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In general quantum mechanics we represent the state of a system with a state vector $| \psi \rangle $ in some Hilbert space in some base. Assuming a complete discrete set of bases vectors $ |n \rangle $ we can write the state $\psi $ as:

$$| \psi \rangle = \sum_n ^N c_n |n \rangle $$

Having by definition an inner product we can see that $c_n = \langle \psi | n \rangle $. This is the projection of the state vector on the base vector $n$. This coefficient squared, we say, gives us the probability of finding out system in the state $n$.

As I understand so far $ c_n $ is a complex number. Also, as a function, $\psi$ is a function of variable same as the variable of $n$, where $n$ are also functions.

In position representation we have a base denoted as $| x \rangle $. So to the question.

I would say that:

$$| \psi \rangle = \sum_n ^N c_n |x \rangle $$

$$c_n = \langle x| \psi \rangle $$ where $|x \rangle =f(x) $ so that, in this bases $| \psi \rangle = \psi (x)$, x meaning position.

My problem is that from what I read, what holds is this:

$$c_n = c_n (x) = \langle x | \psi \rangle = \psi (x) ,$$ $ \psi (x)$ being the wavefunction- the projection of $|\psi \rangle $ on $|x \rangle $ and

$$\psi(x) = \sum_n ^N c' _n u_n (x) $$, where $ u_n (x)$ constitute a base for the position representation. So, the probability of finding the particle in position $x$ is $| \psi (x) |^2 $. But I thought that $|x \rangle $ was already our base in space.

Are $u(x) $ and $| x \rangle $ the same or not? If not, why do we need $u(x)$?

Is $\psi (x) $ as given in the end by the inner product a function that can be expressed in functions of $x$ that constitute a base or is it a number as a result of an inner product giving the coefficient in a sum? Can the coefficients be functions and if yes, when? Is it correct to say that the wavefunction in secondary to the state vector and if so, how is it a function if it is a coefficient?

There is certainly something here I don't get and I would really like your help and clarifications.

Note: I wrote the question with sums and not integrals for convenience. An answer might have as well integrals, there is clearly no problem. Also I' ve had a look in these questions and answers:

General wavefunction and Schrödinger Equation

State of a system in Quantum Mechanics and state vectors

Representations in quantum mechanics

Vector representation of wavefunction in quantum mechanics?

but I'm not sure they address my problem.

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  • $\begingroup$ The object $\lvert x\rangle$ is not actually a state in the Hilbert space, but lies outside of it. There are no eigenvecturs of the position operator inside the space since its spectrum is fully continuous. The proper way to deal with what the $\lvert x \rangle$ are rigged Hilbert spaces $\endgroup$ – ACuriousMind Nov 8 '15 at 21:04
  • $\begingroup$ @ACuriousMind . Thanks for the reply. I thought that $|x \rangle $ define a base and also that the inner product is defined inside the space. Could you elaborate on these two? $\endgroup$ – Constantine Black Nov 8 '15 at 21:08
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Your confusion stems from the fact that $\lvert x \rangle$ is not inside the Hilbert space of states. It cannot be because $\langle x \vert x \rangle = \delta(x-x) = \delta(0)$ is not an allowed value for an inner product in a Hilbert space to have. There are several things to say about $\lvert x \rangle$:

  1. If you want to make precise what kind of objects $\langle x \rvert$ and $\lvert x \rangle$ are, you need the notion of rigged Hilbert spaces. A nice answer about them is here by user1504.

  2. In general, the "eigenstates" associated to eigenvalues in the continuous spectrum of an unbounded operator do not lie inside the space of states. Only the discrete eigenvalues have proper eigenvectors.

  3. The set of objects $\lvert x \rangle$ is uncountable and hence not a basis in the usual sense. In particular, writing $\sum_x c_x \lvert x \rangle$ doesn't make sense because series over uncountable sets do not converrge. The analogon for $\lvert \psi =\sum_n c_n\lvert n \rangle$ for a countable Hilbert basis $\lvert n \rangle$ is writing $$ \lvert\psi\rangle = \int \psi(x)\lvert x \rangle \mathrm{d}x\tag{1}$$ where writing $\psi(x) = \langle x \vert \psi \rangle$ leads to $$ 1 = \int \lvert x \rangle \langle x \rvert \mathrm{d}x$$ It is in the sense of $(1)$ that the wavefunction gives the coefficients in the position basis.

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  • $\begingroup$ Does the coefficient $\psi (x) $ inside the integral becomes a function because of having an integral and not a sum ( where the coefficients are numbers)? Is that correct? If so, what are the vectors x inside the integral? If the second question can be answered by the link proposed please tell. Thank you. $\endgroup$ – Constantine Black Nov 8 '15 at 21:42
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    $\begingroup$ @ConstantineBlack: Yes, what the $\lvert x \rangle$ are is that they're elements of the rigged space. They're defined by $\vert x_0 \rangle$ being the linear functional that maps a function $\psi\in S$ (where $S$ is the Schwartz space) to the value $\psi(x_0)$. $\psi(x)$ is the value of the function $\psi$ at $x$ - you need $\psi$ to be a Schwartz function to be able to meaningfully talk about $\lvert x \rangle$ acting on it. $\endgroup$ – ACuriousMind Nov 8 '15 at 21:47
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    $\begingroup$ @ACuriousMind Did Dirac know about the rigged space when he set up the formalism his Principles? The reason I interfered after you already commented is because I don't think it is productive to confuse an obvious beginner at this point in his introduction to QM. It is one thing to keep him curious (pun intended) and another to overwhelm his intellectual appetite for the sake mathematical refinement. $\endgroup$ – udrv Nov 8 '15 at 22:49
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    $\begingroup$ @udrv: I don't know whether Dirac knew about this. Yet that the $\lvert x \rangle$ aren't actual states leads repeatedly to confusions when students try to calculate something and do operations that'd be fine with actual vectors, but aren't with position kets. I'm really not a fan of the "let's pretend this is linear algebra until something goes wrong" approach to QM. If you think that you have a better approach to answering such questions, that's why you can to post your own answers - I certainly don't claim that mine are perfect, or the only way to adress the issue at hand. $\endgroup$ – ACuriousMind Nov 8 '15 at 22:59
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    $\begingroup$ @ACuriousMind No, Dirac didn't know. And nobody has to play the linear algebra game badly if we set up the correct basic rules of calculating with the $|x\rangle$'s and $\delta$'s just as he did. It's not "my way", to be sure. But it goes awry especially when said basic rules are not properly emphasized or get drenched and obscured by finer details. $\endgroup$ – udrv Nov 9 '15 at 0:14
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Assuming everything is defined in the correct Hilbert spaces, project the decomposition $$ |\psi\rangle = \sum_n {c_n |n\rangle} $$ onto the position kets ("states") $|x\rangle$ and obtain $$ \psi(x) = \langle x |\psi\rangle = \sum_n {c_n \langle x |n\rangle} = \sum_n {c_n u_n(x)} $$ where the $u_n(x) = \langle x |n\rangle$ are the wavefunctions corresponding to the $|n\rangle$ states. So the result is a decomposition of a wave function into basis wave functions with coefficients corresponding to the decomposition of the state into basis states. What happened is that we subtly changed the representation space.

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  • $\begingroup$ Thanks for the answer. If I understand correctly, the vectors $| x \rangle $ don' t constitute a base for the state vector $| \psi \rangle $? $\endgroup$ – Constantine Black Nov 8 '15 at 21:32
  • $\begingroup$ It's a bit complicated. The $|x\rangle$ were introduced as formal basis kets of the position representation by Dirac, who then had to invent the $\delta$ function distribution and reconstruct a consistent mathematics around it and various singularities. Dirac wrote the expansion of $|\psi\rangle$ as $|\psi\rangle = \int{dx \;\psi(x) |x\rangle}$, but as ACuriousMind has pointed out, on closer examination one has to carefully define the Hilbert spaces where various objects live. For beginner's purposes though, the issue can wait and you can use the $|x\rangle$ as any formal basis for now. $\endgroup$ – udrv Nov 8 '15 at 22:41

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