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The state of a quantum system is represented by a wavefunction usually in some specific Hilbert space, .e.g of position, spin, momentum etc.

  1. But before deciding in which of these bases to decompose the wavefunction in, what form does the wavefunction have? Of course not having decomposed it yet, it does not yet have a component-wise definition. To clarify with example, we write: $$|\psi(x)\rangle = \langle x|\psi\rangle|x\rangle$$ But what form does this mysterious $\psi$ have before the x-basis representation? How is the inner product above even allowed if $\psi$ is not yet defined in some other basis?

  2. On a related note: does the Schrodinger equation have other representations other than its position and momentum forms? I mean e.g. can one obtain the spin wavefunction $\psi_S$ of a system directly from the Schrodinger equation? Or first $\psi_x$ is obtained and then decomposed in a spin basis?

  3. A last question regarding the wavefunction for more than 1 dimension: When one writes $\psi(x,y)$ do $x$ and $y$ correspond to different Hilbert spaces $|x\rangle$ and $|y\rangle$? In which case we'd write $\psi(x,y) \in |x\rangle \otimes |y\rangle $ does this mean that a single particle in 2 dimensions looks very much like 2 particles each in one dimension? Or one should never mix multidimensionality with multi-body systems in QM?

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    $\begingroup$ Your equation should read: $|\psi(x)\rangle = \int dx\langle x|\psi\rangle|x\rangle$. $\endgroup$ – march Sep 3 '15 at 20:51
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    $\begingroup$ @march Actually, the correct expression would be writing $| \psi \rangle$ at the left hand side (without the $x$ in $\psi(x)$, because $| \psi \rangle$ does not depend on x, but when you project at the position space you have: $\psi(x):=\langle x| \psi \rangle$). $\endgroup$ – Ana S. H. Sep 4 '15 at 5:03
  • $\begingroup$ @Timaeus is right, there aren't lots of different Hilbert spaces, but there are lots of different basis, say $\{|x\rangle\}$ or $\{|p\rangle\}$. And BTW, $\psi(x,y):=\langle x, y |\psi \rangle$. $\endgroup$ – Ana S. H. Sep 4 '15 at 5:10
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    $\begingroup$ @Anuar. All too true! That's an important point. Didn't pay enough attention to what I was writing. It reminds me of a (unintentionally nonsensical) homework problem I wrote once about using the 2nd Fundamental Theorem of Calculus. $\endgroup$ – march Sep 4 '15 at 5:30
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Was typing this halfway through when Timaeus posted his answer. There is some overlap, but also some potentially useful additional details. Hope it helps.

  1. One fundamental assumption (axiom, postulate, principle, etc) of Quantum Mechanics is that the state of a system is represented by a state vector $|\psi \rangle$ in a unique Hilbert space ${\mathcal H}$ that is "rich enough" to comprise all possible states of the system. Note that a Hilbert space comes equipped with an inner product between its vectors, in this case the famed $\langle \phi | \psi \rangle$ for $|\phi\rangle, |\psi \rangle$ any two state vectors in ${\mathcal H}$. This answers your question about the usage of the inner product. As for the state vector, there is nothing else to "the form" of the state vector "before any decomposition". Think about it as a formal object that encapsulates all the information about the state of the system. The real deal is how we do the encapsulation and how we can use it to do some real physics. It sounds a bit circular, and we need to cut the Gordian knot. In QM this is done by defining observables, eigenvalues, and eigenstates, and characterizing the Hilbert space of states in terms of a complete set of observables and their common eigenstates. If it sounds very formal, it is. But it works.

So, before we get to the part on $\psi(x)$, we need to clarify the part about observables in general. The fundamental assumption of QM regarding observables is that every observable O is represented by a self-adjoint operator ${\hat O}$ on ${\mathcal H}$, that is ${\hat O}:{\mathcal H} \rightarrow {\mathcal H}$, ${\hat O} = {\hat O}^\dagger$. It is further postulated that the average value of O in any state $|\psi \rangle$ is given by the matrix element $\langle \psi | {\hat O} | \psi \rangle = \langle \psi | {\hat O} \psi \rangle$. The fact that ${\hat O}$ is postulated self-adjoint, has several dramatic consequences:

1) If ${\hat O}$ is self-adjoint, it is already implied that there exists in ${\mathcal H}$ a basis of eigenstates $|\omega \rangle$ of ${\hat O}$, for which ${\hat O}|\omega \rangle = \omega |\omega \rangle$.

2) If the $| \omega \rangle$-s form a basis, it follows that any state vector $|\psi \rangle$ can be expressed as a superposition $|\psi \rangle = \sum_{\omega} {c_\omega |\omega \rangle }$.

3) From the existence of decompositions $|\psi \rangle = \sum_{\omega} {c_\omega |\omega \rangle }$ it follows that for an arbitrary $|\psi \rangle$ the standard deviation associated to the average value of O on $|\psi \rangle$ is in general non-zero, $\langle \psi | (\Delta {\hat O})^2 | \psi \rangle = \langle \psi | {\hat O}^2 | \psi \rangle - \langle \psi | {\hat O} | \psi \rangle^2 ≠ 0$, but for the eigenstates of O we find $\langle \psi | (\Delta {\hat O})^2 | \psi \rangle = \omega^2 - \omega^2 = 0$. In other words, just from the self-adjointness of ${\hat O}$ we have that in any state $|\omega \rangle$ observable O has a well-defined, sharp value $\langle \omega | {\hat O} | \omega \rangle = \omega$.

4) Given the above, it follows further that two observables A and B cannot produce sharp values simultaneously unless their corresponding operators commute and admit a common set of eigenstates. It is not difficult to prove, but I will not go into details. Suffice it to emphasize that this is an expression of the uncertainty principle and everything follows from the mere self-adjointness of observables on the Hilbert space ${\mathcal H}$.

Now, how does this help us with the $\psi(x)$ problem and everything? It is actually the crux of it, since the Hilbert space of states ${\mathcal H}$ is always defined in terms of the system's degrees of freedom, which are nothing but a complete set of commuting observables. In general, if a system is completely characterized by degrees of freedom (or observables) $Q_1$, $Q_2$, ..., $Q_n$, then each of its states is necessarily labeled by a corresponding set of values $\{q_1, q_2, ..., q_n \}$ that can be measured simultaneously. According to the fundamental assumptions of QM, this means that $Q_1$, $Q_2$, ..., $Q_n$ must necessarily be represented on ${\mathcal H}$ by mutually commuting self-adjoint operators ${\hat Q_1}$, ${\hat Q_2}$, ..., ${\hat Q_n}$ that admit a common set of eigenstates labeled by $\{q_1, q_2, ..., q_n \}$, say $|q_1, q_2, ..., q_n \rangle$. All is good so far, but how do we get such $|q_1, q_2, ..., q_n \rangle$ in the first place? Simple: we postulate them. There is literally nothing else that we can do. But once we do this, we are left with an airtight definition of the Hilbert space of states ${\mathcal H}$ and we can even do physics.

To get to the actual heart of the matter: Let's take a spinless particle on a 1D line. It has one degree of freedom, which we can take to be the position $x$ along the line. Observable $x$ must be represented in the Hilbert space of the particle by a self-adjoint operator ${\hat x}$ that generates a basis set $|x\rangle$. The states $|x \rangle$ must be states in which the particle is found with certainty at position $x$, such that ${\hat x} | x \rangle = x | x \rangle$. Conversely, if we postulate the states $|x \rangle$ as a basis set, we have defined ${\mathcal H}$. The important thing is that now any state $|\psi \rangle$ of the particle can be represented as a superposition

$$ | \psi \rangle = \int{dx\; \psi(x) |x \rangle}, \;\; \text{with}\;\;\psi(x) = \langle x | \psi \rangle $$

Of course, we can also define the Hilbert space in terms of the particle's momentum $p$ and obtain a similar decomposition in terms of momentum eigenstates $|p \rangle$. Is this Hilbert space different from the one before? No. Since the Hilbert space comprises all possible states of the particle, the momentum eigenstates $|p \rangle$ must admit a decomposition in terms of position eigenstates $|x \rangle$ and vice-versa. Once the plane-wave form of the wavefunction $\langle x | p \rangle$ is defined, all rules about decomposing a state as a superposition of basis states apply as before.

  1. Are there other representations than position or momentum? Absolutely! Think energy eigenstates in the hydrogen atom. The bound states are labeled by energy, angular momentum, azimuthal angular momentum, and spin. The corresponding wave functions are decompositions in the position representation. However, the spin component comes as an additional degree of freedom, not from the Schroedinger equation itself. Once the existence of spin is acknowledged, the Hilbert space is extended to account for spin states. And this brings us to the last question.

  2. The essence of your question is: given a Hilbert space for a certain number of degrees of freedom, how is it extended if we need to account for additional degrees of freedom? Take for instance the 1D particle. What if the particle has spin? It obviously keeps the $x$ degree of freedom, but we need to account in addition for the spin degree of freedom. We can build new states $|x, \sigma \rangle$ that locate a particle with spin $\sigma$ (along some given direction) at position $x$, rebuild the Hilbert space, and think about $x$ and $\sigma$ as labels, the way Timaeus suggested. But the reality is that the Hilbert space so extended is isomorphic to the direct product of the spinless Hilbert space by a new Hilbert space corresponding to the spin degree of freedom. In fact, the direct product form is the preferred one for a variety of reasons. This is why we can and do commonly write something like $|\Psi \rangle = |\psi \rangle \otimes |\sigma \rangle$, where $|\psi \rangle$ is a spinless state vector.

So can we think the same way about spatial degrees of freedom like $x$ and $y$? Formally yes. Although you can view $x$ and $y$ as labels for a 2D particle, there is also an isomorphism with a system of two distinguishable 1D particles. Similarly, a system of two distinguishable 3D particles is isomorphic to one particle living in a 6D configuration space. But unlike for spin, it is commonly more convenient to use the label view, especially when dealing with coordinate changes to curvilinear coordinates. The isomorphism is still there, but is not invoked.

Additional info based on comments:

Are Hilbert spaces for different degrees of freedom (DOFs) isomorphic? If so, don't we need just one copy in the total Hilbert space? The Hilbert spaces corresponding to dissimilar DOFs, like position and spin, may have very different cardinality (dimension) and thus need not be isomorphic to each other. In general, even if some DOFs do generate formally isomorphic Hilbert spaces, like the $x$ and $y$ coordinates, the total Hilbert space of the system must account for each DOF individually. It is always a direct product of the Hilbert spaces for each of the DOFs. For a 1D particle with spin, it is ${\mathcal H}_x \otimes {\mathcal H}_\sigma$. For a 2D particle with spin, it is ${\mathcal H}_x \otimes {\mathcal H}_y \otimes {\mathcal H}_\sigma$ or ${\mathcal H}_{x,y} \otimes {\mathcal H}_\sigma$, but no longer just ${\mathcal H}_x \otimes {\mathcal H}_\sigma$.

How do we account for non-commuting observables? Take $x$ and $p$ for the 1D particle. There are no common $|xp \rangle$ eigenstates ($[x,p] ≠0$), but the spinless ${\mathcal H}$ is completely characterized by means of either the $\{ |x \rangle \}$ or the $\{ |p \rangle \}$ basis. Since ${\mathcal H}$ must contain all possible (spinless) states, if we choose the $\{ |x \rangle \}$ basis then we must express the $ |p \rangle $ states in terms of the $ |x \rangle $ states, and conversely. This is done by specifying the (canonical) commutation relation between $x$ and $p$, $[x, p] = i\hbar$, which leads to an expression for the momentum wave functions $\langle x | p \rangle$.

For a complete set of commuting observables, why do we keep only the common eigenstates? This is simply because the "common eigenstates" include all the eigenstates for the every observable in the complete set. In other words, given any observable ${\hat O}$ from the complete set, there are no eigenstates of ${\hat O}$ that are not in the set of "common eigenstates". For an observable ${\hat A}$ that is not among those in the complete set, the situation depends on whether it commutes or not with those in the complete set, but the result is the same. If it commutes with the entire set (think any $f({\hat O})$ ), its eigenstates are still those of the complete set. If it doesn't commute, (at least some of) its eigenstates are not in the set's common eigenstates, but can always be expressed as superpositions of those common eigenstates.

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  • $\begingroup$ Thanks a lot, clarified tons for me. (1): So the Hilbert space is spanned by the common set of eigenstates of all the commuting observables of a system at hand: meaning if we have position x and spin S to consider, the complete Hilbert space is $H = H_x \otimes H_S$, correct? But now I'm confused again, because you said that e.g. the x Hilbert space comprises all possible states, so we can transform $H_x \to H_S$ and obtain all the spin states! Doesn't this make $H = H_x \otimes H_S$ redundant? because I can always use just $H_x$ and if I need another DOF $x'$ I transform $H_x \to H_{x'}$ $\endgroup$ – user098876 Sep 4 '15 at 13:39
  • $\begingroup$ Anothere closely related question, (2): How is the complete Hilbert space defined when we have non-commuting observables, like x and p? Surely one cannot find a common set of eigenstates anymore! Thanks a bunch in advance for any further clarifications. $\endgroup$ – user098876 Sep 4 '15 at 13:43
  • $\begingroup$ On ${\mathcal H}_x \rightarrow {\mathcal H}_S$ isotropy: It most definitely does not exist! ${\mathcal H}_x$ and ${\mathcal H}_S$ concern not only different, but dissimilar DOFs and have different cardinality (dimension). ${\mathcal H}_x$ does contain all possible states of the particle as far as dependence on $x$ is concerned, but the defining basis states $| x \rangle$ do not depend on or even refer to spin. To cover spin we must define spin states $|\sigma \rangle$ and extend the total Hilbert space as a tensor product ${\mathcal H}_x \otimes {\mathcal H}_S$. $\endgroup$ – udrv Sep 4 '15 at 23:35
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    $\begingroup$ On non-commuting observables: Very good question. Take $x$ and $p$ for the 1D particle. There are no common $|xp \rangle$ eigenstates ($[x,p] ≠0$), but the spinless ${\mathcal H}$ is completely characterized by means of either the $\{ |x \rangle \}$ or the $\{ |p \rangle \}$ basis. Since ${\mathcal H}$ must contain all possible (spinless) states, if we choose the $\{ |x \rangle \}$ basis then we must express the $ |p \rangle $ states in terms of the $ |x \rangle $ states, and conversely. $\endgroup$ – udrv Sep 4 '15 at 23:38
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    $\begingroup$ Please note that the "complete set of observables" is usually not unique, there may be many ways to choose different sets, all equivalent to each other. For instance, a 3D spin 1/2 particle can be formally characterized in terms of position and spin projection along x, ${\hat {\bf r}}$ and ${\hat \sigma_x}$. But it is often convenient to characterize it in eigenstates of momentum and spin along z, ${\hat {\bf p}}$ and ${\hat \sigma_z}$. The two sets, ${\hat {\bf r}}$, ${\hat \sigma_x}$ and ${\hat {\bf p}}$, ${\hat \sigma_z}$, are mutually non-commuting, but provide equivalent descriptions. $\endgroup$ – udrv Sep 5 '15 at 17:54
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The state of a quantum system is represented by a wavefunction usually in some specific Hilbert space, .e.g of position, spin, momentum etc.

I'm not sure it is helpful to think of lots of different Hilbert Spaces. This would be like thinking of having lots of different 3d vector spaces, one for each choice of basis. Just think of a nice Hilbert space with vectors that can be added and scaled and have an inner product.

But before deciding in which of these bases to decompose the wavefunction in, what form does the wavefunction have?

Would you ask the same thing about a vector in 3d space? It would be a vector. And addition is a function that take two vectors and gives a third. And scaling is a function that takes a a scalar and a vector and gives a vector. And the scalar product is a function that takes two vectors and gives a scalar. Same with a Hilbert Space. A Hilbert Space is just a Vector Space with a particular property that automatically holds for finite dimensional spaces (that it lacks holes, that if a sequence acts like it has a limit then there is a thing it limits to). So they are designed to be exactly the closest thing possible to tour intuitions about vector spaces. Not every intuition will carry over but it's the best you can do.

But what form does this mysterious $\psi$ have before the x-basis representation?

It is a vector. For instance you might say that a vector is invariant under a rotation which means it is eigen to that rotation. So you might describe a vector by what it is eigen to. Or you might say how it is a particular linear combination of vectors eigen to particular operators. And you can do the same in a Hilbert space.

How is the inner product above even allowed if $\psi$ is not yet defined in some other basis?

Is the scalar product not defined in 3d until after you pick a basis? Or is it that some vectors are already orthogonal before you picked a basis. And some of are unit length? Well now you can compute inner products fr knowing that.

On a related note: does the Schrodinger equation have other representations other than its position and momentum forms?

Yes. You could pick any reasonable operator and then express things as linear combinations of some eigenvectors of that operator. Just like you could do in 3d and just like you can do with the position operator or the momentum operators.

I mean e.g. can one obtain the spin wavefunction $\psi_S$ of a system directly from the Schrodinger equation?

Maybe Schrödinger-Pauli you need something, like a Hamiltonian with a spin dependence to have it show up in the Schrödinger equation at all.

Or first $\psi_x$ is obtained and then decomposed in a spin basis?

I don't see two choices. You can pick basis before or afterwards. Just like for a rotation, you could pick a basis and write a matrix or you can have an operator that is a function that takes vectors and gives other vectors. If you do the latter then you can have operators first and a basis later. Then you can have spin operators and pick your basis later.

A last question regarding the wavefunction for more than 1 dimension: When one writes $\psi(x,y)$ do $x$ and $y$ correspond to different Hilbert spaces $|x\rangle$ and $|y\rangle$?

Not really. And not just because as I've pointed out that these are all the same Hilbert space, just with different basis. And part of the objection is that it is a bit vague. When you write $|x\rangle$ what you mean is that you have a particular eigenvector of the operator $\hat x.$ And you should have had a specific one in mind before you assigned that label to it because even if you pick a unit length one any unit scalar multiple will still be eigen and will still be unit length. But worse there could be a whole 2+ dimensional subspace of eigenvectors with that eigenvalue.

One thing you can do is choose a bunch of operators hat all commute with each other such that any two distinct vectors that are eigen to all of the operators has a distinct set of eigenvalues. Then you could label those vectors by all the eigenvalues.

So $|x,y,z\rangle$ could be eigen to $\hat x,$ to $\hat y,$ and to $\hat z.$ It is one vector (sort of, but it might be easiest to think it is) in one Hilbert space that is eigen to multiple operators. And then you write $\left|\psi\right\rangle =\int dxdydz\left\langle x,y,z\right|\psi\rangle\left|x,y,z\right\rangle.$ And that can be confusing. The integral has nothing to do with the inner product it is just to indicate that there are too many eigenvectors to do a sum. So really it is like a linear combination where the $\left\langle x,y,z\right|\psi\rangle$ are the coefficients. And those (the $\left\langle x,y,z\right|\psi\rangle$) do come from the inner product. And finally, you can then write $\psi(x,y,z)=\left\langle x,y,z\right|\psi\rangle$

Or one should never mix multidimensionality with multi-body systems in QM?

You should definitely not confuse them. But there are some similarities.

when one talks of a Hilbert space in QM, it is always assumed that it is the Hilbert space of some system's observable in mind?

Yes and no. In principle there is a state and the state is the same regardless of the observables you consider. However sometimes you don't write the whole state. For instance there is a spatial state and a spin state. But sometime people don't write one or the other. And for instance a Stern-Gerlach "measures" by changing the state so that the spatial state becomes entangled with the spin state (which then allows other things to become entangled with the spatial part, and then other things to become entangled with that thing and so on). But if someone is just writing the spin state you ignore that. And then just project the spin states onto the eigenspaces and ignore the time it took or the change in the spatial part.

and not an abstract observable-independent space!

The state is generally sufficient for any observable. Picking an observable is like picking a basis. It can give you coefficients, but the same state can give you coefficients for any basis and for any observable.

Finally I didn't understand your comment to the "Or first $\psi_x$ is obtained and then decomposed in a spin basis?" question, would you be so kind to expand on it a bit?

The full state includes the spatial part and the spin part. But then works for any observable. After you have the state you could decompose it into position or momentum (which just depend on the spatial part) or spin components for any axis (which just depends on the spatial part) or on energy (which could depend on both).

You can't get spin from position and you can't get position from spin. When you have both however, you have everything. And you start with everything. Your everything is a vector in a Hilbert space. You can have the vector and then decompose the spatial part in a position basis or a momentum basis. And you could decompose the spin part in a z basis or an x basis or in a y basis or a basis for any other fixed direction in 3d.

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  • $\begingroup$ Thanks for writing this. One question that comes to mind: so when one talks of a Hilbert space in QM, it is always assumed that it is the Hilbert space of some system's observable in mind? and not an abstract observable-independent space! Finally I didn't understand your comment to the "Or first $\psi_x$ is obtained and then decomposed in a spin basis?" question, would you be so kind to expand on it a bit? $\endgroup$ – user098876 Sep 4 '15 at 13:49
  • $\begingroup$ @user098876 Edited. $\endgroup$ – Timaeus Sep 6 '15 at 10:36

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