Find inner product of two states given inner product of an orthogonal state

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!

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.)

• From the first condition, $\langle i|\psi\rangle=\frac{1}{\sqrt{2}}$, $\langle i|j\rangle=0$ is always hold. – Yitian Wang Dec 7 '20 at 7:22
• Hi @Adam, could you please comment on how did you arrive at the quantity <i|phi> = e^(i theta)/sqrt(2)? Thanks! – QuestionEverything Dec 17 '20 at 18:59
• $\langle i|\phi\rangle = \frac{e^{i\theta}}{\sqrt{2}} (\langle i|i\rangle - \langle i|j\rangle) = \frac{e^{i\theta}}{\sqrt{2}} (1 - 0) = \frac{e^{i\theta}}{\sqrt{2}}$. – Adam Zalcman Dec 17 '20 at 19:11
• 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. – Adam Zalcman Dec 17 '20 at 19:55
• 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. – Adam Zalcman Dec 17 '20 at 20:02