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?

  • $\begingroup$ You understand that the norm is a function on vectors not pairs of vectors right? $\endgroup$
    – rschwieb
    Jul 30, 2020 at 13:45
  • $\begingroup$ 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 $\mathbb C^n$. $\endgroup$
    – rschwieb
    Jul 30, 2020 at 13:50
  • 1
    $\begingroup$ 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 $. $\endgroup$
    – Rammus
    Jul 30, 2020 at 13:52
  • $\begingroup$ 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 :/ $\endgroup$ Jul 30, 2020 at 13:55
  • $\begingroup$ @rschwieb what is the difference between the norm and the modulus? Generally I think I mix them up is due to notation. $\endgroup$ Jul 30, 2020 at 13:58

1 Answer 1


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}

  • $\begingroup$ Sorry, fixed that in the question $\endgroup$ Jul 30, 2020 at 13:14
  • $\begingroup$ @GaussStrife also the inner product is wrong, it should be $\langle a|b\rangle=\sum_{i}a^*_{i}b_{i}$ $\endgroup$ Jul 30, 2020 at 13:17
  • $\begingroup$ yes I have made those syntax corrections $\endgroup$ Jul 30, 2020 at 13:26
  • $\begingroup$ Yeah just got there myself. Always slips my mind that the bra is the conjugate transpose of the ket. $\endgroup$ Jul 30, 2020 at 13:52

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