# Confused regarding norm and why is $\langle a|b\rangle\langle b|a\rangle = ||\langle a|b\rangle||^2$

Given $$\langle v|v\rangle=\sum_{i}a_{i}^2$$ then $$|||v\rangle||=\sqrt{\langle v|v\rangle}=\sqrt{\sum_{i}|a_{i}|^2}$$.

Also, $$\langle a|b\rangle=\langle b|a\rangle^{*}$$.

So how then is $$\langle a|b\rangle\langle b|a\rangle=||\langle a|b\rangle||^{2}=(\sqrt{\sum_{i}a_{i}^{*}b_{i}})^{2}$$?

I think I have misunderstood what the norm is.

The only way I can see the above working is if $$||\langle a|b\rangle||^{2}=(\sqrt{\sum_{i}a_{i}^{*}b_{i}})^{2}=\sqrt{\langle a|b\rangle\langle b|a\rangle}$$

If this is the case, then I believe I have gotten confused regarding what the norm is actually doing, as I normally only come across it for either complex numbers or just single vectors. If the above does hold, then I take it to mean I need to take the sum of the product of the coefficients and their conjugate for vectors and complex numbers, but for inner products I must take the product of the innerproduct and it's transpose?

• You understand that the norm is a function on vectors not pairs of vectors right? Jul 30, 2020 at 13:45
• It feels a little like you might be conflating norms and inner products with each other, not understanding they do different things. You also might be conflating the modulus in $\mathbb C$ with the norm in $\mathbb C^n$. Jul 30, 2020 at 13:50
• I think you are confusing the definition of a norm on your inner product space and the norm on $\mathbb{C}$ (by your question I'm assuming your vector space is over $\mathbb{C}$. Note that $z = \langle a | b \rangle$ is just a complex number and so $\| \langle a | b \rangle \|^2 = \| z\|^2 = z z^* = \langle a | b \rangle \langle b | a \rangle$. Jul 30, 2020 at 13:52
• Yeah I forgot that to get the conjugate all I need do is swap the bras and kets. For some reason I convinced myself that doing was getting the conjugate transpose :/ Jul 30, 2020 at 13:55
• @rschwieb what is the difference between the norm and the modulus? Generally I think I mix them up is due to notation. Jul 30, 2020 at 13:58

Given the inner product $$\langle a | b \rangle = \sum_ia^*_ib_i$$ we have $$\langle b | a \rangle = \langle a | b \rangle^* = \sum_ia_ib^*_i$$ then \begin{align} |\langle a | b \rangle|^2 &=\langle b | a \rangle \langle a | b \rangle = \langle a | b \rangle^*\langle a | b \rangle = \\ &= \sum_i a_i b^*_i \sum_j a^*_j b_j = \left( \sum_i a^*_i b_i \right)^* \sum_j a^*_j b_j = \left|\sum_j a^*_j b_j\right|^2 \end{align}
• @GaussStrife also the inner product is wrong, it should be $\langle a|b\rangle=\sum_{i}a^*_{i}b_{i}$ Jul 30, 2020 at 13:17