High School Superposition principle: clarifications

[Karolus]

What is the "superposition principle"?

I have a confusion about one of the fundamental concepts of quantum mechanics, the principle of superposition
This sounds, more or less: a linear combination with arbitrary coefficients of different quantum states is a new quantum state

If I have (for maximum simplicity) two quantum states $|a_1\rangle$ and $|a_2\rangle$, I can "build" a new state
$|a_3\rangle = c_1|a_1\rangle + c_2|a_2\rangle$
Where $c_1$ and $c_2$ are arbitrary coefficients

But what does $|a_3\rangle$ represents?

Things become embarrassed if we think in terms of eigenvectors.
Let's recap the problem in these terms:
Let A be a observable generic (hermitian matrix) and $|a_k\rangle$ the set of eigenvectors and eigenvalues $\lambda_k$.
Let's exclude cases of "degenerate" for simplicity.
We know that $\{|a_k\rangle\}$ is a complete set in Hilbert's space.
What does it mean?.
It means that $\{|a_k\rangle\}$is an orthonormal base of Hilbert space, so any "vector" of Hilbert space
Is represented by a suitable linear combination of $|a_k\rangle$.

Now, it's not true at all that a linear combination of $|a_k\rangle$ is still an eigenvector of A
On the other hand, we think of the hydrogen atom.

The eigenfunctions of the hydrogen atom $|\psi_i\rangle$ (suppose in its simplest form) with eigenvalues $E_i$
Let's suppose they answer the equation $H|\psi_i\rangle = E_i|\psi_i\rangle$

In what way a linear combination of hydrogen eigenstates is a quantum state?
It is usually represented as a generic quantum state of the hydrogen atom as a combination
Linear of all its eigenstates like $|\psi\rangle = \sum_{i}c_i|\psi_i\rangle$. A possible measure will collapse the generic $|\psi\rangle$
in one of the $|\psi_i\rangle$ with probability $|c_i|^2$

How does this result prove?

 

[DrClaude]

If I have (for maximum simplicity) two quantum states $|a_1\rangle$ and $|a_2\rangle$, I can "build" a new state
$|a_3\rangle = c_1|a_1\rangle + c_2|a_2\rangle$
Where $c_1$ and $c_2$ are arbitrary coefficients

But what does $|a_3\rangle$ represents?

Be careful with your notation. Since you are later taking $\{|a_k\rangle\}$ as a basis, then $|a_3\rangle \neq c_1|a_1\rangle + c_2|a_2\rangle$. Rather, use $|\psi\rangle = c_1|a_1\rangle + c_2|a_2\rangle$. In that case, $|\psi\rangle$ is simply the state of the quantum system, which doesn't correspond to any particular eigenstate of $\hat{A}$. Take as an analogy a normal vector in 3-space: it doesn't have to point along any particular axis, but can point in any direction.

Now, it's not true at all that a linear combination of $|a_k\rangle$ is still an eigenvector of A

If the eigenvalues are not degenerate, then no linear combination of $|a_k\rangle$ is an eigenvector of A.

In what way a linear combination of hydrogen eigenstates is a quantum state?

Nothing forces a quantum system to be in a particular eigenstate of any observable. In fact, all states are superposition states in some basis.

Linear of all its eigenstates like $|\psi\rangle = \sum_{i}c_i|\psi_i\rangle$. A possible measure will collapse the generic $|\psi\rangle$
in one of the $|\psi_i\rangle$ with probability $|c_i|^2$

How does this result prove?

Collapse is interpretation-dependent. The Born rule itself is a postulate of QM, not "provable" from something more fundamental.

 

[Karolus]

mmmm... that's right on point one and two. (missing notation)
no linear combination of eigenvectors is eigenvector ... right.
I'll think on...
Thanks

 

[kith]

In what way a linear combination of hydrogen eigenstates is a quantum state?

The energy eigenstates are stationary states which means that they don't change under time evolution (except for an overall phase rotation which doesn't affect the probabilities you calculate with the Born rule). So whenever you have a nontrivial time evolution in your system (for example, when a wavepacket is moving), the system has to be in a superposition of energy eigenstates.

For the hydrogen atom, see for example http://web.uvic.ca/~chem347/hsuperposition/hsuperposition.htm.

 

[Orodruin]

no linear combination of eigenvectors is eigenvector ... right.

Unless, as pointed out in #2, the eigenvectors with non-zero coefficients have degenerate eigenvalues (i.e., the same eigenvalues). Then the new vector will also be an eigenvector with that eigenvalue.

Note that a special case is when only one of the coefficients are non-zero. A basis vector is also a linear combination of basis vectors, it is just that all coefficients except one are zero.

 

[Karolus]

Let's suppose the scenery of $\hat A$ and the eigenvectors $|a_k\rangle$
We build a generic quantum state given by:
$|b\rangle = c_1|a_1\rangle + c_2|a_2\rangle$

Let's suppose now to "rotate" the axis $|a_k\rangle$ ( imagine the 3-dimensional case) so that one of $|a_k\rangle$ axes coincides with $|b\rangle$.
We can express new $|a_k\rangle$ rotated in terms of old $|a_k\rangle$.
We made a transformation in which Hilbert's space has as $|b\rangle$ one of its base vectors. I then expect that there exists an operator $\hat B$ such that $|b\rangle$ is an his eigenvector.
If this argument was correct, the concept of superposition would seem more meaningful to me. A mixed state in system $\hat A$ is a pure state in system $\hat B$.
Where there is a transformation to go from $\hat A$ to $\hat B$ and vice versa
Is it possible?

 

[Orodruin]

Given any vector, there exists an operator which has it as its eigenvector. In particular, all vectors are eigenvectors of the identity operator.

Also, I think you have a false notion of what a pure state is. A pure state is any state that can be written as a single vector (i.e., a superposition of vectors is a pure state). In order to describe a mixed state properly you need to work with density matrices.

 

[DrClaude]

If this argument was correct, the concept of superposition would seem more meaningful to me. A mixed state in system $\hat A$ is a pure state in system $\hat B$.

You are employing incorrect terms here. These are all pure states (mixed states require a density operator approach). And what you are talking about are representations in different bases.

Where there is a transformation to go from $\hat A$ to $\hat B$ and vice versa
Is it possible?

It is not only possible, but mandatory. There is a theorem that proves that given two basis, there exists a unitary operator that transforms from one basis to the other.

 

[Karolus]

Great.
Mixed and pure state are my erroneous lexical interpretation.
Eventually, I will post in another thread