In literature on an introduction to quantum mechanics which I am working through, there is a section which explains that a vector has different representations based on the basis you choose and then relates this to quantum mechanics. It does this by making giving the following explanation:
Imagine an ordinary vector $A$ in two dimensions. How would you describe this vector to someone? The most convenient way is to set up cartesian coordinates axes, $x$ and $y$, and specify the components of $A$: $A_x = \hat{i} \cdot A$, $A_y = \hat{j}\cdot A$. Of course you could have a different set of axis $x'$ and $y'$ and the components would then be: $A_x' = \hat{i}'\cdot A$, $A_y = \hat{j}'\cdot A$. But it's the same vector.
The same is true for the state of a system in quantum mechanics. It is represented by a vector, $\lvert\mathcal{S(t)}\rangle$, that lives "out there in Hilbert space," but we can express it with respect to any number of different bases. The wave function $\Psi(x,t)$ is actually the coefficient in the expansion of $\mathcal{S(t)}$ in the basis of position eigenfunctions: $$\Psi(x,t) = \langle x | \mathcal{S(t)} \rangle$$
(with $|x\rangle$ standing for the eigenfunction of $\hat{x}$ with eigenvalue $x$), whereas the momentum space wavefunction $\Phi(p,t)$ is the expansion of $| \mathcal{S} \rangle$ in the basis of momentum eigenfunctions: $$\Phi(p,t) = \langle p | \mathcal{S}(t) \rangle$$
(with $|p \rangle$ standing for the eigenfunction of $\hat{p}$ with eigenvalue $p$).
It then states that $\Psi$ and $\Phi$ contain the same information and describe the same state.
Question:
In the description, when it is stated that "$A$: $A_x = \hat{i} \cdot A$, $A_y = \hat{j}\cdot A$" and "$A_x' = \hat{i}'\cdot A$, $A_y = \hat{j}'\cdot A$" for different basis. In this description $A$ is the same but it's representation is different when working out components.
Why is it then that when describing the components of a system, $| \mathcal{S} \rangle$, in quantum mechanics, $| \mathcal{S} \rangle$, that the representation of $|\mathcal{S} \rangle$ does not change when working out the components in different basis (specifically the basis consisting of position operator eigenvectors and momentum operators eigenvectors), as can be seen by the following
$$\Psi(x,t) = \langle x | \mathcal{S(t)} \rangle = \int^{\infty}_{-\infty}\delta(x-y)\Psi(y,t)dy$$ and
$$\Phi(p,t) = \langle p | \mathcal{S(t)} \rangle= \int^{\infty}_{-\infty}\frac{e^{\frac{-ipy}{\hbar}}}{\sqrt{2 \pi \hbar}}\Psi(y,t)dy$$
where we have basis vectors (eigenvetors) $\delta(x-y)$ and $\frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{ipy}{\hbar}}$ respectively.
Why is the vector $|\mathcal{S} \rangle$ represented as $\Psi$ for both bases when working out the components for the quantum mechanics case above?
Best Answer
Let me rephrase those precise equations in the language of finite-dimensional linear algebra. You have a vector $A$ and two bases $\beta=\{e_i\}_i$ and $\beta'=\{e_i'\}_i$. This means you can write the components of $A$ with respect to $\beta$ as $$ A_i=e_i·A=\sum_j\delta_{ij}e_j·A $$ and the components with respect to $\beta'$ as $$ A_i'=e_i'·A=\sum_j(e_i'·e_j)e_j·A. $$ Here the $\delta_{ij}=e_i·e_j$ is the direct equivalent of the $\delta(x-y)=⟨x|y⟩$, and the $e_i'·e_j$ is the direct equivalent of $e^{-ipx/\hbar}/\sqrt{2\pi\hbar}=⟨x|p⟩$.
Simply put, the presence of $\Psi(y,t)$ on the right-hand side of both expressions is simply because you're doing the inner products themselves in the position representation. You could equally write $$\Psi(x,t) = \langle x | \mathcal{S(t)} \rangle = \int^{\infty}_{-\infty} \frac{e^{\frac{+iqx}{\hbar}}}{\sqrt{2 \pi \hbar}} \Phi(q,t)dq$$ and $$\Phi(p,t) = \langle p | \mathcal{S(t)} \rangle= \int^{\infty}_{-\infty}\delta(p-q)\Phi(q,t)dq,$$ or any combination of the two representations for the inner products.