Let's suppose, that applying $\mathbf{H}$ (Hadamard operator) to the first register of the state $c \cdot \sum_{x}|x\rangle|f(x)\rangle$ ($f$ is a permutation, $c$ is a normalization factor), and measuring the first register gives $|0\rangle$ with some probability $p$.

Now consider the state $c \cdot |y\rangle \sum_{x}|x\rangle|f(x) + g(y)\rangle$, where $|x\rangle$ and $|y\rangle$ are not entangled, $c$ is again a normalization factor. Will it be correct, if I "throw away" $|y\rangle$ and apply the statement above to the second and third register treating $g(y)$ as a random value with some distribution, that is writing the state as $\sum_{x}|x\rangle|f(x) + s\rangle$, where s is random, but fixed.

In other words, am I able to consider only a subspace of the whole space treating values that involve $y$ as random values?

  • $\begingroup$ Welcome to QCSE! What is $c$? Is it just a normalization factor? And also, what exactly is $\mathbf{H}$? Also in the state $$c \cdot \sum_x |y\rangle |x\rangle |f(x) + g(y)\rangle \,,$$ the $|y\rangle$ does not seem to have any dependence on $x$. Then why are you keeping it inside the summation? The question is very confusing. Please add details and clarify things. $\endgroup$
    – FDGod
    Feb 7 at 0:01
  • $\begingroup$ @FDGod $c$ is a normalization factor, $\mathbf{H}$ is Hadamard operator, $|y\rangle$ does not depend on $x$ (kept inside the sum just to write three registers together). Edited. $\endgroup$ Feb 7 at 0:21

2 Answers 2


After throwing the first register away the state of the second and third registers becomes $$|\psi_{23}\rangle=\sum_x|x\rangle|f(x)+s\rangle\tag1$$ where $s$ is a fixed value. More precisely, $s:=g(y)$ where $y$ is whatever was in the first register.

Partial trace

Mathematically, discarding a register corresponds to the partial trace, so to prove $(1)$, we perform the following calculation $$ \begin{align} \rho_{23}&=\mathrm{tr}_1\rho_{123}\tag2\\ &=\mathrm{tr}_1\left[|y\rangle\sum_x|x\rangle|f(x)+g(y)\rangle\langle y|\sum_{x'}\langle x'|\langle f(x')+g(y)|\right]\tag3\\ &=\mathrm{tr}_1\left[|y\rangle\langle y|\otimes\sum_x|x\rangle|f(x)+g(y)\rangle\sum_{x'}\langle x'|\langle f(x')+g(y)|\right]\tag4\\ &=\mathrm{tr}_1\left[|y\rangle\langle y|\right]\cdot\sum_x|x\rangle|f(x)+g(y)\rangle\sum_{x'}\langle x'|\langle f(x')+g(y)|\tag5\\ &=1\cdot|\psi_{23}\rangle\langle\psi_{23}|\tag6 \end{align} $$ where subscripts indicate registers.

The conclusion actually applies more generally. If the register $1$ is not entangled with other registers, then discarding the register $1$ corresponds to merely dropping the corresponding ket from the initial state. Here, register $1$ is not entangled with registers $2$ and $3$, because we can write our initial state $|y\rangle\sum_x|x\rangle|f(x)+g(y)\rangle$ as a product $$ |y\rangle\sum_x|x\rangle|f(x)+g(y)\rangle=|y\rangle\otimes|\phi(y)\rangle\tag7 $$ where $|\phi(y)\rangle=\sum_x|x\rangle|f(x)+g(y)\rangle$.

Randomness of $s$

Note that $s$ does not become random as a result of the partial trace in the calculation above (as it would if the registers were entangled as in for example $\sum_{x,y}|y\rangle|x\rangle|f(x)+g(y)\rangle$). You can treat $s$ as a random variable, but this is a modeling choice, not a necessity arising from quantum theory. For example, $y$ might be unknown or uncertain and you may choose to model that by declaring $y$ to be a random variable. In this case, $s$ is a random variable, too (unless of course $g$ is constant).


If $y$ is a bitstring, meaning $|y\rangle$ is a basis state, yes, you can just throw it away since it is not entangled with the rest of the state.

However, if you had something like $\sum_y|y\rangle\sum_x|x\rangle|f(x)+g(y)\rangle$, this wouldn't work anymore, since the first register would be entangled with the last one.


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