A What happens when an operator maps a vector out of the Hilbert space?

[gaiussheh]

This question is closely related to my previous thread mentioning that a linear operator can map a ket out of the original Hilbert space. That example was about infinite squares well, so it may be seen as an artificial example. More recently, I came up with a more "natural" example that does not involve any potential at all.

Consider a wave function $\psi[x]=\frac{1}{\sqrt{1+|x|^3}}$. It is surely integrable, $\displaystyle \psi[x]=\int \frac{\mathrm{d} x}{1+|x|^3}<+\infty$. With the position operator $x$ on it, one have $\displaystyle x\psi[x]=\frac{x}{\sqrt{1+|x|^3}}$, which is not in the Hilbert space anymore $\displaystyle \int \frac{x^2\mathrm{d} x}{1+|x|^3}=+\infty$. In this example, it seems that $\psi$ is not in the domain of the position operator. I find this explanation somewhat funny as it was telling me that its position is no longer obseravle. This can't be right. You can integrate $\displaystyle \langle x \rangle = \int \psi^* x \psi \mathrm{d} x = \int \frac{x\mathrm{d} x}{1+|x|^3}$ and the integration is regular.

I just wonder why QM requires the result of an operator acting on a ket always be a ket, and for this sake even kicking some very regular states out of the domain of the operator.

 

[PeroK]

That's simply not a valid wave function. Square integrability is not sufficient.

An operator on the set of square integrable functions must produce a square integrable function. If it does not, then it is not an operator on that Hilbert space.

You must respect the mathematical definitions.

 

[PeroK]

I just wonder why QM requires the result of an operator acting on a ket always be a ket, and for this sake even kicking some very regular states out of the domain of the operator.

The mathematics on which QM is based demands it.

 

[martinbn]

You are missing the point that an operator need not be defined on the hole space, but just on a dense subspace. Look up unbounded operators.

 

[PeroK]

You are missing the point that an operator need not be defined on the hole space, but just on a dense subspace. Look up unbounded operators.

I think the real issue here is that the original wave function is not physically viable. The space of physically viable wave-functions needs stronger criteria than being square integrable. Essentially, a wavefunction must tend to zero faster than any power of $x$ as $|x| \to \infty$. It's generally assumed that there is some subspace of the square integrable functions that meets the criteria of being physically viable. And, as Griffiths puts it, "any good math major" can find exceptions. I don't think this can be a dense subspace of $L^2$. It must be strictly smaller than that.

Note that the function in the OP has an unbounded first derivative - which is also not a viable property.

The difference for the original question in the previous thread is that the boundary conditions for the infinite square well prevent the momentum operator from being well-defined on the space of possible wavefunctions. This makes the infinite square well a physically impossible system - much like an infinite uniformly charged wire or plate is ultimately physically impossible, even if we can usefully study it.

 

[martinbn]

I think the real issue here is that the original wave function is not physically viable. The space of physically viable wave-functions needs stronger criteria than being square integrable. Essentially, a wavefunction must tend to zero faster than any power of $x$ as $|x| \to \infty$. It's generally assumed that there is some subspace of the square integrable functions that meets the criteria of being physically viable. And, as Griffiths puts it, "any good math major" can find exceptions. I don't think this can be a dense subspace of $L^2$. It must be strictly smaller than that.

Note that the function in the OP has an unbounded first derivative - which is also not a viable property.

The difference for the original question in the previous thread is that the boundary conditions for the infinite square well prevent the momentum operator from being well-defined on the space of possible wavefunctions. This makes the infinite square well a physically impossible system - much like an infinite uniformly charged wire or plate is ultimately physically impossible, even if we can usefully study it.

The Schwartz space is dense in the $L^2$ space.

 

[gentzen]

I don't think this can be a dense subspace of $L^2$. It must be strictly smaller than that.

I don't understand why you "guess around" in this case, where there is a well developped mathematical theory, which is not overly complicated and reasonably easy to understand. And which seems quite appropriate for the physical situations that it does cover (i.e. I exclude QFT here).

 

[PeroK]

I don't understand why you "guess around" in this case, where there is a well developped mathematical theory, which is not overly complicated and reasonably easy to understand. And which seems quite appropriate for the physical situations that it does cover (i.e. I exclude QFT here).

What is the exact definition of the Hilbert space and the set of operators that are relevant to QM?

 

[Morbert]

I just wonder why QM requires the result of an operator acting on a ket always be a ket, and for this sake even kicking some very regular states out of the domain of the operator.

We use a Hilbert space $\mathcal{H}$ to model all possible states of a system*. However we model a measurement - from the highly specific (PVM and the projection postulate) to the highly generalized (quantum instruments) - the state of the system after measurement should be in the same space, as it is a space of all possible states.

There might be a further generalization that relaxes this condition if papers like this are to be believed.

* Really we want the space $\Sigma (\mathcal{H})$ of both pure and mixed states.

 

[jbergman]

What is the exact definition of the Hilbert space and the set of operators that are relevant to QM?

To answer this probably requires an insight article. But like @martinbn stated both the position and momentum operators are unbounded and need special care in handling their domains. This is discussed in detail in Brian Hall's book.

 

[gaiussheh]

I think the real issue here is that the original wave function is not physically viable. The space of physically viable wave-functions needs stronger criteria than being square integrable. Essentially, a wavefunction must tend to zero faster than any power of $x$ as $|x| \to \infty$. It's generally assumed that there is some subspace of the square integrable functions that meets the criteria of being physically viable. And, as Griffiths puts it, "any good math major" can find exceptions. I don't think this can be a dense subspace of $L^2$. It must be strictly smaller than that.

Note that the function in the OP has an unbounded first derivative - which is also not a viable property.

The difference for the original question in the previous thread is that the boundary conditions for the infinite square well prevent the momentum operator from being well-defined on the space of possible wavefunctions. This makes the infinite square well a physically impossible system - much like an infinite uniformly charged wire or plate is ultimately physically impossible, even if we can usefully study it.

I wonder why one should exclude this wavefunction simply for this reason - It seems I can talk about $\langle x | \psi \rangle^2$ as the probability (density) of measurement outcome $x$, I can get $\langle x \rangle$ out of $\langle x | x|\psi \rangle$, I can measure it energy $\langle x |\hat{H}|\psi \rangle$ and momentum $\langle x |\mathrm{i}\hbar \partial_x|\psi \rangle$. Everything is well-defined. In any sense, you can't directly feel the wavefunction but only approach it with measurements, and it seems there's nothing stopping me from measuring it, and the wavefunction feels very real.

 

[gentzen]

I just wonder why QM requires the result of an operator acting on a ket always be a ket, and for this sake even kicking some very regular states out of the domain of the operator.

QM does not restrict operators in that way. It probably cannot restrict them even if it wanted, because otherwise the canonical commutation relation could no longer be realized. So operators are allowed to be unbounded.

The next question could be why only square integrable wavefunction are allowed, instead of arbitrary locally nice functions or even tempered distributions (i.e. elements of the dual of the Schwartz space). This is because they cannot be normed otherwise, so you wouldn't be able to define probabilities. But don't worry if the solution to some concrete problem should turn out to not be square integrable: All you need to do is form wavepackets of such "forbidden" functions to get back some square integrable function. Then you take the limit of an arbitrary sharp wavepacket, and basically go with the forbidden solution anyway.

 

[PeroK]

QM does not restrict operators in that way. It probably cannot restrict them even if it wanted, because otherwise the canonical commutation relation could no longer be realized. So operators are allowed to be unbounded.

The next question could be why only square integrable wavefunction are allowed, instead of arbitrary locally nice functions or even tempered distributions (i.e. elements of the dual of the Schwartz space). This is because they cannot be normed otherwise, so you wouldn't be able to define probabilities. But don't worry if the solution to some concrete problem should turn out to not be square integrable: All you need to do is form wavepackets of such "forbidden" functions to get back some square integrable function. Then you take the limit of an arbitrary sharp wavepacket, and basically go with the forbidden solution anyway.

This is mathematical hocus pocus!

 

[gaiussheh]

We use a Hilbert space $\mathcal{H}$ to model all possible states of a system*. However we model a measurement - from the highly specific (PVM and the projection postulate) to the highly generalized (quantum instruments) - the state of the system after measurement should be in the same space, as it is a space of all possible states.

There might be a further generalization that relaxes this condition if papers like this are to be believed.

* Really we want the space $\Sigma (\mathcal{H})$ of both pure and mixed states.

I think PVM is actually not stopping me from talking about position measurement outcomes (and associated probabilities) of my (strange) example $\langle\psi|E_{\Omega}|\psi\rangle$? However, the operator refuses to act on it.

 

[PeroK]

I wonder why one should exclude this wavefunction simply for this reason - It seems I can talk about $\langle x | \psi \rangle^2$ as the probability (density) of measurement outcome $x$, I can get $\langle x \rangle$ out of $\langle x | x|\psi \rangle$, I can measure it energy $\langle x |\hat{H}|\psi \rangle$ and momentum $\langle x |\mathrm{i}\hbar \partial_x|\psi \rangle$. Everything is well-defined. In any sense, you can't directly feel the wavefunction but only approach it with measurements, and it seems there's nothing stopping me from measuring it, and the wavefunction feels very real.

The problem with the wavefunction is that it doesn't tend to zero fast enough. So, it can't be associated with a realistic physical system. That's the simplest answer to this dilemma. You could try to calculate the expected momentum or kinetic energy of a particle with that wavefunction. If the integral diverges, then something is wrong.

That the position and momentum operators have unphysical eigenfunctions is a further issue. That's not the same issue as you have here. That issue is resolved by the theory of "rigged" Hilbert spaces and functions that are distributions.

 

[gentzen]

This is mathematical hocus pocus!

It was not my intention to create even more confusion by "mathematical hocus pocus".

I had the impression that you seriously confused @gaiussheh now, so I decided to tell him some of the mathematical background. He can still decide for himself whether this makes everything even more confusing, or whether it becomes more plausible und understandable.

 

[gaiussheh]

QM does not restrict operators in that way. It probably cannot restrict them even if it wanted, because otherwise the canonical commutation relation could no longer be realized. So operators are allowed to be unbounded.

The next question could be why only square integrable wavefunction are allowed, instead of arbitrary locally nice functions or even tempered distributions (i.e. elements of the dual of the Schwartz space). This is because they cannot be normed otherwise, so you wouldn't be able to define probabilities. But don't worry if the solution to some concrete problem should turn out to not be square integrable: All you need to do is form wavepackets of such "forbidden" functions to get back some square integrable function. Then you take the limit of an arbitrary sharp wavepacket, and basically go with the forbidden solution anyway.

Not sure if I am confusing you! I am not talking about the form of the operator but the domain. I mean, QM always wants to pick a subspace of the Hilbert space such that an operator always maps a square-integrable function to a square-integrable function. Why is this necessary? I can always compute $\displaystyle \int \psi^*[x] x \psi[x] \mathrm{d} x$, however if $\psi$ is not in the domain then since $x |\psi\rangle$ is not a vector in the Hilbert space the inner product $\langle \psi | x |\psi \rangle$ is not readily defined, but can only be seen as a short-hand notation of the former?

 

[gentzen]

I mean, QM always wants to pick a subspace of the Hilbert space such that an operator always maps a square-integrable function to a square-integrable function. Why is this necessary?

It is not necessary. What is actually necessary is that the operator is self-adjoint.

I can always compute $\displaystyle \int \psi^*[x] x \psi[x] \mathrm{d} x$, however if $\psi$ is not in the domain then since $x |\psi\rangle$ is not a vector in the Hilbert space the inner product $\langle \psi | x |\psi \rangle$ is not readily defined, but can only be seen as a short-hand notation of the former?

My guess is that the inner product $\langle \psi | x |\psi \rangle$ can be defined on the space of square integrable functions (i.e. $L^2$), but I would have to look-up the details. (And if the integral should happen to be defined too, then one should be able to show that its value coincides with that definition.) The operator $x$ itself can be defined on the space of tempered distributions by transposition, whenever the operator $x$ is defined on the Schwartz space. (And it is precisely for that case that I believe that your inner product can be defined on $L^2$.)

 

[PeterDonis]

My guess is that the inner product $\langle \psi | x |\psi \rangle$ can be defined on the space of square integrable functions

I'm not sure it can, or that it needs to be. The operator under discussion is not physically realizable: it corresponds to an infinitely precise measurement of position, which is not possible. Any real measurement has some finite error. So the operator $\hat{x}$ is not what we are actually realizing physically when we "measure the position" of something; we are realizing some other operator, which would be much more complicated to write down explictly, but which basically corresponds to measuring the position to within some finite error bar.

The reason the operator $\hat{x}$ is often discussed in textbooks, even though it's not physically realizable, is that, with appropriate mathematical tools, it's easy to work with, and it's easy to understand what it's doing. That makes it suitable for pedagogy, at least to many textbook authors, even though it's useless for practical experiments.

 

[Albertus Magnus]

I wonder why one should exclude this wavefunction simply for this reason - It seems I can talk about $\langle x | \psi \rangle^2$ as the probability (density) of measurement outcome $x$, I can get $\langle x \rangle$ out of $\langle x | x|\psi \rangle$, I can measure it energy $\langle x |\hat{H}|\psi \rangle$ and momentum $\langle x |\mathrm{i}\hbar \partial_x|\psi \rangle$. Everything is well-defined. In any sense, you can't directly feel the wavefunction but only approach it with measurements, and it seems there's nothing stopping me from measuring it, and the wavefunction feels very real.

Quantum mechanics is more than a handful of seemingly relevant quantities. Consider the following thought experiment. Given a quantum system, one should be able to perform a measurement corresponding to an operator $A$ and then perform a subsequent measurement on the same system corresponding to another operator $B$. However, if the operator $A$ maps the state to some other Hilbert space $\mathcal B$ or even outside of Hilbert space, then it is not possible to perform the subsequent measurement corresponding to $B$ given that $B$ is only defined on $\mathcal H$. Thus, when working with a given system whose state is in $\mathcal H$, all the relevant operators must be definable on $\mathcal H$. So given a set of relevant operators, one must exclude functions which are not mapped from $\mathcal H\rightarrow\mathcal H$.

 

[PeroK]

It is not necessary. What is actually necessary is that the operator is self-adjoint.

My guess is that the inner product $\langle \psi | x |\psi \rangle$ can be defined on the space of square integrable functions (i.e. $L^2$), but I would have to look-up the details.

The problem with functions like the one in the OP is that the inner product involving the operator $\hat x ^2$ is not defined (even if the integral is finite for the operator $\hat x$). This is why the function must eventually decay like a Gaussian - i.e. faster than any power of $x$.

Eventually the mathematical theory that supports QM gets murky and goes beyond what would be taught in a linear algebra/functional analysis class. All I was really trying to say, however, is that the initial lesson to learn is that not all functions in $L^2$ are possible wavefunctions and we must be dealing with a subspace of $L^2$. Not least, because there are "pathological" functions in $L^2$ that would not submit to much of the theory of QM. This is true of physics more generally - it can assume that functions are "well-behaved" - and anything else can be excluded on the grounds of being "unphysical".

 

[PeroK]

Not sure if I am confusing you! I am not talking about the form of the operator but the domain. I mean, QM always wants to pick a subspace of the Hilbert space such that an operator always maps a square-integrable function to a square-integrable function. Why is this necessary? I can always compute $\displaystyle \int \psi^*[x] x \psi[x] \mathrm{d} x$, however if $\psi$ is not in the domain then since $x |\psi\rangle$ is not a vector in the Hilbert space the inner product $\langle \psi | x |\psi \rangle$ is not readily defined, but can only be seen as a short-hand notation of the former?

My point is simply that if you try to include all the functions in $L^2$ in a physical theory like QM, then you are going to come unstuck.

 

[PeroK]

One final mathematical point. We can study linear operators from one Hilbert Space to another: $T: \mathcal{H_1} \to \mathcal{H_2}$. But, we don't have the operation of functional composition. Whereas, when we study linear operators from a fixed Hilbert space into itself, we can compose operators:

If $T_1, T_2: \mathcal{H} \to \mathcal{H}$, then $T = T_1 \circ T_2$, defined by $T(\psi) = T_1(T_2(\psi))$ is a linear operator from $\mathcal{H} \to \mathcal{H}$.

The property of operator composition underpins much of the theory used in QM and means that you can't consider operators that map functions into a larger Hilbert space.

 

[martinbn]

Not sure if I am confusing you! I am not talking about the form of the operator but the domain. I mean, QM always wants to pick a subspace of the Hilbert space such that an operator always maps a square-integrable function to a square-integrable function. Why is this necessary?

It ensures that the inner product that you need to compute is finite by Cauchy-Schwartz. It is also needed if you want to do more than one measurement in succession . If the image of the ket is not another ket, how do you apply an operator to it!

 

[gaiussheh]

The problem with functions like the one in the OP is that the inner product involving the operator $\hat x ^2$ is not defined (even if the integral is finite for the operator $\hat x$). This is why the function must eventually decay like a Gaussian - i.e. faster than any power of $x$.

Eventually the mathematical theory that supports QM gets murky and goes beyond what would be taught in a linear algebra/functional analysis class. All I was really trying to say, however, is that the initial lesson to learn is that not all functions in $L^2$ are possible wavefunctions and we must be dealing with a subspace of $L^2$. Not least, because there are "pathological" functions in $L^2$ that would not submit to much of the theory of QM. This is true of physics more generally - it can assume that functions are "well-behaved" - and anything else can be excluded on the grounds of being "unphysical".

My concern on this is that even if you have a wavefunction like $\mathrm{e}^{-x^2}$, you can still construct $\hat{O}=\mathrm{e}^{x^2}$, and again your wave function got kicked outside by this operator. Once you start to constrain the behaviour of operators, it becomes a rabbit hole, and you end up with a specific theory instead of a theory for any self-adjoint operator.

 

[martinbn]

My concern on this is that even if you have a wavefunction like $\mathrm{e}^{-x^2}$, you can still construct $\hat{O}=\mathrm{e}^{x^2}$, and again your wave function got kicked outside by this operator. Once you start to constrain the behaviour of operators, it becomes a rabbit hole, and you end up with a specific theory instead of a theory for any self-adjoint operator.

This is not an operator. You cannot just pick a function and say multiplying by it is an operator.

What is the definition of operator that you use?

 

[PeroK]

My concern on this is that even if you have a wavefunction like $\mathrm{e}^{-x^2}$, you can still construct $\hat{O}=\mathrm{e}^{x^2}$, and again your wave function got kicked outside by this operator. Once you start to constrain the behaviour of operators, it becomes a rabbit hole, and you end up with a specific theory instead of a theory for any self-adjoint operator.

The problem, as highlighted by @martinbn above, is that you are trying to build without proper mathematical foundations. Self-adjoint has no meaning outside of precise, well-defined functional analysis.

 

[martinbn]

The problem, as highlighted by @martinbn above, is that you are trying to build without proper mathematical foundations. Self-adjoint has no meaning outside of precise, well-defined functional analysis.

I think his question is more basic than this. He asks if an operator sends vectors from one space to the same space. And I think the problem may be the text he uses. If it is a QM textbook, then there will be a lot to grok and a lot left out as they do in physics.

 

[PeroK]

I think his question is more basic than this. He asks if an operator sends vectors from one space to the same space. And I think the problem may be the text he uses. If it is a QM textbook, then there will be a lot to grok and a lot left out as they do in physics.

I understand that. An operator more generally can map vectors from one space into another. But, whether the text book says so explicitly or not, QM uses operators than map vectors from one space into the same space.

For example, in Modern QM by Sakurai (who is definitely not the most rigorous), he says explicitly that an operator maps a ket to a ket. This should be quite clear.

Further to the points above, this requirement is essential for the composition (multiplication) of operators, which is essential for the fundamental commutation relations.

You cannot get started with QM without having a single, fixed Hilbert Space (for each system in question) on which you can study operator commutation.

This is all on pages 12-16 of Sakurai.

 

[gaiussheh]

This is not an operator. You cannot just pick a function and say multiplying by it is an operator.

What is the definition of operator that you use?

Well, that is actually where my confusion starts. I take an operator to be a map from a function to a function. It may not be closure in the Hilbert space, so people usually restrict its domain to, for example, the Schartz space. I just do not understand the rationale behind this. As long as your function decays faster than $\mathrm{e}^{-x^2}$, in integrate $\displaystyle \int f[x] \mathrm{e}^{-x^2} g[x] \mathrm{d} x$ converge and it is still self adjoint. So, similar to restricting wavefunctions to the Schwartz space, you can say that the operator $\mathrm{e}^{-x^2}$ has a smaller domain and is still self-adjoint. This never comes to an end, so why would one ever do this at the beginning?

 

[martinbn]

Well, that is actually where my confusion starts. I take an operator to be a map from a function to a function. It may not be closure in the Hilbert space, so people usually restrict its domain to, for example, the Schartz space. I just do not understand the rationale behind this. As long as your function decays faster than $\mathrm{e}^{-x^2}$, in integrate $\displaystyle \int f[x] \mathrm{e}^{-x^2} g[x] \mathrm{d} x$ converge and it is still self adjoint. So, similar to restricting wavefunctions to the Schwartz space, you can say that the operator $\mathrm{e}^{-x^2}$ has a smaller domain and is still self-adjoint. This never comes to an end, so why would one ever do this at the beginning?

Which book/source do you follow?

 

[PeroK]

Well, that is actually where my confusion starts. I take an operator to be a map from a function to a function. It may not be closure in the Hilbert space, so people usually restrict its domain to, for example, the Schartz space. I just do not understand the rationale behind this. As long as your function decays faster than $\mathrm{e}^{-x^2}$, in integrate $\displaystyle \int f[x] \mathrm{e}^{-x^2} g[x] \mathrm{d} x$ converge and it is still self adjoint. So, similar to restricting wavefunctions to the Schwartz space, you can say that the operator $\mathrm{e}^{-x^2}$ has a smaller domain and is still self-adjoint. This never comes to an end, so why would one ever do this at the beginning?

QM is a physical theory. It's supposed to model actual physical systems. It's not a mathematical game. If you were studying pure functional analysis you might have other objectives. The operators $\hat x$ and $\hat p \equiv -i\hbar \frac d {dx}$ arise out of the attempt to find the laws of nature for microscopic systems. These operators (and their linear combinations) must be accommodated by the theory. This stretches the theory of linear spaces somewhat and eventually leads to the "rigged Hilbert Space" formulation. Alternatively, you could try to formulate QM differently without these operators - I believe this can be done, although I don't know the details.

With these operators in the theory, not every function in $L^2$ is a valid wavefunction. This makes some physical sense anyway.

There is no physical justification to accommodate an operator like $e^{x^2}$. So, you don't have to do that. You're wasting your time. That is a rabbit hole. You either find a pragmatic solution to allow linear algebra to model physical systems or you give up physics. As Griffiths mentions more than once in his Introduction to QM "a good math major can find exceptions to this ...".

Those of us who come from a mathematics background accept this (and see the benefits of this leniency).

Using the theory as it stands, you generally begin by producing the spectrum of hydrogen (Griffiths) or modelling electron spin (Sakurai). Either way, you demonstrate that the mathematical machinery you have developed (to whatever level of rigour and with whatever necessary assumptions) is useful in studying physical phenomena. That is what physics is all about.

Whereas, if you produce a mathematical model that cannot do this, then it is useless - whatever its mathematical niceties.

 

[Albertus Magnus]

The problem with functions like the one in the OP is that the inner product involving the operator $\hat x ^2$ is not defined (even if the integral is finite for the operator $\hat x$). This is why the function must eventually decay like a Gaussian - i.e. faster than any power of $x$.

Eventually the mathematical theory that supports QM gets murky and goes beyond what would be taught in a linear algebra/functional analysis class. All I was really trying to say, however, is that the initial lesson to learn is that not all functions in $L^2$ are possible wavefunctions and we must be dealing with a subspace of $L^2$. Not least, because there are "pathological" functions in $L^2$ that would not submit to much of the theory of QM. This is true of physics more generally - it can assume that functions are "well-behaved" - and anything else can be excluded on the grounds of being "unphysical".

Yes, one of the early mistakes I made when studying quantum mechanics was to confuse the state space or Hilbert space of quantum mechanics with the whole of $L^2$. The space of square integrable functions is larger by far than the relevant quantum mechanical space, further sometimes as physicists we are interested in eigenstates that aren't even in $L^2$ such as the $|x\rangle$ and $|p\rangle$ states. $L^2$ should be understood as necessary but not sufficient.

 

[gaiussheh]

There is no physical justification to accommodate an operator like $e^{x^2}$. So, you don't have to do that. You're wasting your time.

Can you not construct a $\mathrm{e}^{x^2}$ potential and measure it?

 

[gentzen]

My guess is that the inner product $\langle \psi | x |\psi \rangle$ can be defined on the space of square integrable functions (i.e. $L^2$), but I would have to look-up the details. (And if the integral should happen to be defined too, then one should be able to show that its value coincides with that definition.) The operator $x$ itself can be defined on the space of tempered distributions by transposition, whenever the operator $x$ is defined on the Schwartz space. (And it is precisely for that case that I believe that your inner product can be defined on $L^2$.)

The problem with functions like the one in the OP is that the inner product involving the operator $\hat x ^2$ is not defined (even if the integral is finite for the operator $\hat x$). This is why the function must eventually decay like a Gaussian - i.e. faster than any power of $x$.

This is a good argument why my guess that the inner product $\langle \psi | x |\psi \rangle$ can be defined on the space of square integrable functions (in the sense I had in mind) is wrong.
To confirm this, I also tried to look-up that stuff in the lecture notes of the course where I learned that stuff in 1999 (in Chapitre III. La transformation de Fourier). It is not mentioned there, and the inner product doesn't seem to have the same "you can see and understand when it works and when it fails" properties like the convolution.

 

[PeroK]

Can you not construct a $\mathrm{e}^{x^2}$ potential and measure it?

That would be similar to an infinite well or deep finite well. Any laboratory system is finite in any case. A real potential that grew as fast as that would have to be truncated at some finite point.

Finally, this basic QM is manifestly non-relativistic. There's little point in accommodating potentials and wavefunctions that ultimately are incompatible with a relativistic theory in any case.

 

[gentzen]

Can you not construct a $\mathrm{e}^{x^2}$ potential and measure it?

No, you cannot, because the "boundary behavior" is wrong. You can only control stuff local to some "finite" region. Outside that region, it should decay to zero sufficiently fast, otherwise it is simply not a good approximation to anything you can construct.

But the mathematical theory for that specific example is not complicated either. The multiplication by $\mathrm{e}^{x^2}$ is simply not defined as an operator on the Schwartz space.

There is no physical justification to accommodate an operator like $e^{x^2}$. So, you don't have to do that. You're wasting your time. That is a rabbit hole. You either find a pragmatic solution to allow linear algebra to model physical systems or you give up physics.

I find it hard to predict how whether such "simple" stuff which is well understood will turn into a rabbit hole for some specific student like @gaiussheh. It is certainly true that if he won't stop at some point, he risks to waste enormous amounts of time. On the other hand, digging a bit into that stuff could be helpful, also in case he should later need to read some "numerics" paper which freely makes use of known stuff that "works". (But I admit that this is typically only "much" later, so a student is well adviced to not stray too far from the physics course initially.)

 

[gaiussheh]

No, you cannot, because the "boundary behavior" is wrong. You can only control stuff local to some "finite" region. Outside that region, it should decay to zero sufficiently fast, otherwise it is simply not a good approximation to anything you can construct.

But the mathematical theory for that specific example is not complicated either. The multiplication by $\mathrm{e}^{x^2}$ is simply not defined as an operator on the Schwartz space.

I find it hard to predict how whether such "simple" stuff which is well understood will turn into a rabbit hole for some specific student like @gaiussheh. It is certainly true that if he won't stop at some point, he risks to waste enormous amounts of time. On the other hand, digging a bit into that stuff could be helpful, also in case he should later need to read some "numerics" paper which freely makes use of known stuff that "works". (But I admit that this is typically only "much" later, so a student is well adviced to not stray too far from the physics course initially.)

Thanks for these suggestions. It was not my first course in QM but my first TA of it. I got quite good scores back in my undergraduate and didn't find this course to be that much of abstract formulation. I was just amazed by how much math I have neglected!

P.S. I asked the lecturer of this course and his answer was like "this question is completely unphysical, you should not think of it", and I wasn't very convinced. QM use the term axioms, and my understanding is that it should be mathematically regorous itself.

 

[gaiussheh]

This is a good argument why my guess that the inner product $\langle \psi | x |\psi \rangle$ can be defined on the space of square integrable functions (in the sense I had in mind) is wrong.

Does it make sense to "define" the "inner product" as $\langle \psi | x \rangle \psi \rangle = \lim_{n\to\infty} \langle \psi | x | \psi_n \rangle$, where $\psi_n$ is in the Schwartz space as it is dense? This way, one can handle $x | \psi \rangle$ even if it is not in the Hilbert space, and this is consistent with the integration $\displaystyle \int \psi^* x \psi \mathrm{d}x$?

 

[PeroK]

Does it make sense to "define" the "inner product" as $\langle \psi | x \rangle \psi \rangle = \lim_{n\to\infty} \langle \psi | x | \psi_n \rangle$, where $\psi_n$ is in the Schwartz space as it is dense?

That limit will be $+\infty$. Which essentially follows from the definition of dense.

 

[gaiussheh]

That limit will be $+\infty$. Which essentially follows from the definition of dense

Not sure if I have understood this correctly, however if I expend my wavefunction using the eigenstates of harmonic oscillator, the result is compatible with the integration result (=0). In this sense I have constructed an extension of the operator $x$ to my wavefunction $\psi$ such that the inner product can still be defined even if $x | \psi \rangle$ is not. (Note: $\displaystyle \sum_{n} \alpha_n |n\rangle$ is in the schwartz space for any finite $n$.)

 

[Albertus Magnus]

Can you not construct a $\mathrm{e}^{x^2}$ potential and measure it?

You could have a Hamiltonian with an $e^{-x^2}$ term. Potentials such as $e^{x^2}$ are unphysical as they do not go to zero at infinity.

 

[PeroK]

Not sure if I have understood this correctly, however if I expend my wavefunction using the eigenstates of harmonic oscillator, the result is compatible with the integration result (=0). In this sense I have constructed an extension of the operator $x$ to my wavefunction $\psi$ such that the inner product can still be defined even if $x | \psi \rangle$ is not. (Note: $\displaystyle \sum_{n} \alpha_n |n\rangle$ is in the schwartz space for any finite $n$.)

I have very little idea what you are trying to do there. With all respect, I probably can't help you any further with this.

 

[gaiussheh]

I have very little idea what you are trying to do there. With all respect, I probably can't help you any further with this.

Thank you indeed for helping, and apologize if I've been annoying.

It is obviously up to you whether you wish to further this.

My point is that it is not true to say $\displaystyle \lim_{n\to\infty} \langle \psi_n| \hat{x}| \psi_n\rangle =+\infty$, at least for the case I constructed.

My construction is to expand $\psi=\frac{1}{\sqrt{1+|x|^3}}$ to series on the basis of harmonic oscillator energy eigenvectors $|m \rangle \sim \mathrm{e}^{-x^2/2}H_m[x]$, where $H_m$ is the Hermitian Polynomials. This can always be done by $| \psi \rangle = \sum_{m=0}^{\infty} \alpha_m | m \rangle$, where $\alpha_m = \langle m | \psi \rangle$. As $|m \rangle$ is in the Schwartz space, for any finite n, $\displaystyle \psi_n = \sum_{m=0}^{n} \alpha_m | m \rangle$ is in the also in the Schwartz space. Thus, the position operator can act on it. Because the Schwartz space is dense, the sequence $\left\{ \displaystyle \psi_n \right\}$ converge to $|\psi \rangle$ under $L^2$ norm.

Obviously one can not define $\hat{x}|\psi \rangle$ because $\psi \not\in \mathrm{Dom}[\hat{x}]$, thus there is no definition of $\langle \psi | \hat{x} | \psi \rangle$ as an inner product in the $L^2[\mathbb{R}]$ space. My question is whether it is reasonable to define it as $\displaystyle \lim_{n\to\infty} \langle \psi_n| \hat{x}| \psi_n\rangle$, such that the result is consistent with $\displaystyle \int \psi^* x \psi \mathrm{d} x$. My construction seems to support this. Here is the reason.

Because for harmonic oscillator the matrix element for the position operator $\langle i |\hat{x} | j \rangle$ is only non-zero for $i=j\pm 1$ , when summing $\displaystyle \lim_{n \to \infty} \langle \psi_n| \hat{x}| \psi_n\rangle = \sum_{i,j} \alpha_i^* \alpha_j \langle i |\hat{x} | j \rangle$, only these tern will contribute. However $\frac{1}{\sqrt{1+|x|^3}}$ is an even function. It is only non-zero for even $n$. Therefore $\forall i, \alpha_i \alpha_{i+1}=0$. Thus $\displaystyle \lim_{n\to\infty} \langle \psi_n| \hat{x}| \psi_n\rangle =0 = \int \psi^* x \psi \mathrm{d} x$, not $+\infty$.

I really hope that I have not annoyed you or anyone in the physics forum. And I apologies if I have been!

 

[PeroK]

I really hope that I have not annoyed you or anyone in the physics forum. And I apologies if I have been!

Who says that the span of the eigenfunctions of the QHO are dense in $L^2$? If they were, then we could show that for all $\psi \in L^2$, we have $\langle \psi|x|\psi \rangle = 0$.

Note also that earlier I said:

You cannot get started with QM without having a single, fixed Hilbert Space (for each system in question) on which you can study operator commutation.

The QHO has a smaller set of possible wavefunctions than the free particle, say. In particular, the pathological function you started with is not in the the Hilbert Space for the QHO. Schwarz space is irrelevant here.

The underlying problem is that you lack the mathematical rigour to prove stuff like this. You are largely wasting your time, IMO. You post some dodgy mathematics and we have to figure out where you've gone wrong. I don't have the time or the inclination for this, sorry.

 

[bhobba]

This stretches the theory of linear spaces somewhat and eventually leads to the "rigged Hilbert Space" formulation.

For the full technical details see:

http://galaxy.cs.lamar.edu/~rafaelm/webdis.pdf

Beware - it is not for the faint-hearted in its mathematical sophistication. I took a few years from my study of QM to fully come to grips with it. And I am a mathematician self-taught in physics. Even the great von Neumann couldn't figure it out - it took the combined efforts of mathematicians like Gelfland, Schwartz and Grothendieck (when he was working with Schwartz).

Thanks
Bill

 

[WWGD]

You are missing the point that an operator need not be defined on the hole space, but just on a dense subspace. Look up unbounded operators.

And a good example would be the Laplacian defined in a dense subspace of $L^2$, as $L^2$ functions may not even be (once)-differentiable.

 

[A. Neumaier]

An operator on the set of square integrable functions must produce a square integrable function. If it does not, then it is not an operator on that Hilbert space.

The concept of operator relevant for QM is that of a 'densely defined operator', i.e., a linear operator A from a dense subspace of the Hilbert space into the Hilbert space; ; see https://en.wikipedia.org/wiki/Densely_defined_operator . The maximal such subspace is called the domain of A.

All (Hermitian) unbounded operators (and in particular position x and momentum p) are only densely defined; their domain is not the full Hilbert space. Therefore there are always some states in which the expectation is undefined.

Physicists usually don't bother about domains until they run into ill-defined calculations. This is the reason why mathematical physics (which imposes the mathematical standards of rigor) is significantly more complex than Dirac-style physics.

 

[dextercioby]

[...]
All unbounded operators (and in particular position x and momentum p) are only densely defined; their domain is not the full Hilbert space. [...]

I've always had an issue with that, because you see, the theorem of Hellinger & Toeplitz precisely says that a symmetric operator everywhere defined on a Hilbert space must be bounded/continuos. Which from the elementary logic perspective means: A \wedge B \Rightarrow C. Negate it \neg C \Rightarrow \neg A \vee \neg B. So an unbounded linear operator can be be either non-symmetric, or not defined on whole Hilbert space. But we have an "or" relationship. I can choose it to be non-symmetric and couldn't care about the domain. So, can one have an unbounded operator fully defined on all vectors in a separable Hilbert space? Clearly not, if we want it symmetric (or even s-a), but what about the general case, in which we don't want symmetric operators.

 

[SergejMaterov]

If you drop symmetry, you get pathological, discontinuous, unbounded maps—but they’re of almost no use in analysis or physics.