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Say I have a quantum state $|\psi\rangle$ written as a linear combination of some basis vectors $ \{ | \varphi \rangle _i \}_{i \in \mathbb N}$ of $\scr H$.

Goal. Rewrite $|\psi\rangle$ in terms of another basis $ \{ | \phi \rangle _i \}_{i \in \mathbb N}$ of $\scr H$.

I can apply a unitary operator $U$, in more ways.

  1. Passive transformation (AKA keep fixed the state and transform the basis).

$ \displaystyle | \psi_\varphi \rangle = \sum_i f_i |\varphi_i\rangle \longrightarrow \displaystyle | \psi_\phi \rangle = \sum_i g_i |\phi_i\rangle $, in the transformed basis $|\phi_i\rangle = U | \varphi_i \rangle $. The problem is to find these new coefficients. The $j$-th coefficient $g_j$ can be found by projecting the state $|\psi_\phi \rangle$ over its $j$-th component $|\phi_j\rangle$:

$\displaystyle \langle \phi_j | \psi\rangle = \langle \phi_j |\sum_i g_i |\phi_i\rangle = \sum_i g_i \delta_{ji} = g_j$

Let's repeat the same procedure with $|\psi_\varphi \rangle$, since the two formulations must be equal:

$\displaystyle \langle \phi_j | \psi\rangle = \langle \phi_j |\sum_i f_i |\varphi_i\rangle = \sum_i f_i \langle \phi_j |\varphi_i\rangle = \sum_i f_i \langle \varphi_j | U^\dagger |\varphi_i\rangle = \sum_i f_i \langle \varphi_i | U |\varphi_j\rangle^* = \sum_i f_i U^{*}_{ij}$

That means:

$$\boxed {\displaystyle g_j = \sum_i f_i U^*_{ij} }$$

  1. Active transformation (AKA keep fixed the basis and transform the state).

$ \displaystyle | \psi \rangle = \sum_i f_i |\varphi_i\rangle \longrightarrow \displaystyle | U \psi \rangle = \sum_i h_i |\varphi_i\rangle $, in the transformed state $U | \psi \rangle $.

As in 1., the $j$-th coefficient $h_j$ can be found projecting the state $|U \psi⟩$ over its j-th component $|\varphi_j⟩$:

$\displaystyle ⟨\varphi_j|Uψ⟩=⟨\varphi_j|\sum_i h_i|\varphi_i⟩=\sum_ih_i\delta_{ji}=h_j$

On the other hand:

$\displaystyle \langle \varphi_j | U \psi\rangle = \langle \varphi_j |\sum_i f_i U |\varphi_i\rangle = \sum_i f_i \langle \varphi_j | U |\varphi_i\rangle = \sum_i f_i U_{ji}$

We hence conclude:

$$\boxed {\displaystyle h_j = \sum_i f_i U_{ji} }$$


Problem. $U_{ji} = U^*_{ij}$ means $U = U^\dagger$, but $U$ is supposed to be unitary, not self-adjoint. Hence the two transformations are not equivalent, even though they should be. What did I do wrong?

Addendum. The only way I found to obtain $h_j = g_j$ is to swap $U$ with $U^\dagger$ in the passive transformation (1.). That means writing $U |\phi_i\rangle =| \varphi_i \rangle $ instead of $|\phi_i\rangle = U | \varphi_i \rangle $, but that's a bit incoherent. I start with the basis $\{| \varphi_i \rangle\}$, so I should apply $U$ to what I start with.

Solution. Actually, the change-of-basis formula prescribes exactly the opposite. Citing Wikipedia:

Such a conversion results from the change-of-basis formula which expresses the coordinates relative to one basis in terms of coordinates relative to the other basis. Using matrices, this formula can be written ${\displaystyle \mathbf {x} _{\mathrm {old} }=A\,\mathbf {x} _{\mathrm {new} },}$ where $\operatorname{old}$ and $\operatorname{new}$ refer respectively to the firstly defined basis and the other basis, ${\displaystyle \mathbf {x} _{\mathrm {old} }}$ and ${\displaystyle \mathbf {x} _{\mathrm {new} }}$ are the column vectors of the coordinates of the same vector on the two bases, and ${\displaystyle A}$ is the change-of-basis matrix (also called transition matrix), which is the matrix whose columns are the coordinate vectors of the new basis vectors on the old basis.

In this context $| \varphi _i\rangle $ plays the role of $\operatorname{old}$, so I should have written $| \varphi _i\rangle = U |\phi_i \rangle$ in (1.).

I'm sorry for my blunder. Thank you all.

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I think you are mixing up what $f,g,h$ represent. I think you actually derived this:

$$g_j = \sum_i f_i U_{ij}^*$$ $$f_i = \sum_j g_j U_{ij}$$ which implies

$$f_i = \sum_i \sum_j f_i U_{ij}^*U_{ij}$$ which suggests $U^*U = I$ as desired.

Here is the fixed proof. Let $\mid \theta \rangle$ represent some arbitrary state in a Hilbert space. Note that $\mid \theta \rangle$ is just a function in the space and is not defined relative to any basis. Let $\{\mid \psi_i \rangle: i \in \mathbb{N} \}$ and $\{\mid \phi_i \rangle: i \in \mathbb{N} \}$ be two orthonormal basis' of the Hilbert space. Then, we can write $$\mid \theta \rangle = \sum_i f_i \mid \psi_i \rangle$$ and $$\mid \theta \rangle = \sum_i g_i \mid \phi_i \rangle$$ for two different sets of basis coefficients $\{g_i: i \in \mathbb{N}\}$ and $\{f_i: i \in \mathbb{N}\}$.

Since the basis' are orthonormal, we have $$f_j = \langle \psi_j \mid \theta\rangle$$ and $$g_j = \langle \phi_j \mid\theta\rangle.$$ The above arguments actually hold for any $\theta$, so we actually have $$\mid \psi_j \rangle = \sum_i \langle \phi_i\mid \psi_j \rangle \mid \phi_i \rangle $$ and $$\mid \phi_j \rangle = \sum_i \langle \psi_i\mid \phi_j \rangle \mid \psi_i \rangle $$ Define, $U_{ij} = \langle \phi_i\mid \psi_j \rangle $ then $(U)^*_{ij} =\langle \psi_i\mid \phi_j \rangle $. Plug these formulas into the earlier expressions for $\mid \theta \rangle$ and you should be able to derive the desired result.

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  • $\begingroup$ @ric.san Note that $|\Psi\rangle = \sum\limits_i f_i |\varphi_i\rangle = \sum\limits_i g_i |\phi_i\rangle$. But since $|\phi_i\rangle = U |\varphi_i$, we find that $f_j = \langle \varphi_j|\Psi\rangle = \sum\limits_i g_i \langle \varphi_j |U |\phi_i \rangle$. The similar result holds for $g_j$, as shown in the answer. $\endgroup$
    – Jakob
    Jul 18 at 21:00
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    $\begingroup$ @ric.san It is the same story: Write $|\varphi_i\rangle = U^\dagger |\phi_i\rangle$ and compute $\langle \phi_j|\Psi\rangle$. Btw. You asked in the comments that you wouldn't see where $f_i$ comes from. This is what I've explained. $\endgroup$
    – Jakob
    Jul 18 at 21:05

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