I What does the state vector mean?

1. Mar 31, 2017

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.

Last edited: Mar 31, 2017
2. Mar 31, 2017

Staff: Mentor

This doesn't make sense to me.

3. Mar 31, 2017

mike1000

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.

Last edited: Mar 31, 2017
4. Apr 1, 2017

stevendaryl

Staff Emeritus
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.

5. Apr 1, 2017

mike1000

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

6. Apr 1, 2017

Staff: Mentor

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.

7. Apr 1, 2017

stevendaryl

Staff Emeritus
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 $|d\rangle$)

8. Apr 1, 2017

tomdodd4598

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.

9. Apr 1, 2017

Staff: Mentor

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.

10. Apr 1, 2017

mike1000

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

11. Apr 1, 2017

mike1000

I really like this response. Specially the last sentence.

12. Apr 2, 2017

Staff: Mentor

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.

13. Apr 2, 2017

mike1000

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.)

14. Apr 2, 2017

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.

15. Apr 2, 2017

mike1000

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

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

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

Last edited: Apr 2, 2017
16. Apr 2, 2017

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,
$$P(a)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2.$$

17. Apr 2, 2017

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.

18. Apr 2, 2017

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.

19. Apr 2, 2017

mike1000

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?

Last edited: Apr 2, 2017
20. Apr 2, 2017

mike1000

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?

21. Apr 2, 2017

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.

22. Apr 2, 2017

stevendaryl

Staff Emeritus
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 $|C_n|^2$.

23. Apr 2, 2017

mike1000

When I said in the previous post
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

24. Apr 2, 2017

vanhees71

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!

25. Apr 2, 2017

Staff: Mentor

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.