- #1
Unkraut
- 30
- 1
A friend of mine who studies physics recently tried to explain some basic quantum mechanics to me. He told me quite a lot of interesting stuff but it got me confused somehow and I think I really didn't get it all. Thus I am here to recapitualte my newly gained knowledge and ask questions.
What I have learned is this:
1. The state of a quantum mechanical system is described as a vector in a Hilbert space, written in this weird notation |v>
2. For each physically measurable variable (like energy, momentum or the like) there exists an operator which describes the results of measurement of that variable in this strange way:
If the system is in the state |v>, and |n> is an eigenvector to the operator H with eigenvalue E_n the probability that E_n will be the result of measurement is |<n|v>|². By measurement the system's state is changed into the corresponding eigenvector to the result.
What I do not yet understand is what exactly that has to do with wave functions. My friend wasn't able to make that clear to me.
Let's say we have an electron in free space and we are only interested in its position and momentum. So we have a complex-valued wave function for the position, [itex]\phi: \mathbb{R}^{3+1} \to \mathbb{C}[/itex], and one for the momentum [itex]\varphi: \mathbb{R}^{3+1} \to \mathbb{C}[/itex], which we can put together to one function depending on one time coordinate and three spatial coordinates each for momentum and position: [itex]\psi: \mathbb{R}^{6+1} \to \mathbb{C}, (x_p, y_p, z_p, x_m, y_m, z_m, t) \mapsto \phi(x_p, y_p, z_p, t)\varphi(x_m, y_m, z_m, t)[/itex]. The state at a particular time t is then described by a wave function obtained from [itex]\psi[/itex] by setting t=t_0 to be some constant: [itex]\psi_{t_0}:=\psi_{|t=t_0}[/itex].
I am guessing now that the Hilbert space discussed above is just the space of wave functions like the latter. Am I right? Clearly [itex]\int_{\mathbb{R}^{6n}} |\psi_{t_0}|^2 d(x_p, y_p, z_p, x_m, y_m, z_m)=1[/itex] must hold for all wave functions since their squares are meant to describe probability distributions. Thus the space of such wave functions is a Hilbert space with the inner product given by [itex]<\phi, \psi>=\int_{\mathbb{R}^{6n}} \phi(x)\overline\psi(x)dx[/itex].
If I look at it this way all that arcane stuff looks quite straightforward to me and gives me a picture consistent with many things I have heard of but never understood.
So for example, if I perform a measurement of, say, the position of the electron in our system, the wave function "collapses" and the state [itex]\psi_{t_0}[/itex] is suddenly transformed into a different state [itex]\psi'_{t_0}[/itex] which is an eigenvector of the position operator. But what about the time evolution of the changed wave function now? Where do we get that from? Or can't we? Just having determined the position of our electron, its momentum must be completely unknown according to Heisenberg's relation of uncertainty. So how does our new wave_function [itex]\psi'_{t_0}[/itex] look actually? It can't be a wave anymore because the probability of the particle being at the position we measured is 1 and the probability of it being anywhere else is 0. And the momentum has to be completely unknown with equal probability for every possible value. The wave function of the momentum alone would have to be a constant function then, but which constant? The integral of the momentum's absolute squared "wave function" over whole 3-space would have be infinity then, and the integral of the positions absolute squared wave function would be zero because the wave function has to be zero for every single value except the one we measured. Is it here where Dircac's "distributions" kick in, which I have heard of but don't understand?
But let's go back to the stuff my friend told me: The measurement puts the system in an eigenstate of the measured observable's operator. This state certainly has to be an element of the Hilbert space we are talking about all the time. But obviously if I think of the elements of that Hilbert space just as wave functions like I did above there occur problems, because after measurement our wave function is not even a function anymore. I'm really getting confused here. It looked so straightforward when I thought the vectors in our Hilbert space were just the wave functions. But now I see that performing a measurement which, according to what my friend told me, should transform the former state of the system to an eigenstate of the measurement operator, in fact would transform such a wave function to something which is not even a function anymore.
Can someone try to explain the dirty details to me, please?
Regards,
Unkraut
What I have learned is this:
1. The state of a quantum mechanical system is described as a vector in a Hilbert space, written in this weird notation |v>
2. For each physically measurable variable (like energy, momentum or the like) there exists an operator which describes the results of measurement of that variable in this strange way:
If the system is in the state |v>, and |n> is an eigenvector to the operator H with eigenvalue E_n the probability that E_n will be the result of measurement is |<n|v>|². By measurement the system's state is changed into the corresponding eigenvector to the result.
What I do not yet understand is what exactly that has to do with wave functions. My friend wasn't able to make that clear to me.
Let's say we have an electron in free space and we are only interested in its position and momentum. So we have a complex-valued wave function for the position, [itex]\phi: \mathbb{R}^{3+1} \to \mathbb{C}[/itex], and one for the momentum [itex]\varphi: \mathbb{R}^{3+1} \to \mathbb{C}[/itex], which we can put together to one function depending on one time coordinate and three spatial coordinates each for momentum and position: [itex]\psi: \mathbb{R}^{6+1} \to \mathbb{C}, (x_p, y_p, z_p, x_m, y_m, z_m, t) \mapsto \phi(x_p, y_p, z_p, t)\varphi(x_m, y_m, z_m, t)[/itex]. The state at a particular time t is then described by a wave function obtained from [itex]\psi[/itex] by setting t=t_0 to be some constant: [itex]\psi_{t_0}:=\psi_{|t=t_0}[/itex].
I am guessing now that the Hilbert space discussed above is just the space of wave functions like the latter. Am I right? Clearly [itex]\int_{\mathbb{R}^{6n}} |\psi_{t_0}|^2 d(x_p, y_p, z_p, x_m, y_m, z_m)=1[/itex] must hold for all wave functions since their squares are meant to describe probability distributions. Thus the space of such wave functions is a Hilbert space with the inner product given by [itex]<\phi, \psi>=\int_{\mathbb{R}^{6n}} \phi(x)\overline\psi(x)dx[/itex].
If I look at it this way all that arcane stuff looks quite straightforward to me and gives me a picture consistent with many things I have heard of but never understood.
So for example, if I perform a measurement of, say, the position of the electron in our system, the wave function "collapses" and the state [itex]\psi_{t_0}[/itex] is suddenly transformed into a different state [itex]\psi'_{t_0}[/itex] which is an eigenvector of the position operator. But what about the time evolution of the changed wave function now? Where do we get that from? Or can't we? Just having determined the position of our electron, its momentum must be completely unknown according to Heisenberg's relation of uncertainty. So how does our new wave_function [itex]\psi'_{t_0}[/itex] look actually? It can't be a wave anymore because the probability of the particle being at the position we measured is 1 and the probability of it being anywhere else is 0. And the momentum has to be completely unknown with equal probability for every possible value. The wave function of the momentum alone would have to be a constant function then, but which constant? The integral of the momentum's absolute squared "wave function" over whole 3-space would have be infinity then, and the integral of the positions absolute squared wave function would be zero because the wave function has to be zero for every single value except the one we measured. Is it here where Dircac's "distributions" kick in, which I have heard of but don't understand?
But let's go back to the stuff my friend told me: The measurement puts the system in an eigenstate of the measured observable's operator. This state certainly has to be an element of the Hilbert space we are talking about all the time. But obviously if I think of the elements of that Hilbert space just as wave functions like I did above there occur problems, because after measurement our wave function is not even a function anymore. I'm really getting confused here. It looked so straightforward when I thought the vectors in our Hilbert space were just the wave functions. But now I see that performing a measurement which, according to what my friend told me, should transform the former state of the system to an eigenstate of the measurement operator, in fact would transform such a wave function to something which is not even a function anymore.
Can someone try to explain the dirty details to me, please?
Regards,
Unkraut