# On change of basis in quantum mechanics [closed]

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.

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.
• @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. Jul 18 at 21:00
• @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. Jul 18 at 21:05