\endgroup$
    – Mark S
    Feb 24, 2021 at 23:48
  • $\begingroup$ @MarkS I have changed the way I wrote things. So the book has many errors. So sad I have studied this one for so long. I am scared now. $\endgroup$
    – user27286
    Feb 24, 2021 at 23:53
  • $\begingroup$ @MarkS I do think it is important to keep easily findable errata lists about knowledge books, because that can comfuse many people. I find this initiative very commendable $\endgroup$ Feb 25, 2021 at 8:17

1 Answer 1


You are right. The statement is wrong. Tensor product is not commutative. In fact,

$$ |\phi\rangle\otimes|\chi\rangle \ne |\chi\rangle\otimes|\phi\rangle $$

unless $|\phi\rangle = c|\chi\rangle$ for some scalar $c$ and so the order of the factors certainly is relevant.

Moreover, if the first and second systems are associated with different Hilbert spaces then one of the expressions may fail to denote a valid state. For example, if the first system is a qubit and the second system is a qutrit then $|1\rangle\otimes|2\rangle$ is a valid state, but $|2\rangle\otimes|1\rangle$ is not.

I suppose what the author might have in mind is that the order is arbitrary. In other words, when describing two systems $A$ and $B$ we can choose to write the state of $A$ first and the state of $B$ second or we can choose the opposite order. Once we have chosen the order we must stick to it. In particular, changing the order in the middle of calculations or in an expression such as $|\phi\rangle\otimes|\chi\rangle =|\chi\rangle\otimes|\phi\rangle$ is an error.

Mathematically, the arbitrariness of the choice above reflects the fact that $\mathcal{H}_A\otimes\mathcal{H}_B$ is isomorphic to $\mathcal{H}_B\otimes\mathcal{H}_A$ for any two vector spaces $\mathcal{H}_A$ and $\mathcal{H}_B$.

Note that sometimes people use subscripts rather than position to indicate subsystems as in $|\phi_A\chi_B\rangle$. In this case, changing the order is not strictly speaking an error, but even then it can surprise and mislead the reader. In any case, the author is not using subscript labels.

Another way this could be true is if the author was talking about the symmetric tensor product $\odot$ which does have the property that

$$ \phi\odot\chi = \chi\odot\phi $$

for all $\phi, \chi$ in some vector space $V$. Again, this does not appear to be the case since the author uses standard tensor product symbol $\otimes$, not $\odot$.

  • 1
    $\begingroup$ Scary that a book that is this popular is holding these kinds of wrong things. Beginners would get really get bad concepts. Thanks for your help. $\endgroup$
    – user27286
    Feb 24, 2021 at 23:35
  • 2
    $\begingroup$ Indeed. Hopefully, confused readers will find your question! Also, I encourage you to send an email to the author so that the next edition is fixed :-) $\endgroup$ Feb 24, 2021 at 23:38
  • 1
    $\begingroup$ I will do that. Thanks Adam, $\endgroup$
    – user27286
    Feb 24, 2021 at 23:38

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