What does the state vector mean?

[mike1000]

To me the state vector represents the following...

1) The number of elements in the state vector is the number of possible outcomes. Call that number n.
2) The value of each element in the state vector is the probability amplitude associated with that outcome.

If that is true, then it seems to me, that a state vector which is a linear combination of the eigenstates, cannot be interpreted as creating a new possible outcome. It can only be interpreted as a way to calculate the amplitude of one of the n, apriori, possible outcomes.

 

[DrClaude]

To me the state vector represents the following...

1) The number of elements in the state vector is the number of possible outcomes. Call that number n.
2) The value of each element in the state vector is the probability amplitude associated with that outcome.
[\quote]
You are making the same mistake as in another thread. You are confusing a state vector and its representation. A state vector doesn't have elements. It "lives" in a Hilbert with a certain dimension, and it is that that determines the number of basis states required to fully represent it.

[QUOTE="mike1000, post: 5730239, member: 615603"
If that is true, then it seems to me, that a state vector which is a linear combination of the possible outcomes, cannot be interpreted as creating a new possible outcome. It can only be interpreted as a way to calculate the amplitude of one of the n, apriori, possible outcomes.

This doesn't make sense to me.

 

[mike1000]

This doesn't make sense to me.

Thanks. I am trying to say that a state which is a linear combination of eigenstates cannot be interpreted as creating a new possible outcome, where the new possible outcome, for instance, would be a coin in the head and tail state at the same time. That state would represent a new possible outcome. But it is more logical to interpret it as a way to calculate the probability amplitudes of the two existing possible outcomes/states.

There cannot be more than n possible outcomes.

 

[stevendaryl]

To me the state vector represents the following...

1) The number of elements in the state vector is the number of possible outcomes. Call that number n.

The number of possible outcomes depends on what it is that you're measuring. If you measure the component of the spin of an electron along the z-axis, there are two possible outcomes. If you measure momentum, there are infinitely many possible outcomes.

 

[mike1000]

The number of possible outcomes depends on what it is that you're measuring. If you measure the component of the spin of an electron along the z-axis, there are two possible outcomes. If you measure momentum, there are infinitely many possible outcomes.

I thought you would have said there were 3 possible outcomes because of this superposition state $S_z = \frac{1}{\sqrt{2}}\big(\uparrow+\downarrow\big)$

 

[Nugatory]

I thought you would have said there were 3 possible outcomes because of this superposition state $S_z = \frac{1}{\sqrt{2}}\big(\uparrow+\downarrow\big)$

No matter which spin measurement you make, there are two possible outcomes: up and down. That's another way of saying that the Hilbert space containing the spin states is two-dimensional. However, that Hilbert space is only a subspace of the infinite-dimensional Hilbert space that contains the state vectors of the particle in question.

 

[stevendaryl]

I thought you would have said there were 3 possible outcomes because of this superposition state $S_z = \frac{1}{\sqrt{2}}\big(\uparrow+\downarrow\big)$

No, the number of outcomes, as I said, is determined by the observable that you are measuring. Every observable corresponds to an operator, and the number of possible outcomes is equal to the number of eigenvalues for the operator.

For the observable of "z-component of spin", there are two eigenvalues, +1/2 and -1/2, which means that those are the only two possible outcomes of a measurement.

There are infinitely many different state vectors, though (although every state vector is a linear combination of two possibilities: $|u\rangle$ and $|d\rangle$)

 

[tomdodd4598]

I thought you would have said there were 3 possible outcomes because of this superposition state $S_z = \frac{1}{\sqrt{2}}\big(\uparrow+\downarrow\big)$

I guess you mean a state $\left| \psi \right> =\frac { \left| \uparrow \right> +\left| \downarrow \right> }{ \sqrt { 2 } }$ of a spin-1/2 particle. What you have written is the spin state using a certain representation (The z-direction spin basis), and there are two possible outcomes when measuring that spin in the z-direction, those being up or down, with a 50% probability of either. The same is true for measuring the spin in the y-direction, but since that state is an eigenstate of the x-direction of spin operator with eigenvalue ħ/2, then measuring the spin in the x-direction will always yield an 'up' result.

 

[PeterDonis]

To me the state vector represents the following

You're confusing the state vector with its representation in a particular basis. It seems like we already went over this in a previous thread.

 

[mike1000]

You're confusing the state vector with its representation in a particular basis. It seems like we already went over this in a previous thread.

No, I am not confusing the state vector with its representation in a particular basis.

 

[mike1000]

No, the number of outcomes, as I said, is determined by the observable that you are measuring. Every observable corresponds to an operator, and the number of possible outcomes is equal to the number of eigenvalues for the operator.

For the observable of "z-component of spin", there are two eigenvalues, +1/2 and -1/2, which means that those are the only two possible outcomes of a measurement.

There are infinitely many different state vectors, though (although every state vector is a linear combination of two possibilities: $|u\rangle$ and $|d\rangle$)

I really like this response. Specially the last sentence.

 

[PeterDonis]

I am not confusing the state vector with its representation in a particular basis.

Yes, you are, in the OP, when you say this:

1) The number of elements in the state vector is the number of possible outcomes. Call that number n.
2) The value of each element in the state vector is the probability amplitude associated with that outcome.

These two statements are statements about the representation of the state vector in a particular basis--the one corresponding to the measurement whose outcomes you are describing.

 

[mike1000]

Yes, you are, in the OP, when you say this:
These two statements are statements about the representation of the state vector in a particular basis--the one corresponding to the measurement whose outcomes you are describing.

As others have said, the number of possible outcomes is equal to the number of eigenvalues of the particular operator. That was actually what I was thinking when I wrote the original post. Unless you are saying that changing the basis changes the number of eigenvalues of a particular operator (which I do not think is the case when you apply a unitary operation to an operator.)

 

[Adrian59]

State vector |A> = c1 |A1> + c2 |A2> + c3 |A3> + …. + Cn |An>

ie n terms which I think is what you are saying.

If all the base vectors (|Am> ) are normalised ,that is their magnitude is equal to 1, then the coefficients (cm ) are equal to the amplitudes, so that the square of | cm | are the probabilities.

If the system is definitely in state 3 then the probability for this is | c3 | - squared . However, if the system is in a state which is a superposition of two or more of the base states then several coefficients will need to be added to find this probability.

At measurement there is only one outcome, despite the various probabilities for different possible states prior to measurement.

I hope that helps.

 

[mike1000]

State vector |A> = c1 |A1> + c2 |A2> + c3 |A3> + …. + Cn |An>

ie n terms which I think is what you are saying.

If all the base vectors (|Am> ) are normalised ,that is their magnitude is equal to 1, then the coefficients (cm ) are equal to the amplitudes, so that the square of | cm | are the probabilities.

If the system is definitely in state 3 then the probability for this is | c3 | - squared . However, if the system is in a state which is a superposition of two or more of the base states then several coefficients will need to be added to find this probability.

At measurement there is only one outcome, despite the various probabilities for different possible states prior to measurement.

I hope that helps.

First I am going to express your equation for $|A\rangle$ .$$C_n = \langle n|C\rangle$$ Where $\langle n$ is the orthonormal basis($A$) and $|C\rangle$ is the state vector.

I think I can then write

\begin{equation}C=\begin{bmatrix}
\\{C_1}\\
\\{C_2}\\
\\{\vdots}\\
\\{C_n}
\end{bmatrix}\end{equation}

and would each $C_n$ represent the probability amplitude of measuring the nth possible outcome?

 

[vanhees71]

Again: An observable in quantum theory is represented by a self-adjoint operator on the Hilbert space. The possible values of the observable are the eigenvalues (more precisely the spectral values) of this self-adjoint operator. So let $A$ be the observable you measure and $\hat{A}$ the corresponding self-adjoint operator and $|a,\beta \rangle$ an orthonormalized set of eigenvalues of $\hat{A}$ to the eigenvalue $a$. For simplicity let's assume that all eigenvalues and the labels $\beta$ are all discrete (the generalization to continuous parts of the spectrum are straight forward; everywhere where I write sums you have to add integrals, and where I write Kronecker-$\delta$'s write Dirac-$\delta$-distributions):
$$\langle a,\beta|a',\beta' \rangle=\delta_{a,a'} \delta_{\beta,\beta'}, \quad \sum_{a,\beta} |a,\beta \rangle \langle a,\beta|=\hat{1}.$$
This means that the eigenvectors of $\hat{A}$ can be chosen as a complete orthonormalized set, i.e., an orthonormal basis of the Hilbert space (which except for academic examples is the separable infinite-dimensional Hilbert space).

Pure states are represented by rays in Hilbert space, which are themselves again represented by normalized state vectors. If the system is in the pure state, represented in this sense by the state vector $|\psi \rangle$, then the probality to measure the value $a$ of the observable $A$ is given by Born's rule,
$$P(a)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2.$$

 

[Adrian59]

That doesn't look right. If I was writing it that way I would have the C's as a row vector or even as a diagonal of a square matrix with all other coefficients zero and then the base | A > 's as the column vector. Dirac introduced the ket vector as a column vector and the operator as a matrix. A bra vector is represented as a row vector to the left, if required.

 

[vanhees71]

No, Dirac introduced the ket vector as an abstract vector. There are no columns and rows. It's just an abstract vector! You get columns and rows, if you choose a descrete basis of the Hilbert space and calulate the corresponding components of the vectors.

 

[mike1000]

That doesn't look right. If I was writing it that way I would have the C's as a row vector or even as a diagonal of a square matrix with all other coefficients zero and then the base | A > 's as the column vector. Dirac introduced the ket vector as a column vector and the operator as a matrix. A bra vector is represented as a row vector to the left, if required.

No, Dirac introduced the ket vector as an abstract vector. There are no columns and rows. It's just an abstract vector! You get columns and rows, if you choose a descrete basis of the Hilbert space and calulate the corresponding components of the vectors.

Ok. Please, just simply, what is the name of a vector that just contains the probability amplitudes? And how do I express that vector in Dirac notation? Is that vector a state vector?

 

[mike1000]

That doesn't look right. If I was writing it that way I would have the C's as a row vector or even as a diagonal of a square matrix with all other coefficients zero and then the base | A > 's as the column vector. Dirac introduced the ket vector as a column vector and the operator as a matrix. A bra vector is represented as a row vector to the left, if required.

No, Dirac introduced the ket vector as an abstract vector. There are no columns and rows. It's just an abstract vector! You get columns and rows, if you choose a descrete basis of the Hilbert space and calulate the corresponding components of the vectors.

Maybe I am using the wrong word. Maybe the word vector is throwing me off. Let me call it the n-tuple that represents the probability amplitudes. What do we call the n-tuple which contains the probability amplitudes? How do we represent that n-tuple using Dirac notation? Can I think of it as a column vector?

 

[vanhees71]

Have you read my posting #16? There, everything is explained, and I use the notation introduced there in the following.

If you have the state vector $|\psi \rangle$ and the complete orthonormal set $|a,\beta \rangle$ of eigenvectors of $\hat{A}$, then the probability amplitudes are given by the components of the state vector with repsect to this basis,
$$\psi_{a,\beta}=\langle a,\beta|\psi \rangle.$$
From the completeness relation given in #16, it's clear that you get back the vector via
$$|\psi \rangle=\sum_{a,\beta} |a,\beta \rangle \langle a,\beta|\psi \rangle=\sum_{a,\beta} \psi_{a,\beta} |a,\beta \rangle.$$
You have a one-to-one mapping from the abstract separable Hilbert space $\mathcal{H}$ to the Hilbert space $\ell^2$ of square-summable sequences.

 

[stevendaryl]

Ok. Please, just simply, what is the name of the vector that just contains the probability amplitudes? And how do I express that vector in Dirac notation?

I would say that in the case that your Hilbert space is n-dimensional, then what you wrote is okay. You're talking about a Hilbert space with basis vectors $|A_1\rangle, |A_2\rangle, ... |A_n\rangle$. Then you can represent the basis vectors this way:

$|A_1\rangle = \left( \begin{array} \\ 1 \\ 0 \\ 0 \\ . \\ . \\ . \end{array} \right)$
$|A_2\rangle = \left( \begin{array} \\ 0 \\ 1 \\ 0 \\ . \\ . \\ . \end{array} \right)$
$|A_1\rangle = \left( \begin{array} \\ 0 \\ 0 \\ 1 \\ . \\ . \\ . \end{array} \right)$
etc.

Then in this representation, the state vector $|\psi\rangle = C_1 |A_1\rangle + C_2 |A_2\rangle + ...$ would be represented by:

$|\psi\rangle = \left( \begin{array} \\ C_1 \\ C_2 \\ C_3 \\ . \\ . \\ . \end{array} \right)$

That's correct. What's not correct is to say that "each $C_n$ represents the probability amplitude of measuring the outcome $|A_n\rangle$. In this case, $|A_n\rangle$ is not an outcome, it's a basis vector. If you represent the state vectors as column matrices, then observables are represented by square matrices and outcomes are just numbers: the eigenvalues of the matrices corresponding to the observables.

What we can do in this case is define an operator $\hat{N}$ via the eigenvalue equation

$|\hat{N}\rangle |A_n\rangle = n |A_n\rangle$

This would be represented by the square matrix:

$\hat{N}= \left( \begin{array} \\ 1 & 0 & 0 & ... \\ 0 & 2 & 0 & ... \\ 0 & 0 & 3 & ... \\ . \\ . \\ . \end{array} \right)$

Then the possible outcomes of a measurement of $\hat{N}$ would be 1, 2, 3, etc. And with $|\psi\rangle$ as above, the probability of getting result $n$ would be given by $|C_n|^2$.

 

[mike1000]

I would say that in the case that your Hilbert space is n-dimensional, then what you wrote is okay. You're talking about a Hilbert space with basis vectors $|A_1\rangle, |A_2\rangle, ... |A_n\rangle$. Then you can represent the basis vectors this way:

$|A_1\rangle = \left( \begin{array} \\ 1 \\ 0 \\ 0 \\ . \\ . \\ . \end{array} \right)$
$|A_2\rangle = \left( \begin{array} \\ 0 \\ 1 \\ 0 \\ . \\ . \\ . \end{array} \right)$
$|A_1\rangle = \left( \begin{array} \\ 0 \\ 0 \\ 1 \\ . \\ . \\ . \end{array} \right)$
etc.

Then in this representation, the state vector $|\psi\rangle = C_1 |A_1\rangle + C_2 |A_2\rangle + ...$ would be represented by:

$|\psi\rangle = \left( \begin{array} \\ C_1 \\ C_2 \\ C_3 \\ . \\ . \\ . \end{array} \right)$

That's correct. What's not correct is to say that "each $C_n$ represents the probability amplitude of measuring the outcome $|A_n\rangle$. In this case, $|A_n\rangle$ is not an outcome, it's a basis vector. If you represent the state vectors as column matrices, then observables are represented by square matrices and outcomes are just numbers: the eigenvalues of the matrices corresponding to the observables.

What we can do in this case is define an operator $\hat{N}$ via the eigenvalue equation

$|\hat{N}\rangle |A_n\rangle = n |A_n\rangle$

This would be represented by the square matrix:

$\hat{N}= \left( \begin{array} \\ 1 & 0 & 0 & ... \\ 0 & 2 & 0 & ... \\ 0 & 0 & 3 & ... \\ . \\ . \\ . \end{array} \right)$

Then the possible outcomes of a measurement of $\hat{N}$ would be 1, 2, 3, etc. And with $|\psi\rangle$ as above, the probability of getting result $n$ would be given by $|C_n|^2$.

When I said in the previous post

represent the probability amplitude of measuring the outcome $A_n$?

I was actually thinking the possible outcome n as you stated in your last sentence. I was thinking that $A_n$ was the eigenvector for the nth possible outcome.

Thanks

 

[vanhees71]

I would say that in the case that your Hilbert space is n-dimensional, then what you wrote is okay. You're talking about a Hilbert space with basis vectors $|A_1\rangle, |A_2\rangle, ... |A_n\rangle$. Then you can represent the basis vectors this way:

$|A_1\rangle = \left( \begin{array} \\ 1 \\ 0 \\ 0 \\ . \\ . \\ . \end{array} \right)$
$|A_2\rangle = \left( \begin{array} \\ 0 \\ 1 \\ 0 \\ . \\ . \\ . \end{array} \right)$
$|A_1\rangle = \left( \begin{array} \\ 0 \\ 0 \\ 1 \\ . \\ . \\ . \end{array} \right)$
etc.

Then in this representation, the state vector $|\psi\rangle = C_1 |A_1\rangle + C_2 |A_2\rangle + ...$ would be represented by:

$|\psi\rangle = \left( \begin{array} \\ C_1 \\ C_2 \\ C_3 \\ . \\ . \\ . \end{array} \right)$

That's correct. What's not correct is to say that "each $C_n$ represents the probability amplitude of measuring the outcome $|A_n\rangle$. In this case, $|A_n\rangle$ is not an outcome, it's a basis vector. If you represent the state vectors as column matrices, then observables are represented by square matrices and outcomes are just numbers: the eigenvalues of the matrices corresponding to the observables.

What we can do in this case is define an operator $\hat{N}$ via the eigenvalue equation

$|\hat{N}\rangle |A_n\rangle = n |A_n\rangle$

This would be represented by the square matrix:

$\hat{N}= \left( \begin{array} \\ 1 & 0 & 0 & ... \\ 0 & 2 & 0 & ... \\ 0 & 0 & 3 & ... \\ . \\ . \\ . \end{array} \right)$

Then the possible outcomes of a measurement of $\hat{N}$ would be 1, 2, 3, etc. And with $|\psi\rangle$ as above, the probability of getting result $n$ would be given by $|C_n|^2$.

Please, don't do that! I've seen such notation from time to time even in textbooks, but this thread shows, how confusing that is! It's of course not wrong, but it's confusing. My experience is that it is much better to keep to bra-ket notation strictly for the abstract vectors. It's the great advantage of the Dirac notation that it is representation independent, and you can do many general calculations in a much clearer way without introducing a basis/representation.

If you have an application, where matrix-vector notation in $\mathbb{C}^d$ or $\ell^2$ are needed, it's very easy to introduce them. I suggest then you write
$$\psi_n=\langle n|\psi \rangle,$$
where the $|n \rangle$ are an orthonormal basis and then define
$$\psi=\begin{bmatrix}\psi_1 \\ \psi_2\\ \vdots \end{bmatrix},$$
and for the Operators you introduce matrix elements
$$A_{mn}=\langle m |\hat{A}|n \rangle$$
and matrices
$$\boldsymbol{A}=\begin{pmatrix} A_{11} & A_{12} & \ldots \\
A_{21} & A_{22} & \ldots & \\
\vdots & \vdots & \ddots & \end{pmatrix}.$$
This, perhaps somewhat overpedantic, notation helps a lot to avoid the kind of confusion we try to cure in this thread!

 

[PeterDonis]

As others have said, the number of possible outcomes is equal to the number of eigenvalues of the particular operator.

Yes, that is true. But notice that it says nothing about the state vector. It's a property of the operator.

Please, just simply, what is the name of a vector that just contains the probability amplitudes?

There isn't one, if you mean a vector independent of a choice of basis, i.e., of a specific operator. There are no probability amplitudes unless you choose what measurement you are going to make. That means choosing an operator, which means choosing a basis: the probability amplitudes are then the components of the state vector in that basis. You are having problems because you keep trying to think of the state vector as giving amplitudes without choosing an operator/basis. You can't do that; it doesn't work.

 

[mike1000]

Yes, that is true. But notice that it says nothing about the state vector. It's a property of the operator.
There isn't one, if you mean a vector independent of a choice of basis, i.e., of a specific operator. There are no probability amplitudes unless you choose what measurement you are going to make. That means choosing an operator, which means choosing a basis: the probability amplitudes are then the components of the state vector in that basis. You are having problems because you keep trying to think of the state vector as giving amplitudes without choosing an operator/basis. You can't do that; it doesn't work.

No. You keep projecting onto me something that is not in my mind.The problem I am having now, is with getting the notation correct. I have always realized that the components of the vector were the probability amplitudes.

 

[mike1000]

Please, don't do that! I've seen such notation from time to time even in textbooks, but this thread shows, how confusing that is! It's of course not wrong, but it's confusing. My experience is that it is much better to keep to bra-ket notation strictly for the abstract vectors. It's the great advantage of the Dirac notation that it is representation independent, and you can do many general calculations in a much clearer way without introducing a basis/representation.

If you have an application, where matrix-vector notation in $\mathbb{C}^d$ or $\ell^2$ are needed, it's very easy to introduce them. I suggest then you write
$$\psi_n=\langle n|\psi \rangle,$$
where the $|n \rangle$ are an orthonormal basis and then define
$$\psi=\begin{bmatrix}\psi_1 \\ \psi_2\\ \vdots \end{bmatrix},$$
and for the Operators you introduce matrix elements
$$A_{mn}=\langle m |\hat{A}|n \rangle$$
and matrices
$$\boldsymbol{A}=\begin{pmatrix} A_{11} & A_{12} & \ldots \\
A_{21} & A_{22} & \ldots & \\
\vdots & \vdots & \ddots & \end{pmatrix}.$$
This, perhaps somewhat overpedantic, notation helps a lot to avoid the kind of confusion we try to cure in this thread!

I will try to adhere to this notation. I think this post should be somehow bookmarked so people like me can always readily refer to.

 

[Adrian59]

An aside for Vanhees71, can I refer you to 'The Principles of Quantum Mechanics' by PAM Dirac, bottom of page 71 in the 4th Ed, "we may look upon the representative of a ket |P> as a matrix with a single column by setting all the numbers <ξ1 -ξu | P> which form this representative one below the other." He makes a similar suggestion for the bra vector as a row on page 72. I know this text is a little old now so whether the misunderstanding has to do with some later edition that I do not possess.

 

[Adrian59]

To mike1000, my apologies we seem to have veered off; the best simple explanation of a state vector is to consider them as like i, j and k vectors which I hope you are familiar with. They are of unit length and just represent a state of going in the direction of the x, y and z axes. This analogy can be expanded to see how other states can be thought of as linear combinations of those pure x, y and z states. The probability amplitude is the magnitude of the vector not its direction. When the magnitude is squared that gives the probability of being in that state, whether a pure i,j or k state or some superposition of these. The n-tuple as you put it is just the number of base states which in my analogy is three and the state vector in Dirac notation is written as in my first response as:
State vector |A> = c1 |A1> + c2 |A2> + c3 |A3> + …. + Cn |An>
The state vector is really just a label for a particular state the system is in and when normalised does not contain any value. Values and thereby observables are contained in the operator coefficients, that is the parts of the equations that are not the vectors.

 

[mikeyork]

Your problem, mike1000, as people have tried to explain to you several times, is that you don't adequately distinguish states from eigenstates or your chosen observational basis from any other basis. The number of states n you keep referring to is the number of eigenstates not the number of possible states. (And as long as you insist on talking about "elements" rather than eigenstates you'll probably stay confused about this. There is a reason for the language we use in QM.) You get only n possible outcomes only if that basis is the chosen basis for your observation. If you have a superposition in basis A, then yes, you'll have n possibilities if your observation basis is A. If your observation basis is B, then you'll have another maximum n independent possible measurement results. That means you have up to 2n possible states but only n eigenstates in any given basis. In general you can add another n eigenstates for every possible basis you might choose to make your measurement and the result is an infinite number of states in an n-dimensional Hilbert space. It is only in a specific basis that the number of possible results is n.

The key point I think you are missing is that when you make an observation you are explicitly choosing a basis but it doesn't have to be the same basis as that of your initial state representation.

 

[mike1000]

Your problem, mike1000, as people have tried to explain to you several times, is that you don't adequately distinguish states from eigenstates or your chosen observational basis from any other basis. The number of states n you keep referring to is the number of eigenstates not the number of possible states. (And as long as you insist on talking about "elements" rather than eigenstates you'll probably stay confused about this. There is a reason for the language we use in QM.) You get only n possible outcomes only if that basis is the chosen basis for your observation. If you have a superposition in basis A, then yes, you'll have n possibilities if your observation basis is A. If your observation basis is B, then you'll have another maximum n independent possible measurement results. That means you have up to 2n possible states but only n eigenstates in any given basis. In general you can add another n eigenstates for every possible basis you might choose to make your measurement and the result is an infinite number of states in an n-dimensional Hilbert space. It is only in a specific basis that the number of possible results is n.

The number n, I have been referring to, represents the number of possible outcomes. If you go back and read my posts you will see that I have always referred to it as the number of possible outcomes, not the number of states. The number of possible outcomes is fixed to whatever the number of eigenvalues for the operator happen to be. The number of states is infinite.

 

[mike1000]

Please, don't do that! I've seen such notation from time to time even in textbooks, but this thread shows, how confusing that is! It's of course not wrong, but it's confusing. My experience is that it is much better to keep to bra-ket notation strictly for the abstract vectors. It's the great advantage of the Dirac notation that it is representation independent, and you can do many general calculations in a much clearer way without introducing a basis/representation.

If you have an application, where matrix-vector notation in $\mathbb{C}^d$ or $\ell^2$ are needed, it's very easy to introduce them. I suggest then you write
$$\psi_n=\langle n|\psi \rangle,$$
where the $|n \rangle$ are an orthonormal basis and then define
$$\psi=\begin{bmatrix}\psi_1 \\ \psi_2\\ \vdots \end{bmatrix},$$
and for the Operators you introduce matrix elements
$$A_{mn}=\langle m |\hat{A}|n \rangle$$
and matrices
$$\boldsymbol{A}=\begin{pmatrix} A_{11} & A_{12} & \ldots \\
A_{21} & A_{22} & \ldots & \\
\vdots & \vdots & \ddots & \end{pmatrix}.$$
This, perhaps somewhat overpedantic, notation helps a lot to avoid the kind of confusion we try to cure in this thread!

This is also for vanhees71.

You show us what you mean by $\psi$ but you do not show us what you mean by $|\psi\rangle$

If $$\psi=\begin{bmatrix}\psi_1 \\ \psi_2\\ \vdots \end{bmatrix},$$ then what does $$|\psi\rangle$$ represent?

 

[mikeyork]

The number n I have been referring to represents the number of possible outcomes. If you go back and read my posts you will see that I have always referred to it as the number of outcomes, not the number of states. The number of outcomes is fixed to whatever the number of eigenvalues for the operator happen to be.

Only in a specifically chosen basis.

 

[mike1000]

The number n I have been referring to represents the number of possible outcomes. If you go back and read my posts you will see that I have always referred to it as the number of outcomes, not the number of states. The number of outcomes is fixed to whatever the number of eigenvalues for the operator happen to be.

Only in a specifically chosen basis.

Are you implying that if I rotated the coordinate system, the number of eigenvalues will change which implies that the number of possible outcomes will change?

As long as the dimension is n, then the operator matrix will be nxn so there will be n eigenvalues. The only way I can see your point of view is if you change the dimension of the operator.

 

[mikeyork]

Are you implying that if I rotated the coordinate system, the number of eigenvalues will change which implies that the number of possible outcomes will change?

The number of eigenvalues will not change. But for a given superposition in the unrotated frame, the rotated superposition may involve a different number of eigenstates and so perhaps a different number of "possible outcomes". (Fewer or more eigenstates may have vanishing probabilities.)