I have a homework problem in nuclear magnetic resonance. After a bunch of calculations, I have arrived at the expression: $$\langle M_1(t)\rangle = {\rm tr}\left [\rho(0)\sigma_+^{(1)}\exp\left(i\frac{Jt}{2}\sigma_z^{(2)}\right)\bigotimes\left(e_\uparrow^{(2)}+e_\downarrow^{(2)}\right)\right]$$

where (1) refers to qubit 1 and (2) refers to qubit 2

$M$ is the magnetization

$\rho$ is the density matrix

$\sigma_{x/y/z}$ is the Pauli matrix

$\sigma_+ = \sigma_ x+ i \sigma_y$

$e_\uparrow=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}; e_\downarrow= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

I was next thinking of writing the above as: $$\langle M_1(t)\rangle = {\rm tr}\left [\rho(0)\sigma_+^{(1)}\exp\left(i\frac{Jt}{2}\sigma_z^{(2)}\right)\bigotimes e_\uparrow^{(2)}+ \rho(0)\sigma_+^{(1)}\exp\left(i\frac{Jt}{2}\sigma_z^{(2)}\right)\bigotimes e_\downarrow^{(2)}\right]$$

But is this mathematically valid?

  • $\begingroup$ Is your trace in this case a real number? The tensor product is bilinear so this would not be valid. $\endgroup$
    – gabe
    Feb 13, 2018 at 20:39
  • $\begingroup$ I assumed so because $M_1$ is the magnetization of qubit $1$ and is a measurable quantity. I updated my question to elaborate on the terms. $\endgroup$
    – elt23
    Feb 13, 2018 at 23:39
  • 1
    $\begingroup$ @gabe: This is valid because the tensor product is bilinear. $\endgroup$
    – WillO
    Nov 14, 2022 at 7:17
  • 2
    $\begingroup$ Downvoted for cruelty to the reader, using about two dozen characters (including superscripts, subscripts, etc etc) for something you could just as well have called $w$. $\endgroup$
    – WillO
    Nov 14, 2022 at 7:22
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    $\begingroup$ WillO has a point. The question could have been just: "is it true that $u\otimes (v+w) = u\otimes v + u\otimes w$?". The answer is then yes, by definition of the tensor product. Look for the definition of the tensor product in terms of the so called universal property. $\endgroup$
    – Gold
    Jul 19, 2023 at 14:41

1 Answer 1


I can't quite tell if your first term is a scalar (since it is a trace) or an operator (since you are using a tensor product). I'll answer for both cases:

For $v \in V$ and $w_1, w_2 \in W$, the tensor product distributes as such: $$ v \otimes (w_1 + w_2) = v \otimes w_1 + v \otimes w_2 $$

So if your trace is an operator, you are correct in your assumption and your value will be in the space $O \otimes E$ where $O$ is the space containing your operator and $V$ is the space containing $e_{\uparrow(\downarrow)}$.

If your trace is a real number, you can omit the tensor product and distribute as you would with scalar multiplication (since that is what you are doing).

Without knowing more context, I suspect yours is the former case as it looks like you might be tracing over the space of one of your qubits (leaving you with a reduced density matrix operator).


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