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I have a pure quantum state $|i\rangle$ and another state $|\psi\rangle = \frac{1}{\sqrt{2}}(|i\rangle + |j\rangle)$. A state orthogonal to $|\psi\rangle$ is $|\phi\rangle$. Among these states, I know the following:

$$ \langle i | \psi \rangle = \frac{1}{\sqrt{2}} \\ \langle \phi | \psi \rangle = 0. \\ $$ Then, what can I say about the inner product of $|i\rangle$ and $|\phi\rangle$? I.e., is there a way to find:

$$ \langle i | \phi \rangle $$. Thanks!

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The answer depends on the dimension $d$ of the Hilbert space.

If $d = 2$ then $|\phi\rangle = \frac{e^{i\theta}}{\sqrt{2}} (|i\rangle - |j\rangle)$ for some $\theta \in [0, 2\pi)$ and so $\langle i | \phi \rangle = \frac{e^{i\theta}}{\sqrt{2}}$. In other words, it can be any complex number of absolute value $\frac{1}{\sqrt{2}}$.

If $d > 2$ then $|\phi\rangle = \frac{a}{\sqrt{2}} (|i\rangle - |j\rangle) + b|k\rangle$ for some $a, b \in \mathbb{C}$ such that $a^2 + b^2 = 1$ and any $|k\rangle$ orthogonal to $|i\rangle$ and $|j\rangle$. Therefore, $\langle i | \phi \rangle = \frac{a}{\sqrt{2}}$. In other words, it can be any complex number whose absolute value is in $[0, \frac{1}{\sqrt{2}}]$.

(I assume that $\langle i|j \rangle = 0$ and all kets are normalized.)

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    $\begingroup$ From the first condition, $\langle i|\psi\rangle=\frac{1}{\sqrt{2}}$, $\langle i|j\rangle=0$ is always hold. $\endgroup$
    – Yitian Wang
    Dec 7, 2020 at 7:22
  • $\begingroup$ Hi @Adam, could you please comment on how did you arrive at the quantity <i|phi> = e^(i theta)/sqrt(2)? Thanks! $\endgroup$
    – QuestionEverything
    Dec 17, 2020 at 18:59
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    $\begingroup$ $\langle i|\phi\rangle = \frac{e^{i\theta}}{\sqrt{2}} (\langle i|i\rangle - \langle i|j\rangle) = \frac{e^{i\theta}}{\sqrt{2}} (1 - 0) = \frac{e^{i\theta}}{\sqrt{2}}$. $\endgroup$
    – Adam Zalcman
    Dec 17, 2020 at 19:11
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    $\begingroup$ Complete $|i\rangle, |j\rangle$ to an orthogonal basis and expand $|\phi\rangle$ in this basis. Let $U = \mathrm{span}(|i\rangle, |j\rangle)$. Collect the terms of the expansion into two groups: $|\phi_U\rangle \in U$ and $|\phi_{U^\perp}\rangle \in U^\perp$. Note that $\langle\psi|\phi_{U^\perp}\rangle = 0$. Therefore, $\langle\psi|\phi_U\rangle = 0$. The form of $|\phi_U\rangle$ now follows from the case $d=2$. The form of $|\phi_{U^\perp}\rangle$ can be taken to be a simple multiple of a basis vector due to the freedom we have in completing $|i\rangle, |j\rangle$ to an orthonormal basis. $\endgroup$
    – Adam Zalcman
    Dec 17, 2020 at 19:55
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    $\begingroup$ Note that the above construction is valid up to normalization and relative phase of $|\phi_U\rangle$ and $\phi_{U^\perp}\rangle$. The constants $a$ and $b$ take care of normalization and relative phase. $\endgroup$
    – Adam Zalcman
    Dec 17, 2020 at 20:02

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