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I'm trying to learn basic quantum mechanics by myself. As I understand now, any quantum system can be described by an abstract state vector $\left\vert \Psi (t) \right\rangle$ which generally can be considered a function of time. Now this can be expressed as a linear combination of different basis eigenvectors. As Susskind states in its "Quantum mechanics: the theoretical minimum" eigenvectors of an hermitian operator form an orthonormal basis. I understand that examples could be position, momentum or energy operators, which can be used to write the state vector as follows:

$$ \{\left\vert x \right\rangle\} \to \left\vert \Psi(t) \right\rangle = \int \left\langle x \vert \Psi(t) \right\rangle \left\vert x \right\rangle dx = \int \psi(x) \left\vert x \right\rangle dx $$

$$ \{\left\vert p \right\rangle\} \to \left\vert \Psi(t) \right\rangle = \int \left\langle p \vert \Psi(t) \right\rangle \left\vert p \right\rangle dp $$

$$ \{\left\vert E \right\rangle\} \to \left\vert \Psi(t) \right\rangle = \sum_i \left\langle E_i \vert \Psi(t) \right\rangle \left\vert E_i \right\rangle = \sum_i C_i \left\vert E_i \right\rangle$$

as it is shown in this video (minute 1:30). Since it is postulated that $\left\vert \Psi(t) \right\rangle$ belongs to a Hilbert space and can be expanded in terms of an infinite number of eigenvectors which form a basis, the dimensionality of its vector space should be infinite.
On the other hand, there are cases in which the state vector can be expressed as a linear combination of a finite number of eigenvectors, such as when speaking about spin. For example, when analyzing the Stern-Gerlach experiment it is useful to express $\left\vert \Psi(t) \right\rangle$ as:

$$ \left\vert \Psi(t) \right\rangle = \alpha_{\downarrow} \left\vert \downarrow \right\rangle + \alpha_{\uparrow} \left\vert \uparrow \right\rangle $$

The dimension of a vector space is defined to be the number of vectors in one of its basis. Probably I'm missing something about the nature of the state vector, but how is it possible for it to be expanded in terms of infinite and finite sets of eigenvectors? Moreover, even if both infinite, the number of eigenvectors in $\{\left\vert x \right\rangle\}$ and $\{\left\vert p \right\rangle\}$ should be greater than the one in $\{\left\vert E \right\rangle\}$, shouldn't this be same inequality that relates the cardinality of the real and the rational numbers?

Thank you!

P.S.: My guess is that when talking about spin, we are talking about a quantum system itself, that thus requires a state vector to be described, different from the one used to describe a whole particle, for example. Therefore, when talking about a particle with spin, it is as if the two system are merged together...

Luke__
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    Related if not duplicate: This, this and this and links therein. – Tobias Fünke Aug 02 '23 at 20:53
  • Yes, I've read the questions you've linked: they are similar, but I still don't understand. As in the first answer to your first link it is stated, ${ \left\vert x \right\rangle } = \delta \left( x - x_0\right)$ is not an element of $L^2 [\mathbb{R}]$, thus they aren't a basis for a Hilbert space. This arises one question: if the state vector has to belong to an Hilbert space, why can it be expressed as a linear combination of external vectors? The difference between the dimensionality of the state vector between the SG experiment and the previous linear combinations still remains. – Luke__ Aug 02 '23 at 21:05
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    The question with the $|x\rangle$ states is mathematically subtle, but for most practical purposes not so relevant, at least in introductory curses! As so often in physics, we just pretend that some things are mathematically possible (without even thinking about if this procedure is correct - the main point is that we want to make the correct predictions). Regarding the other point: The SG experiment (ignoring spatial degrees of freedom, which in almost all intro curses is done) is $H\cong \mathbb C^2$, i.e. two-dimensional. You cannot expand it into a basis with infinite members. – Tobias Fünke Aug 02 '23 at 21:25
  • Therefore my guess does hold, doesn't it? The state vector when talking about the SG experiment is different from the one used for talking about the particle as a whole (with a quantum notion of position, momentum...)? – Luke__ Aug 02 '23 at 21:41
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    Yes! If you want to describe a particle with e.g. spin-1/2 moving in 3dimension, the corresponding Hilbert space is $L^2(\mathbb R^3) \otimes \mathbb C^2$. This space has infinite-dimensions, though! And as you guessed in your last sentence of your question (If I understood that correctly), this space describes both the spatial and the spin degrees of freedom. But again: usually, when first encountering QM and SG is discussed, one ignores the spatial degrees of freedom and only discusses the spin-degree of freedoms, i.e. working with a two-dimensional Hilbert space. – Tobias Fünke Aug 02 '23 at 22:04
  • Okay, thank you again, now it's Mich more clear – Luke__ Aug 03 '23 at 07:08

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