State Vectors vs. Wavefunctions

[AspiringResearcher]

Hi physicsforums,

I am an undergrad currently taking an upper-division course in Quantum Mechanics and we have begun studying $L^2$ space, state vectors, bra-ket notation, and operators, etc.

I have a few questions about the relationship between $L^2$, the space of square-integrable complex-valued functions $\psi$ and Hilbert Space, which is the space of state vectors $|\psi\rangle$. To me, it is obvious that these two spaces are isomorphic. For the remainder of this problem, for clarity's sake, I will define the isomorphism between these two spaces: $$ζ : H \mapsto L^2$$ defined by
$$∀ x ∈ R, ζ(|x \rangle) = \delta(x)$$
where $|x\rangle$ is the eigenvector of $\hat{x}$ with eigenvalue x.
In linear algebra, for a vector space V, the set of linear operators from mapping V to itself is called $L(V;V)$; this forms a vector space.

My question is this - every linear operator in $L^2$ has a corresponding linear operator in $H$, at least from what I've seen so far.

How do you construct an explicit isomorphism $Λ : L(H;H) \mapsto L(L^2,L^2)$ using $ζ$ as defined above? Please help me with this.

 

[Orodruin]

Why do you need to construct an isomorphism? The set of square integrable functions is a Hilbert space. How you choose your notation is irrelevant for this.

 

[AspiringResearcher]

Why do you need to construct an isomorphism? The set of square integrable functions is a Hilbert space. How you choose your notation is irrelevant for this.

The point is that an arbitrary operator $\hat{Q}$ will have a different form in $L^2$ than in $H$. Perhaps I wasn't clear in the OP that $H$ is the vector space of all kets. Kets and wavefunctions (members of $L^2$) are not the same thing; there is an isomorphism that relates them.

I want to explicitly construct an isomorphism between the linear operators on $H$ and the linear operators on $L^2$.

 

[Orodruin]

The point is that an arbitrary operator $\hat{Q}$ will have a different form in $L^2$ than in $H$. Perhaps I wasn't clear in the OP that $H$ is the vector space of all kets. Kets and wavefunctions (members of $L^2$) are not the same thing; there is an isomorphism that relates them.

This is like saying that basis vectors are not a part of a vector space. Your Hilbert space is the space of square integrable functions and therefore, obviously, a square integrable function is a state in that Hilbert space. You may be looking to use a different representation of the same Hilbert space. This would be equivalent to a basis transformation.

 

[AspiringResearcher]

This is like saying that basis vectors are not a part of a vector space. Your Hilbert space is the space of square integrable functions and therefore, obviously, a square integrable function is a state in that Hilbert space. You may be looking to use a different representation of the same Hilbert space. This would be equivalent to a basis transformation.

I know how to execute a basis transformation using the completeness operator. I think you're missing the point - isn't it true, in general, that
$$\psi(x) \neq |\psi\rangle$$

Rather,
$$\psi(x) = \langle e_x | \psi \rangle$$

 

[hilbert2]

I'm not sure if a general abstract operator can be written as a differential operator that acts on wave functions, but if you have an operator $\hat{A}$ with eigenvectors and eigenvalues

$\hat{A}\left.\right|\varphi_i \left.\right> = a_i \left.\right|\varphi_i \left.\right>$,

you can use the completeness relation

$\hat{A}\left.\right|\psi \left.\right> = \sum\limits_{i}a_i \left.\right|\varphi_i \left.\right>\left<\right.\varphi_i \left.\right|\psi \left.\right>$,

to deduce that $\hat{A}$ acts on wavefunction $\psi (x)$ as

$\hat{A}\psi(x) = \sum\limits_i \left[a_i \varphi_i (x) \int_{-\infty}^{\infty}\varphi_{i}^* (x')\psi(x')dx' \right] = \int_{-\infty}^{\infty}\left[\sum\limits_i a_i \varphi_i (x)\varphi_{i}^* (x')\right]\psi(x')dx'$,

which is an expression of the position space operator $\hat{A}$ as an integral transform.

Note that some operators can't be written as differential operators that act on wavefunctions having only position coordinates as their arguments. The spin operator acts on the discrete spin variables that are separate from continuum position coordinates.

 

[Orodruin]

I know how to execute a basis transformation using the completeness operator. I think you're missing the point - isn't it true, in general, that
$$\psi(x) \neq |\psi\rangle$$

Rather,
$$\psi(x) = \langle e_x | \psi \rangle$$

It is a question of notation and writing out the basis explicitly or not. Nothing else. If you insist on using kets, the wave function is nothing but the components of a state in position basis
$$
|\psi\rangle = \int \psi(x)|x\rangle .
$$
This is no stranger than writing $|\psi\rangle = \sum_n \psi_n |n\rangle$ in the case of a countable basis.

 

[AspiringResearcher]

I'm not sure if a general abstract operator can be written as a differential operator that acts on wave functions, but if you have an operator $\hat{A}$ with eigenvectors and eigenvalues

$\hat{A}\left.\right|\varphi_i \left.\right> = a_i \left.\right|\varphi_i \left.\right>$,

you can use the completeness relation

$\hat{A}\left.\right|\psi \left.\right> = \sum\limits_{i}a_i \left.\right|\varphi_i \left.\right>\left<\right.\varphi_i \left.\right|\psi \left.\right>$,

to deduce that $\hat{A}$ acts on wavefunction $\psi (x)$ as

$\hat{A}\psi(x) = \sum\limits_i \left[a_i \varphi_i (x) \int_{-\infty}^{\infty}\varphi_{i}^* (x')\psi(x')dx' \right] = \int_{-\infty}^{\infty}\left[\sum\limits_i a_i \varphi_i (x)\varphi_{i}^* (x')\right]\psi(x')dx'$,

which is an expression of the position space operator $\hat{A}$ as an integral transform.

Note that some operators can't be written as differential operators that act on wavefunctions having only position coordinates as their arguments. The spin operator acts on the discrete spin variables that are separate from continuum position coordinates.

Sorry if there is any confusion. Here, I am only discussing Hermitian operators whose eigenvectors span Hilbert space. Surely the spin operator's eigenfunctions don't span $L^2$, right?

 

[Orodruin]

Sorry if there is any confusion. Here, I am only discussing Hermitian operators whose eigenvectors span Hilbert space. Surely the spin operator's eigenfunctions don't span $L^2$, right?

The spin operators act on a completely different Hilbert space. Just saying ”Hilbert space” does not mean ”square integrable functions”. In the case of spin operators, the Hilbert spaces considered typically are vector spaces on which there is a representation of SU(2).

 

[hilbert2]

Sorry if there is any confusion. Here, I am only discussing Hermitian operators whose eigenvectors span Hilbert space. Surely the spin operator's eigenfunctions don't span $L^2$, right?

If you have a set of state vectors $\left.\right|\phi_i \left.\right>$ that span $L^2$ in position space, and spin states $\left.\right|\uparrow\left.\right>$, $\left.\right|\downarrow\left.\right>$ representing spin up and down, you can form tensor products like $\left.\right|\phi_i \left.\right>\left.\right|\uparrow\left.\right>$ which contain both the spatial and spin information. Note that now the spin up state is degenerate as many times as there are states $\left.\right|\phi_i \left.\right>$.

 

[stevendaryl]

I know how to execute a basis transformation using the completeness operator. I think you're missing the point - isn't it true, in general, that
$$\psi(x) \neq |\psi\rangle$$

Rather,
$$\psi(x) = \langle e_x | \psi \rangle$$

There is an ambiguity in the common notation for functions, which is that we don't distinguish between a function and the value of that function at some arbitrary point. But they are two different mathematical objects. $|\psi\rangle$ is the function whose value at point $x$ is $\psi(x)$, which in bra-ket notation is written as $\langle x | \psi\rangle$. (I'm not sure why you write $e_x$. Does that mean a basis vector in the x-direction? If so, that shouldn't be used.)

 

[AspiringResearcher]

I believe I have established an isomorphism $Λ : L(H,H) \mapsto L(L^2,L^2)$ which satisfies what I was trying to do. I would appreciate a critique of my reasoning.

Here, I am considering the set of quantum states/kets $H$, and the set of wavefunctions, $L^2$ as isomorphic Hilbert spaces.
The isomorphism is $Γ : H \mapsto L^2$ defined by $$Γ(|\psi\rangle) = \langle x|\psi\rangle = \psi(x)$$.
Now, defining the isomorphism between operators of these two respective spaces:
Let $Λ : L(H,H) \mapsto L(L^2,L^2)$ be defined by $$Λ(\hat{Q})(\psi(x)) = Γ(\hat{Q}(Γ^{-1}(\psi(x)))), \forall \text{ Hermitian } \hat{Q} \in L(H,H).$$
Say $\hat{Q}(|x\rangle) = \int_{-\infty}^{\infty}q(x')|x'\rangle dx'$. Since the $|x\rangle$ form a basis for$H$, this uniquely determines $\hat{Q}$.
Proceeding, we have
$$Λ(\hat{Q})(\psi(x)) = Γ(\hat{Q}(Γ^{-1}(\psi(x)))) = Γ(\hat{Q}(|\psi\rangle)) = Γ\hat{Q}\int_{-\infty}^{\infty}\psi(x)|x\rangle dx$$
$$= Γ\int_{-\infty}^{\infty}\psi(x)\int_{-\infty}^{\infty}q(x')|x'\rangle dx' dx = Γ\int_{-\infty}^{\infty}\psi(y)\int_{-\infty}^{\infty}q(x')|x'\rangle dx' dy \text{ since x is a dummy variable}$$
$$= \int_{-\infty}^{\infty}\psi(y)\int_{-\infty}^{\infty}q(x')Γ(|x'\rangle)dx'dy = \int_{-\infty}^{\infty}\psi(y)\int_{-\infty}^{\infty}q(x')\delta(x)dx'dy = \int_{-\infty}^{\infty}\psi(y)q(x)dy = q(x)\int_{-\infty}^{\infty}\psi(x')dx'$$.

Would this not suffice as a way to calculate the equivalent operator in $L^2$?

@Orodruin @hilbert2 thank you guys for your help btw

 

[stevendaryl]

I believe I have established an isomorphism $Λ : L(H,H) \mapsto L(L^2,L^2)$ which satisfies what I was trying to do. I would appreciate a critique of my reasoning.

I think you missed a critical point, which is that $H$ is not a particular set, so it doesn't make sense to talk about a map between some set $L^2$ and $H$.

Instead, $L^2$ is an example of a Hilbert space. A Hilbert space is any vector space $S$ together with an inner product $\langle u, v \rangle$ on that set satisfying certain completeness properties. A vector space in turn is any set with operators for addition and multiplication by a constant satisfying certain properties.

A ket $|\psi\rangle$ used by nonrelativistic quantum mechanics (without spin) simply IS a square-integrable function. There is no need to map it to a square-integrable function.

When you write:

$\Gamma(|\psi\rangle) = \langle x|\psi\rangle$

that doesn't really make sense. The object $\langle x | \psi\rangle$ means "The value of the function $\psi$ at point $x$". People sometimes use the same expression, $\psi(x)$ to mean both the function and its value at point $x$, but the ket $|\psi\rangle$ means the function itself. There is no need to use a mapping $\Gamma$.

 

[AspiringResearcher]

I think you missed a critical point, which is that $H$ is not a particular set, so it doesn't make sense to talk about a map between some set $L^2$ and $H$.

Instead, $L^2$ is an example of a Hilbert space. A Hilbert space is any vector space $S$ together with an inner product $\langle u, v \rangle$ on that set satisfying certain completeness properties. A vector space in turn is any set with operators for addition and multiplication by a constant satisfying certain properties.

A ket $|\psi\rangle$ used by nonrelativistic quantum mechanics (without spin) simply IS a square-integrable function. There is no need to map it to a square-integrable function.

What I am trying to do here is more of a thought experiment involving the construction of an abstract set $H$ whose elements are kets and is a Hilbert Space

 

[stevendaryl]

What I am trying to do here is more of a thought experiment involving the construction of an abstract set $H$ whose elements are kets and is a Hilbert Space

Okay, but it's hard for me to see, for an abstract set $H$, what the notation $\langle x|\psi\rangle$ means. For the concrete example of $L^2$, it means the value of the function $\psi$ at point $x$. But for an abstract set, I don't know what the operation $|\psi\rangle \rightarrow \langle x | \psi\rangle$ means. A slightly misleading impression given by the Dirac notation is that $|x\rangle$ is a ket. It's not. It's not an element of the Hilbert space. So $\langle x|$ isn't equal to $(|\phi\rangle)^\dagger$ for any ket $|\phi\rangle$. We can make sense of the operator $\langle x|$ in the context of $L^2$, but not for an abstract set. As far as I know.

 

[stevendaryl]

Okay, but it's hard for me to see, for an abstract set $H$, what the notation $\langle x|\psi\rangle$ means. For the concrete example of $L^2$, it means the value of the function $\psi$ at point $x$. But for an abstract set, I don't know what the operation $|\psi\rangle \rightarrow \langle x | \psi\rangle$ means. A slightly misleading impression given by the Dirac notation is that $|x\rangle$ is a ket. It's not. It's not an element of the Hilbert space. So $\langle x|$ isn't equal to $(|\phi\rangle)^\dagger$ for any ket $|\phi\rangle$. We can make sense of the operator $\langle x|$ in the context of $L^2$, but not for an abstract set. As far as I know.

Sometimes people treat $\delta(x-x_0)$ as if it were a function, and that sort of works for many purposes, but it's definitely not an element of $L^2$, because it's not square-integrable. ($\int \delta(x-x_0) dx = 1$ but $\int \delta(x-x_0)^2 dx$ is undefined). People sometimes write $\delta(x-x_0)$ as $\langle x|x_0\rangle$, but it's really not the inner product of two kets, despite the notation that suggests that.

 

[Orodruin]

Sometimes people treat $\delta(x-x_0)$ as if it were a function, and that sort of works for many purposes, but it's definitely not an element of $L^2$, because it's not square-integrable. ($\int \delta(x-x_0) dx = 1$ but $\int \delta(x-x_0)^2 dx$ is undefined). People sometimes write $\delta(x-x_0)$ as $\langle x|x_0\rangle$, but it's really not the inner product of two kets, despite the notation that suggests that.

To expand a little bit, it is a distribution on $L^2$, i.e., a linear map from $L^2$ to the complex numbers. Now, any function in $L^2$ defines a distribution on $L^2$ through the inner product, but a distribution on $L^2$ is not necessarily associated to an element in $L^2$, as in the case of $|x\rangle$.

 

[AspiringResearcher]

Okay, but it's hard for me to see, for an abstract set $H$, what the notation $\langle x|\psi\rangle$ means. For the concrete example of $L^2$, it means the value of the function $\psi$ at point $x$. But for an abstract set, I don't know what the operation $|\psi\rangle \rightarrow \langle x | \psi\rangle$ means. A slightly misleading impression given by the Dirac notation is that $|x\rangle$ is a ket. It's not. It's not an element of the Hilbert space. So $\langle x|$ isn't equal to $(|\phi\rangle)^\dagger$ for any ket $|\phi\rangle$. We can make sense of the operator $\langle x|$ in the context of $L^2$, but not for an abstract set. As far as I know.

This is a bizarre problem. I think my professor and Griffiths are not very instructive in this. In class, $\psi(x)$ is always called a wavefunction, not just the evaluation of $|\psi\rangle$ at x; if it were just the evaluation, then it should be called a scalar and not a vector

 

[Orodruin]

This is a bizarre problem. I think my professor and Griffiths are not very instructive in this. In class, $\psi(x)$ is always called a wavefunction, not just the evaluation of $|\psi\rangle$ at x; if it were just the evaluation, then it should be called a scalar and not a vector

This goes back to the notational ambiguity already mentioned by @stevendaryl in post #11 between $\psi(x)$ as an object being used to denote the function itself or the value of that function at $x$.

 

[stevendaryl]

This is a bizarre problem. I think my professor and Griffiths are not very instructive in this. In class, $\psi(x)$ is always called a wavefunction, not just the evaluation of $|\psi\rangle$ at x; if it were just the evaluation, then it should be called a scalar and not a vector

As I said, it's an ambiguity of function notation. People use the same notation, $sin(x)$ to mean the sine function and the value of that function at point $x$. There is a way to make it unambiguous, using, for example, the lambda calculus: $\lambda x . \phi(x)$ means the function, while $\phi(x)$ means its value at a particular point $x$. But people usually don't bother, since it's usually clear from context.

 

[AspiringResearcher]

Okay, but it's hard for me to see, for an abstract set $H$, what the notation $\langle x|\psi\rangle$ means. For the concrete example of $L^2$, it means the value of the function $\psi$ at point $x$. But for an abstract set, I don't know what the operation $|\psi\rangle \rightarrow \langle x | \psi\rangle$ means. A slightly misleading impression given by the Dirac notation is that $|x\rangle$ is a ket. It's not. It's not an element of the Hilbert space. So $\langle x|$ isn't equal to $(|\phi\rangle)^\dagger$ for any ket $|\phi\rangle$. We can make sense of the operator $\langle x|$ in the context of $L^2$, but not for an abstract set. As far as I know.

Couldn't we let $H = C^{\infty}$ and say there's an isomorphism (coordinatization w.r.t. the ${|x\rangle}$ basis) between $C^{\infty}$ and $L^2$?

 

[hilbert2]

Isn't there a lot more $L^2$ functions than there are $\mathcal{C}^\infty$ functions? I don't think there's a bijective mapping between them.

 

[vanhees71]

Hi physicsforums,

I am an undergrad currently taking an upper-division course in Quantum Mechanics and we have begun studying $L^2$ space, state vectors, bra-ket notation, and operators, etc.

I have a few questions about the relationship between $L^2$, the space of square-integrable complex-valued functions $\psi$ and Hilbert Space, which is the space of state vectors $|\psi\rangle$. To me, it is obvious that these two spaces are isomorphic. For the remainder of this problem, for clarity's sake, I will define the isomorphism between these two spaces: $$ζ : H \mapsto L^2$$ defined by
$$∀ x ∈ R, ζ(|x \rangle) = \delta(x)$$
where $|x\rangle$ is the eigenvector of $\hat{x}$ with eigenvalue x.
In linear algebra, for a vector space V, the set of linear operators from mapping V to itself is called $L(V;V)$; this forms a vector space.

My question is this - every linear operator in $L^2$ has a corresponding linear operator in $H$, at least from what I've seen so far.

How do you construct an explicit isomorphism $Λ : L(H;H) \mapsto L(L^2,L^2)$ using $ζ$ as defined above? Please help me with this.

First of all your isomorphism is flawed, because $|x \rangle \neq \mathcal{H}$ since $\mathcal{H}$ is the Hilbert space. The Dirac bra-ket formalism is mathematically rigorously well-defined in the sense of what's usually called "rigged Hilbert space". It's due to Gelfand et al. A very good two-volume book with the math treated rigorously is

A. Galindo and P. Pascual, Quantum Mechanics, Springer Verlag, Heidelberg, 1990. 2 Vols.

If you are satisfied with a minimal treatment for physicists, and excellent source is

L. Ballentine, Quantum Mechanics, World Scientific

The upshot is that the essentially self-adjoint position operator $\hat{x}$ is defined on a dense proper subspace of Hilbert space (the "nuclear space"). The "generalized eigenvectors" $|x \rangle$ of this operator live in the dual of this dense subspace, which is necessarily larger than the Hilbert space, which is dual to itself. They are distributions. That's most obvious by the fact that they are "normalized to a $\delta$ distribution", $\langle x'|x \rangle$.

The isomorphism is
$$\zeta:\mathcal{H} \rightarrow L^2, \quad |\psi \rangle \mapsto \psi(x)=\langle x|\psi \rangle.$$
This induces an isomorphism between operators
$$\hat{A} |\psi \rangle \mapsto \overline{A} \psi(x):=\langle x|\hat{A} \psi \rangle=\int_{\mathbb{R}} \mathrm{d} x' A(x,x') \psi(x'), \quad A(x,x')=\langle x|\hat{A}|x' \rangle.$$

 

[AspiringResearcher]

First of all your isomorphism is flawed, because $|x \rangle \neq \mathcal{H}$ since $\mathcal{H}$ is the Hilbert space. The Dirac bra-ket formalism is mathematically rigorously well-defined in the sense of what's usually called "rigged Hilbert space". It's due to Gelfand et al. A very good two-volume book with the math treated rigorously is

A. Galindo and P. Pascual, Quantum Mechanics, Springer Verlag, Heidelberg, 1990. 2 Vols.

If you are satisfied with a minimal treatment for physicists, and excellent source is

L. Ballentine, Quantum Mechanics, World Scientific

The upshot is that the essentially self-adjoint position operator $\hat{x}$ is defined on a dense proper subspace of Hilbert space (the "nuclear space"). The "generalized eigenvectors" $|x \rangle$ of this operator live in the dual of this dense subspace, which is necessarily larger than the Hilbert space, which is dual to itself. They are distributions. That's most obvious by the fact that they are "normalized to a $\delta$ distribution", $\langle x'|x \rangle$.

The isomorphism is
$$\zeta:\mathcal{H} \rightarrow L^2, \quad |\psi \rangle \mapsto \psi(x)=\langle x|\psi \rangle.$$
This induces an isomorphism between operators
$$\hat{A} |\psi \rangle \mapsto \overline{A} \psi(x):=\langle x|\hat{A} \psi \rangle=\int_{\mathbb{R}} \mathrm{d} x' A(x,x') \psi(x'), \quad A(x,x')=\langle x|\hat{A}|x' \rangle.$$

YOU'RE MY HERO

 

[stevendaryl]

The isomorphism is
$$\zeta:\mathcal{H} \rightarrow L^2, \quad |\psi \rangle \mapsto \psi(x)=\langle x|\psi \rangle.$$
This induces an isomorphism between operators
$$\hat{A} |\psi \rangle \mapsto \overline{A} \psi(x):=\langle x|\hat{A} \psi \rangle=\int_{\mathbb{R}} \mathrm{d} x' A(x,x') \psi(x'), \quad A(x,x')=\langle x|\hat{A}|x' \rangle.$$

This might be a matter of semantics, but it seems to me that if your Hilbert space is $L^2$, then it's not that there is an isomorphism between square-integrable functions $\psi(x)$ and kets $|\psi\rangle$, but the kets ARE square-integrable functions (or the functions modulo the equivalence relation $\psi \approx \phi$ iff $\int dx |\phi(x) - \psi(x)|^2 = 0$).

Or maybe this is one of those cases in mathematics where it's a matter of taste whether you say that this mathematical object is equal to this other mathematical object, or it is isomorphic to it.

But my objection to your isomorphism is the ambiguity between $\langle x | \psi\rangle$ meaning a real number (the value of $\psi$ at point $x$) and the function. The notation makes it seem like you're forming an inner product, so the result should be a real number. But the isomorphism is between kets and functions, not between kets and real numbers.

 

[samalkhaiat]

isomorphism between operators
$$\hat{A} |\psi \rangle \mapsto \overline{A} \psi(x):=\langle x|\hat{A} \psi \rangle=\int_{\mathbb{R}} \mathrm{d} x' A(x,x') \psi(x'), \quad A(x,x')=\langle x|\hat{A}|x' \rangle.$$

An important remark has to be said about the integral kernel $K_{A}(\bar{x}, x) = \langle \bar{x}| A | x \rangle$: Let $V = \{ | \psi \rangle \}$ be the space of all kets (whatever that means) with $\int_{\mathbb{R}} dx \ | x \rangle \langle x | = 1$ being the identity operator on $V$. Let $\mathcal{B}(V)$ be the algebra of all bounded operators on $V$. Take $A \in \mathcal{B}(V)$ and define the function $(A \psi) (\bar{x}) \equiv \langle \bar{x}| A | \psi \rangle$ on $\mathbb{R}$. Using the identity operator on $V$, we may write $$(A \psi) (\bar{x}) = \int_{\mathbb{R}} dx \ \langle \bar{x}| A | x \rangle \langle x | \psi \rangle .$$ Furthermore, assume that $\langle x | \psi \rangle \equiv \psi (x) \in L^{2}(\mathbb{R})$ and write $$A \psi ( \bar{x}) = \int_{\mathbb{R}} dx \ K_{A}(\bar{x} , x ) \psi ( x ) .$$ Now, consider taking the $L^{2}$-norm squared: $$\lVert A \psi \rVert^{2}_{2} = \int_{\mathbb{R}} d \bar{x} \ \lvert A \psi (\bar{x}) \rvert^{2} = \int d \bar{x} \ \lvert \int_{\mathbb{R}} dx \ K_{A}(\bar{x} , x ) \psi ( x ) \rvert^{2} .$$ Using the CS-inequality in the RHS, we get $$\lVert A \psi \rVert^{2}_{2} \leq \int_{\mathbb{R}} d \bar{x} \ \left( \int_{\mathbb{R}} dx \ \lvert K_{A}(\bar{x},x) \ \rvert^{2} \right) \left( \int_{\mathbb{R}} dy \ \lvert \psi (y) \rvert^{2}\right) = \int_{\mathbb{R} \times \mathbb{R}} d \bar{x} dx \ \lvert K_{A}(\bar{x},x) \rvert^{2} \ \lVert \psi \rVert_{2}^{2} .$$ Now, IF (a big if) $$\lVert K_{A} \rVert_{2}^{2} \equiv \int_{\mathbb{R} \times \mathbb{R}} dx dy \ \lvert K_{A}(x,y) \rvert^{2} < \infty ,$$ then $K_{A} \in L^{2} \left( \mathbb{R} \times \mathbb{R}\right)$ and $$\lVert A \psi \rVert^{2}_{2} \ \leq \lVert K_{A} \rVert^{2}_{2} \ \lVert \psi \rVert^{2}_{2} \ \ \Rightarrow \ \ \lVert A \rVert_{2} \leq \lVert K_{A} \rVert_{2} .$$ In this case, and only in this case, $A \in \mathcal{B}\left( L^{2}( \mathbb{R}) \right)$, i.e., it belongs to the algebra of bounded operators on the Hilbert of square-integrable functions on $\mathbb{R}$.

 

[vanhees71]

This might be a matter of semantics, but it seems to me that if your Hilbert space is $L^2$, then it's not that there is an isomorphism between square-integrable functions $\psi(x)$ and kets $|\psi\rangle$, but the kets ARE square-integrable functions (or the functions modulo the equivalence relation $\psi \approx \phi$ iff $\int dx |\phi(x) - \psi(x)|^2 = 0$).

Or maybe this is one of those cases in mathematics where it's a matter of taste whether you say that this mathematical object is equal to this other mathematical object, or it is isomorphic to it.

But my objection to your isomorphism is the ambiguity between $\langle x | \psi\rangle$ meaning a real number (the value of $\psi$ at point $x$) and the function. The notation makes it seem like you're forming an inner product, so the result should be a real number. But the isomorphism is between kets and functions, not between kets and real numbers.

The definition of $L^2$, of course, implies your equivalence class, i.e., two square integrable functions are taken the same, if
$$\int_{\mathbb{R}} \mathrm{d} x |\psi_1(x)-\psi_2(x)|^2=0.$$

Then what I've written down, is indeed an isomorphism. First, the expression $\langle x|\psi \rangle$ is indeed defined only on the "nuclear space", i.e., the subspace $D$ where the position operator is defined as an essentially self-adjoint operator, but this space is dense in $\mathcal{H}$, and thus you can extent the definition to any Hilbert space vector by defining it through appropriate (and unique) limit. Of course to begin with the expression $\langle x|\psi \rangle$ is not a usual Hilbert-space inner product but denotes the application of the linear functional $\langle x| \in D^* \superset \mathcal{H}$ to a vector $|\psi \rangle \in D$, where this functional is well defined.

Last but not least you are right, I should have written that
$$\zeta:\mathcal{H} \rightarrow L^2$$
is defined as the mapping
$$|\psi \rangle \mapsto \psi:\mathbb{R} \rightarrow \mathbb{C}, \quad \text{with} \quad \psi(x)=\langle x|\psi \rangle,$$
but this I consider a bit too much complication for a physics posting. Every physicists understands that $\psi(x)$ is a sloppy notation for both the function and its value at position $x$. Of course $\psi(x)$ is not necessarily real- but in general complex-valued. There's no necessity to complicate things by insisting on real wave functions, in which case you'd need a two-component wave function anyway to admit the representation of the Heisenberg algebra generated by position and momentum operators with $[\hat{x},\hat{p}]=\mathrm{i} \hbar \hat{1}$. The usual complex-valued wave-function approach a la Schrödinger is much more convenient and used in all textbooks I know.

 

[vanhees71]

An important remark has to be said about the integral kernel $K_{A}(\bar{x}, x) = \langle \bar{x}| A | x \rangle$: Let $V = \{ | \psi \rangle \}$ be the space of all kets (whatever that means) with $\int_{\mathbb{R}} dx \ | x \rangle \langle x | = 1$ being the identity operator on $V$. Let $\mathcal{B}(V)$ be the algebra of all bounded operators on $V$. Take $A \in \mathcal{B}(V)$ and define the function $(A \psi) (\bar{x}) \equiv \langle \bar{x}| A | \psi \rangle$ on $\mathbb{R}$. Using the identity operator on $V$, we may write $$(A \psi) (\bar{x}) = \int_{\mathbb{R}} dx \ \langle \bar{x}| A | x \rangle \langle x | \psi \rangle .$$ Furthermore, assume that $\langle x | \psi \rangle \equiv \psi (x) \in L^{2}(\mathbb{R})$ and write $$A \psi ( \bar{x}) = \int_{\mathbb{R}} dx \ K_{A}(\bar{x} , x ) \psi ( x ) .$$ Now, consider taking the $L^{2}$-norm squared: $$\lVert A \psi \rVert^{2}_{2} = \int_{\mathbb{R}} d \bar{x} \ \lvert A \psi (\bar{x}) \rvert^{2} = \int d \bar{x} \ \lvert \int_{\mathbb{R}} dx \ K_{A}(\bar{x} , x ) \psi ( x ) \rvert^{2} .$$ Using the CS-inequality in the RHS, we get $$\lVert A \psi \rVert^{2}_{2} \leq \int_{\mathbb{R}} d \bar{x} \ \left( \int_{\mathbb{R}} dx \ \lvert K_{A}(\bar{x},x) \ \rvert^{2} \right) \left( \int_{\mathbb{R}} dy \ \lvert \psi (y) \rvert^{2}\right) = \int_{\mathbb{R} \times \mathbb{R}} d \bar{x} dx \ \lvert K_{A}(\bar{x},x) \rvert^{2} \ \lVert \psi \rVert_{2}^{2} .$$ Now, IF (a big if) $$\lVert K_{A} \rVert_{2}^{2} \equiv \int_{\mathbb{R} \times \mathbb{R}} dx dy \ \lvert K_{A}(x,y) \rvert^{2} < \infty ,$$ then $K_{A} \in L^{2} \left( \mathbb{R} \times \mathbb{R}\right)$ and $$\lVert A \psi \rVert^{2}_{2} \ \leq \lVert K_{A} \rVert^{2}_{2} \ \lVert \psi \rVert^{2}_{2} \ \ \Rightarrow \ \ \lVert A \rVert_{2} \leq \lVert K_{A} \rVert_{2} .$$ In this case, and only in this case, $A \in \mathcal{B}\left( L^{2}( \mathbb{R}) \right)$, i.e., it belongs to the algebra of bounded operators on the Hilbert of square-integrable functions on $\mathbb{R}$.

Of course, you are right there are mathematical complications with the domains and co-domains again. First everything applies only to the domain $D \subset \mathcal{H}$ of $\hat{x}$. Quantum theory in the usual sense needs unbound operators to realize the Heisenberg algebra of $\hat{x}$ and $\hat{p}$. As I said, all this can be read in Galindo, Pascual in a modern way, where the sloppy hand-waving treatment of all these mathematical complications physicists are used to, are justified mathematically rigorously in a modern way using the rigged-Hilbert-space formulation. You can as well read the famous classic by von Neumann (but skip everything that's related to the physics interpretation, which is flawed).

 

[stevendaryl]

Of course to begin with the expression $\langle x|\psi \rangle$ is not a usual Hilbert-space inner product but denotes the application of the linear functional $\langle x| \in D^* \superset \mathcal{H}$ to a vector $|\psi \rangle \in D$, where this functional is well defined.

This is actually a good example of the power of notation, for good or ill. $\langle x | \phi \rangle$ looks like an example of the Hilbert space inner product, but it's not. Is that good or bad? It's bad in that it's a little misleading, but it's actually good, because a lot of the operations that are done using inner products have analogs for this type of expression, such as writing $\langle \phi|\psi\rangle = \int \langle \phi|x\rangle \langle x|\psi\rangle dx$

 

[bob012345]

This is actually a good example of the power of notation, for good or ill. $\langle x | \phi \rangle$ looks like an example of the Hilbert space inner product, but it's not. Is that good or bad? It's bad in that it's a little misleading, but it's actually good, because a lot of the operations that are done using inner products have analogs for this type of expression, such as writing $\langle \phi|\psi\rangle = \int \langle \phi|x\rangle \langle x|\psi\rangle dx$

My takeaway from this discussion is that the physics sometimes gets lost in the math. Very disheartening when intelligent people can't even agree on the mathematics of quantum theory.

 

[stevendaryl]

My takeaway from this discussion is that the physics sometimes gets lost in the math. Very disheartening when intelligent people can't even agree on the mathematics of quantum theory.

I don't think the disagreement is very deep. It's a disagreement about how best to describe what's going on, but there's no disagreement about how to use the mathematics.

 

[bob012345]

I don't think the disagreement is very deep. It's a disagreement about how best to describe what's going on, but there's no disagreement about how to use the mathematics.

What is going on that the mathematical treatments are not agreeing. What is the physics? Thanks.

 

[Orodruin]

What is going on that the mathematical treatments are not agreeing.

Nothing. Nobody is disagreeing on what the actual mathematics are. The argument is about how to best define one's notation.