# In Schumacher’s noiseless channel coding theorem, how do we get the exponents in $|0\rangle ^{\otimes n(1−p)/2}|1\rangle ^{\otimes n(1−p)/2}$?

On pg. 55 in Nielsen and Chuang, it's said that:

the $$|0\rangle + |1\rangle$$ product can be well approximated by a superposition of states of the form $$|0\rangle ^{\otimes n(1−p)/2}|1\rangle ^{\otimes n(1−p)/2}$$.

I'm confused about how we get the exponents for $$|0\rangle$$ and $$|1\rangle$$, namely, $$\otimes n(1−p)/2$$.

• I would also mention that the passage you are quoting is in the introduction of the book, it is not meant to be rigorous but just meant to give a flavor of what is to come. I think it would be best to look at the portion of the book dedicated to coding theorems for more information. Oct 14 at 14:35

Forget about the whole $$n(1-p)$$ for a minute.

For simplicity, let $$k$$ be even and think of the product $$\tfrac{(|0\rangle+|1\rangle)}{\sqrt{2}}^{\otimes k}$$ like it's the binomial $$(x+y)^k$$ with commuting indeterminates $$x$$ and $$y$$.

When you expand the binomial $$(x+y)^k$$ the monomial $$x^jy^{k-j}$$ with the largest coefficient i.e. the monomials that appears the most frequent number of times in the expansion before collecting like terms are those with $${k/2}$$ $$x$$'s and $$k/2$$ $$y$$'s. To be exact there are exactly $$k\choose k/2$$ such monomials.

So if someone was going to draw randomly picked monomials from $$(x+y)^N$$ with say $$N$$ really really big the most likely outcome is a monomial with $${N/2}$$ $$x$$'s and $${N/2}$$ $$y$$'s.

This is exactly the same thing that's going on with the product state. When you expand the state $$\frac{(|0\rangle+|1\rangle)}{\sqrt{2}}^{\otimes k}$$ the superposition with the highest probability is one with $$N/2$$ $$|0\rangle$$'s and $$N/2$$ $$|1\rangle$$'s, which after reordering we could write as the product state $$|0\rangle^{\otimes k/2}|1\rangle^{\otimes k/2}$$.

The same thing goes on when $$k$$ is odd since, in the grand scheme of things with say $$k$$ really big, it all comes out in the wash.

Now it should be clear what is going on pg.55 of N&C, by replacing $$k$$ with $$n(1-p)$$ for $$0\leq p\leq 1$$.