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Q1:

In Zetilli's book page 166 (ch. "Postulates of QM", eq. 3.1) i encountered an expression $\hat{A}|\psi\rangle = a_n|\psi_n\rangle$. I know this is an eigenvalue equation, but i have seen another form written like this $\hat{A}|\psi\rangle = a_n|\psi\rangle$ so now i am a bit confused. Is $|\psi\rangle$ equal to $|\psi_n\rangle$ or not? Whats the difference?

Q2:

Below the same expression there is a statement $a_n=\langle \psi_n | \psi(t)\rangle$. I interpret this like $a_n$ is an inner product between an $|\psi_n\rangle$ and $| \psi(t)\rangle$, but what is a physicall meaning of this? Maybee i would know this if i knew anwser to Q1.

In the book it says that $a_n$ is a component of $|\psi(t)\rangle$ projected on an $|\psi_n\rangle$ and it states that this can even be seen if we expand $|\psi(t)\rangle$ in terms of eigenvectors of $\hat{A}$ which form a complete basis. Does this mean we calculate matrix multiplication $\hat{A}|\psi\rangle$?

Author solves it like this:

$$|\psi(t)\rangle = \sum_n |\psi_{n}\rangle\underbrace{\langle \psi_n| \psi(t)\rangle}_{a_n}=\sum_n a_n |\psi_n\rangle$$.

Does this mean that the state of a system $|\psi(t) \rangle$ is a linear combination (if we have a finite possible states) of the eigenvectors multiplied by eigenvalues $a_n$? So the point of the QM is that the state of a system or a vector of a system $|\psi(t)$ remains the same but we use operators to change the eigenvectors and eigenvalues which then correspond to the operator (for example $\hat{x}$) associated with an observable (for example position).

Enough for 1 question.

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1 Answer

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Q1: That looks like a typo. If it is supposed to be an eigenvalue equation (it probably is), then it is indeed $|\psi_n\rangle$ on both sides.

Q2: A state is described by a vector in Hilbert space. This is an abstract mathematical object, but it is often convenient to choose a basis and use the components in that basis to describe the state. If you have a hermitian operator you can find a complete set of eigenvectors with real eigenvalues which form an orthonormal basis. If you are interested in the observable which corresponds to said hermitian operator (e.g. energy, angular momentum, etc.) then it is convenient to use this basis to describe the states. The eigenbasis is in some sense a natural basis, adapted for the particular operator you are studying. But you can alway go to some other basis without changing the physics and nobody is forcing you to choose any basis at all. An analogy here is ordinary 3 dimensional spatial vectors. You can represent these as arrows in space, or alternatively list the $x,y,z$ components. But somebody else can use some rotated set of axes $x',y',z'$ just as well. Which you choose depends on convenience. Either way you find the components by taking the dot product with the unit vector in the given directions. This is just the $a_n=\langle \psi_n | \psi(t)\rangle$ you had.

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About your anwser on Q1: But what if i would write $|\psi\rangle$ on the both sides instead of $|\psi\rangle$? Which one is correct? About your anwser on Q2: I can imagine at most 3 eigenvectors being orthogonal to each other and form a basis. All the rest i cannot visualize being orthogonal, but how is it possible? Is this just an abstraction and i shouldnt bother so much with it? I vizualize inner product is an analog to scalar product which is a projection of 1st vector's norm to the other vec. multiplied by a 2nd vector's norm. And vice versa - where order matters at inner pr. –  71GA May 21 '13 at 9:11
    
I understand now, how to get an eigenvalue like $a_n = \langle\psi_n|\psi(t)\rangle$. What i was missing was the fact that eigenvectors are normalized. –  71GA May 21 '13 at 9:13
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First part: just a matter of notation. There $|\psi\rangle$ would represent an eigenvector rather than a general state. You'll see other notations like $|n\rangle$ instead of $|\psi_n\rangle$ sometimes. Second: The three dimensional analogy was just something to give you a visual. The general case in quantum mechanics is that you have a many dimensional (often infinite dimensional) complex vector space, but the math is closely analogous. Third: order does not matter for ordinary scalar products, though in the quantum case changing the order comes with a complex conjugation. –  Michael Brown May 21 '13 at 12:07
    
That's cause the Hilbert space is a complex vector space. With that single caveat the situation is basically the same as for real vectors. Fourth: The eigenvector equation does not demand that you normalise the eigenvectors, but you always can and often do. You can see this in the completeness relation $\mathbb{I}=\sum_n |\psi_n\rangle\langle\psi_n|$. If you didn't normalise this would be $\mathbb{I}=\sum_n \frac{|\psi_n\rangle\langle\psi_n|}{\langle\psi_n|\psi_n\rangle}$ instead (prove this by acting on an eigenvector $|\psi_k\rangle$). –  Michael Brown May 21 '13 at 12:11
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