Does a wavefunction in a cartesian system create a one-dimensional subspace?

[SeM]

Hi, I have heard (or imagined) that a wavefunction, where Psi is on the y-axis and the positions x is naturally on the x-axis, is really a one-dimensional system in Physics (not in mathematics), because the signal or the oscillation of the wavefunction is not really a dimension, and only the position x makes any physical sense in a cartesian system . Is this correct?

If so, how can a wavefunction Psi(x) generate a one-dimensional subspace as such :

\begin{equation}
\mathcal{Y} = [ \phi | \phi = \beta \psi , \beta \in \mathbb{C} ]
\end{equation}
where \phi is a function of Y of norm 1 and beta is an arbitrary constant, thus defining the probability density of \psi?

 

[George Jones]

Hi, I have heard (or imagined) that a wavefunction, where Psi is on the y-axis and the positions x is naturally on the x-axis, is really a one-dimensional system in Physics (not in mathematics), because the signal or the oscillation of the wavefunction is not really a dimension, and only the position x makes any physical sense in a cartesian system . Is this correct?

No. "one-dimensional" refers to the real (as opposed to complex) dimension of domain of $\psi$. As I asked you in another thread, what does "domain" mean?

Also, if $\psi$ is complex-valued, how can it be on the y-axis?

 

[PeterDonis]

I have heard

Where? Do you have a reference?

 

[SeM]

No. "one-dimensional" refers to the real (as opposed to complex) dimension of domain of $\psi$. As I asked you in another thread, what does "domain" mean?

Also, if $\psi$ is complex-valued, how can it be on the y-axis?

Hi George, and thanks for your comment. The reason I ask this is because of a part in Kreyszig's Functional Analysis, in the chapter 11, "Unbounded linear Operators in Quantum mechanics.", p 574, writes (quote):

"Psi defines a state of our physical system at some instant to be an element :

\begin{equation}
\psi \in L^2(-\infty, \infty) \ \ \ \ \ \ \ \ \ (3)
\end{equation}

..., where \psi in (3) generates a one-dimensional subspace:

\begin{equation}
\mathcal{Y} = \phi | \phi = \beta \psi, \beta \in \mathbb{C}
\end{equation}

of $L^2(-\infty, +\infty)$. Hence we could equally well say that a state of our system (Psi) is a ONE-DIMENSIONAL subspace $Y \subset L^2(-\infty,+\infty) and then use a $\phi \in Y$ of norm 1 in defining a probability according to the probability density intergral.

end quoteHaving "heard" or imagined I can't remember which, that Psi on a REAL axis is not really a dimension in physics, and only the x-axis gives some dimensional concept (i.e. distance from nucleus for a bound electron), I thought this was what Kreyszig was talking about here. However, taking Peters word for it, I disregard from this misconception, and want to then ask:

What does Kreyszig really mean here with saying that Psi is a one-dimensional subspace of Y which si defined by the two other wavefunctions \phi ? Does he mean that the integral of some other wavefunction \phi|\phi in a one dimensional space equals to the full space of the wavefunction (L^2) multiplied by some constant \beta?

Thanks!

 

[blue_leaf77]

What does Kreyszig really mean here with saying that Psi is a one-dimensional subspace of Y

That's the dimension of this subspace, this subspace has only one basis which is $\psi$. Any vector in it must be a linear combination of basis vectors in the subspace but since there is only one such basis in the space under consideration, an arbitrary vector $\phi$ there is written as $\beta\phi$. I have not read the quoted book but I guess the author brings up this discussion in order to introduce the so-called normalized vectors contained in the square-integrable vector space.

 

[SeM]

I have not read the quoted book but I guess the author brings up this discussion in order to introduce the so-called normalized vectors contained in the square-integrable vector space.

That is correct.

Thanks for the extensive explanation. I see one recurring point, that a wavefunction can be presented as vectors, so in this case, those are the ones given in $\phi$ and $\beta$ is some arbitrary constant (such as a coefficient) and makes them a linear combination of $\psi$

 

[George Jones]

The title you chose for the thread is "One-dimensional systems".

It is not

Hence we could equally well say that a state of our system (Psi) is a ONE-DIMENSIONAL subspace $Y$

that indicates that the system is one-dimensional, it is

$\psi \in L^2(-\infty, \infty)$

Do you understand why?

 

[SeM]

Do you understand why?

Thanks for asking George. I see that $\psi$ is element of Hilbert space L^2, after Perok explained the nature of Hilbert space and its elements (functions), However, the elements of the functions can be represented as vectors, and are thus in $\mathbb{R}$ or $\mathbb{C}$. In this case, as Kreyszig says, I am not sure! But part of this suggests that a part of $\psi$ (COMPLEX part?) can be considered as one-dimensional and is in$\mathbb{C}$? Considering the post by Blue leaf, it must, in this assumption, be the complex vectors of $\psi$ (and the vectors are called $\phi$) which are in $\mathbb{C}$ and these vectors are one-dimensional, but NOT the function.

 

[thephystudent]

Aren't you just mixing up configuration space and Hilbert space?

A one-dimensional configuration space means that that there is only one spatial dimension/degree of freedom in play, meaning a system that would also be classically considered one-dimensional. This is the case if $\psi=\psi(x)$ without e.g. y or z coordinates.

Think for example of the standard particle in an infinite well. Even if we look at the standard case which is one-dimensional in the configuration space mentioned above, the wavevector-solutions form an infinite-dimensional Hilbert space.

And if you want to draw the wavefunction for such a system as a graph on a sheet of paper(which is 2D), both the real and the imaginary part can indeed be visualised where horizontally comes $x$ and vertically $Re[\psi(x)]$ or $Im[\psi(x)]$. But that is just the drawing, what you draw on the 'y-axis' has nothing to do with dimensionality of the system in any sense!

 

[SeM]

And if you want to draw the wavefunction for such a system as a graph on a sheet of paper(which is 2D), both the real and the imaginary part can indeed be visualised where horizontally comes $x$ and vertically $Re[\psi(x)]$ or $Im[\psi(x)]$. But that is just the drawing, what you draw on the 'y-axis' has nothing to do with dimensionality of the system in any sense!

This is what I "had heard" or "imagined" I heard, but I could not reproduce it accurately. The y-axis is not a physical dimension, as is x?

 

[thephystudent]

This is what I "had heard" or "imagined" I heard, but I could not reproduce it accurately. The y-axis is not a physical dimension, as is x?

It is a dimension on your drawing, but it doesn't describe a dimension. You even don't need QM to see this actually. Eg. if you would plot a graph of the temperature of the Earth's crust as function of depth, there is only one dimension (depth), temperature is not a dimension, but on a graph you would draw this temperature along a y-axis anyway.

 

[SeM]

It is a dimension on your drawing, but it doesn't describe a dimension. You even don't need QM to see this actually. Eg. if you would plot a graph of the temperature of the Earth's crust as function of depth, there is only one dimension (depth), temperature is not a dimension, but on a graph you would draw this temperature along a y-axis anyway.

Thanks, this is what I ment when I said that the oscillation of a wavefunction is not a spatial dimension, such as is the position.