# A Starting with the Schrodinger equation, how do we find the Hamiltonian matrix?

#### lavinia

Gold Member
In his lectures on Quantum Physics, Richard Feynmann derives the Hamiltonian matrix as an instantaneous amplitude transition matrix for the operator that does nothing except wait a little while for time to pass.(Chapter 8 book3)

The instantaneous rate of change of the amplitude that the wave function is in a specific state is the sum or integral of the Hamiltonian times the wave function.

He then says that for a particle in position coordinates (Chapter 16 section 12) the integral of the Hamiltonian matrix times the wave function equals the - h/2m.Laplacian plus the potential. This is Shroedinger's equation and he goes on to say that there is no derivation of this. Shroedinger just came up with it.

My question is: starting with the Shrodinger equation how do we find the Hamiltonian matrix?
I also wonder whether there is a better motivation than Feynmann's explanation.

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#### Greg Bernhardt

@lavinia did you find any more insight on this topic?

#### lavinia

Gold Member
@lavinia did you find any more insight on this topic?
Hi Greg

It's been a while since I asked this question.

Why do you ask?

I never got a detailed answer but the Shroedinger equation is similar to the Heat Equation but with a complex coefficient. Solutions to the Heat equation are Brownian motions and these can be otherwise derived as the limit of transition matrices from states at a given time to states in a future time. These transition matrices represent conditional probabilities that if a particle is in a state at time t what is its probability of being in another state at a future time.

The complex coefficient in the Shroedinger equation probably changes probabilities to amplitudes but with that change everything else seems to be the same and one can describe the time evolution in terms of transition matrices where now one has the conditional amplitude of going from one state to the next rather than the probability.

Also Brownian motions are continuous paths subject to a probability measure and one might wonder whether the Shroedinger equation similarly gives continuous paths of quantum states subject to some analogue of a probability measure.

This is all just a guess on my part and I never worked it through.

More generally mathematically related questions that I have asked about Physics e.g. this and as I remember a question about Jacobi fields produced by a one parameter family of masses in free fall were not answered. I stopped asking.

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#### samalkhaiat

He then says that for a particle in position coordinates (Chapter 16 section 12) the integral of the Hamiltonian matrix times the wave function equals the - h/2m.Laplacian plus the potential. This is Shroedinger's equation and he goes on to say that there is no derivation of this. Shroedinger just came up with it.
My question is: starting with the Shrodinger equation how do we find the Hamiltonian matrix?
I think he means that there is no derivation of the abstract Schrodinger equation: $$i \hbar \partial_{t}|\Psi (t) \rangle = H |\Psi (t) \rangle , \ \ \ \ \ (1)$$ even though one can argue that it follows from the fact that $H$ is the generator of time evolution. However, the Schrodinger wave equation (i.e., the coordinate representation of (1)) is derivable from (1) in exactly the way he described it: One starts by inserting the identity $1 = \int dy \ |y \rangle \langle y |$ and left multiplying (1) by $\langle x |$$$i \hbar \partial_{t} \langle x | \Psi (t) \rangle = \int dy \ \langle x | H | y \rangle \langle y |\Psi (t)\rangle ,$$ or $$i \hbar \partial_{t} \Psi (x,t) = \int dy \ \langle x | H | y \rangle \ \Psi (y,t) . \ \ \ \ (2)$$ Since the matrix element of $H$ (the integral kernel) is a c-number, we may regard it as a “function” of two variables, $H(x,y)$, and equivalently represent it as function of the difference and the average of its arguments: $$\langle x |H|y \rangle = H(x,y) = h \left( \frac{x+y}{2} , x-y \right).$$ Next, we (partially) expand the difference in Fourier integral $$\langle x |H|y \rangle = \frac{1}{2 \pi \hbar} \int dp \ h \left( \frac{x+y}{2} , p \right) e^{\frac{i}{\hbar} p ( x - y)} ,$$ and declare that the function $h(x,p)$ is the classical Hamiltonian: $$\langle x |H|y \rangle = \frac{1}{2 \pi \hbar} \int dp \ \left( \frac{p^{2}}{2m} + V \left( \frac{x+y}{2} \right) \right) e^{\frac{i}{\hbar} p (x-y)} . \ \ \ \ (3)$$ The rest is easy algebra: substitute (3) in (2) and use $$p^{2} \left( e^{- \frac{i}{\hbar}py} \right) = - \hbar^{2} \ \partial^{2}_{y} \left( e^{- \frac{i}{\hbar}py} \right),$$ then integrate by parts twice and use the integral representation of Dirac’s delta: $$\delta ( x - y ) = \frac{1}{2 \pi \hbar} \int dp \ e^{\frac{i}{\hbar}p( x - y )} .$$ This leads to $$i \hbar \partial_{t} \Psi (x,t) = - \frac{\hbar^{2}}{2m} \int dy \ \delta ( x - y ) \ \partial^{2}_{y} \Psi (y,t) + \int dy \ \delta ( x - y ) V \left(\frac{x + y}{2} \right) \Psi (y,t) ,$$ which gives us the Schrodinger wave equation $$i \hbar \frac{\partial}{\partial t} \Psi (x,t) = - \frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}} \Psi (x,t) + V (x) \Psi (x,t) .$$
Having said this, I still don’t know whether or not by “the Hamiltonian matrix” you meant equation (3)?

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#### forcefield

I think Feynman's "the Hamiltonian matrix" in chapter 8 is about Heisenberg's matrix mechanics.

#### A. Neumaier

starting with the Schrödinger equation how do we find the Hamiltonian
In the cases considered by Schrödinger it was found by analogy to classical mechanics, by a not fully specified process called ''quantization''. To find the potential in concrete cases, one may measure the spectrum and then apply inverse scattering methods to recover from the spectrum an approximate potential.

#### vanhees71

Gold Member
I can't make sense of Eq. (3) in #4.

Usually you have a Hamiltonian of the form
$$\hat{H}=\hat{H}_0+V(\hat{x})=\frac{1}{2m} \hat{p}^2+V(\hat{x}),$$
which is not derived but just guessed from classical mechanics and then after used to calculate measurable predictions and then checked against experiment whether it works out (that's one of many ways physics works in finding new "laws").

Then the position representation is indeed given by the matrix element
$$H(x,y)=\langle x|\hat{H}|y \rangle.$$
We can calculate them separately for the kinetic energy and the potential (using natural units with $\hbar=1$)
$$\langle x|\hat{p}^2|y \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|\hat{p}^2|p \rangle \langle p|y \rangle=\int_{\mathbb{R}} \mathrm{d} p p^2 \langle x|p \rangle \langle p|y \rangle = \frac{1}{2 \pi} \int_{\mathbb{R}} \mathrm{d} p p^2 \frac{1}{2 \pi} \exp[\mathrm{i} p(x-y)] = -\frac{1}{2 \pi} \partial_x^2 \int_{\mathbb{R}} \mathrm{d} p \exp[\mathrm{i} p(x-y)]=-\partial_x^2 \delta(x-y).$$
Thus
$$\hat{H}_0 \psi(x):=\frac{1}{2m} \langle x|\hat{H}_0 \psi \rangle = \frac{1}{2m} \int_{\mathbb{R}} \mathrm{d} y \psi(y) (-\partial_x^2) \delta(x-y)) =\frac{1}{2m} \int_{\mathbb{R}} \psi(y) (-\partial_y^2) \delta(x-y) = \frac{1}{2m} \int_{\mathbb{R}} \mathrm{d} y (-\partial_y^2) \psi(y) \delta(x-y)=-\frac{1}{2m} \partial_x^2 \psi(x).$$
For the potential part it's even simpler without the matrix element

$$V(\hat{x}) \psi(x)=\langle x|V(\hat{x})|\psi \rangle=\langle x V(\hat{x})|\psi = V(x) \langle x|\psi \rangle = V(x) \psi(x).$$

#### samalkhaiat

I can't make sense of Eq. (3) in #4.
Even though Eq(3) is nearly as old as QM, I am not surprised to hear you saying that. This is because of the unfortunate fact that many people are still unfamiliar with the great work of Herman Wyle on the use of group theory in QM:
As early as 1925, Weyl considered the projective unitary representation of the abelian group of translations $\mathbb{R}^{2}$ in phase space: \begin{align*} \pi : & \mathbb{R}^{2} \to \mbox{PU}(\mathcal{H}) \\ & ( \alpha , \beta ) \mapsto e^{\frac{i}{\hbar} (\beta Q - \alpha P)} \end{align*} , where $[Q , P] = i \hbar$. Indeed, it is easy to show that $\pi (\alpha , \beta )$ is a projective unitary operator on $\mathcal{H}$: $$\pi (\alpha , \beta) \pi (\alpha^{\prime} , \beta^{\prime}) = e^{- \frac{i}{2 \hbar} (\alpha \beta^{\prime} - \beta \alpha^{\prime})} \ \pi (\alpha + \alpha^{\prime} , \beta + \beta^{\prime}) .$$ Then, Weyl extends the projective representation $\pi$ to a representation of the (twisted) group algebra of $\mathbb{R}^{2}$ on the same Hilbert space $\mathcal{H}$ (group algebra is another unfamiliar concept, I suppose!). In English, for a function $a ( q , p )$ on phase space, Weyl takes the associated operator to be $$A \equiv \hat{\pi} (a) = \int_{\mathbb{R}^{2}} d \alpha d \beta \ \tilde{a}( \alpha , \beta ) \ \pi (\alpha , \beta ) , \ \ \ \ (1)$$ where $$\tilde{a} ( \alpha , \beta ) = \frac{1}{(2 \pi \hbar)^{2}} \int_{\mathbb{R}^{2}} dq dp \ e^{- \frac{i}{\hbar} (\beta q - \alpha p)} \ a ( q , p ) ,$$ is the Fourier transform of $a$. The operator $\hat{\pi}(a)$ is, at least formally, self-adjoint if $a(q,p)$ is real-valued, and is well-defined and of trace class if $a(q,p)$ belongs to the Schwartz space $\mathcal{S}(\mathbb{R}^{2})$, i.e. if $a$ is $C^{\infty}$ and decreases, together with all its derivatives, faster than the reciprocal of any polynomial at infinity. If the function $a$ is $C^{\infty}$ and grows, together with all its derivatives, at most polynomially at infinity, then $\hat{\pi}(a)$ will be (in general) a densely defined unbounded operator. And, of course, if $\tilde{a} \in L^{1}(\mathbb{R}^{2})$, then $\hat{\pi}(a)$ is a bounded (i.e., continuous) operator: $$\lVert \hat{\pi}(a) \rVert \leq \int_{\mathbb{R}^{2}} d \alpha d \beta \ \lvert \tilde{a}( \alpha , \beta ) \rvert .$$
Okay, let us now derive Eq(3) of #4 from the Weyl formula (1) which we now write as $$H \equiv \hat{\pi}(h) = \int_{\mathbb{R}^{2}} d \alpha d \beta \ \tilde{h}( \alpha , \beta ) \ \pi ( \alpha , \beta ) . \ \ \ \ (2)$$ In the coordinate representation, i.e. when we consider the realization of the space $\mathcal{H}$ in the form $L^{2}(\mathbb{R})$, an observable $H : L^{2}(\mathbb{R}) \to L^{2}(\mathbb{R})$ is characterized as an integral operator defined by its kernel (or matrix elements): $$(H \Psi )(q) = \int_{\mathbb{R}} d \bar{q} \ H(q , \bar{q}) \Psi (\bar{q}) , \ \ \ \forall \Psi \in L^{2} (\mathbb{R}) .$$ Alternatively, as Eq(3) of #4 says, we may define the observable $H$ in terms of its (Weyl) symbol, which is a function $h(q,p)$ on phase space. We will now show that the symbol $h(x,p)$ corresponds to the integral kernel $H(x,y) \equiv \langle x | H |y \rangle$ by means of $$\langle x | H | y \rangle = \frac{1}{2 \pi \hbar} \int_{\mathbb{R}} dp \ h \left( \frac{x + y}{2} , p \right) \ e^{\frac{i}{\hbar} p (x - y)} .$$ Okay, let us calculate the following matrix element of the operator $H$ of Eq(2): $$\langle q - \frac{\lambda}{2} | H | q + \frac{\lambda}{2} \rangle = \int d \alpha d \beta \ \tilde{h} ( \alpha , \beta ) \ \langle q - \frac{\lambda}{2} | \pi ( \alpha , \beta ) | q + \frac{\lambda}{2} \rangle . \ \ \ (3)$$ The matrix element of $\pi$ can be calculated easily from the factorized form $$\pi ( \alpha , \beta ) = e^{- \frac{i}{2 \hbar} \alpha \beta} \ e^{\frac{i}{\hbar} \beta Q} \ e^{- \frac{i}{\hbar} \alpha P} .$$ Using $e^{- \frac{i}{\hbar} \alpha P} | q \rangle = |q + \alpha \rangle$ and $e^{\frac{i}{\hbar} \beta Q} |q \rangle = e^{\frac{i}{\hbar} \beta q }|q \rangle$, we obtain $$\langle q - \frac{\lambda}{2} | \pi ( \alpha , \beta ) | q + \frac{\lambda}{2} \rangle = e^{\frac{i}{\hbar} \beta q} \ \delta ( \lambda + \alpha ) .$$ Substituting this in (3) then multiplying the result by $e^{\frac{i}{\hbar} \lambda p}$ and integrating over $\lambda$, we find $$\int_{\mathbb{R}} d \lambda \ e^{\frac{i}{\hbar} \lambda p} \langle q - \frac{\lambda}{2} | H | q + \frac{\lambda}{2} \rangle = \int_{\mathbb{R}^{2}} d \alpha d \beta \ \tilde{h} ( \alpha , \beta ) \ e^{\frac{i}{\hbar}( \beta q - \alpha p )} .$$ We recognize the RHS as the inverse Fourier transform $h (q , p)$ of $\tilde{h}( \alpha , \beta )$: $$\int_{\mathbb{R}} d \lambda \ e^{\frac{i}{\hbar} \lambda p} \langle q - \frac{\lambda}{2} | H | q + \frac{\lambda}{2} \rangle = h (q , p ) .$$ Thus $$\int_{\mathbb{R}} dp \int_{\mathbb{R}} d \lambda \ e^{\frac{i}{\hbar} ( \lambda - \lambda^{\prime} ) p} \langle q - \frac{\lambda}{2} | H | q + \frac{\lambda}{2} \rangle = \int_{\mathbb{R}} dp \ h(q , p) \ e^{- \frac{i}{\hbar} \lambda^{\prime} p} .$$ The p-integral on the LHS produces $2 \pi \hbar \delta ( \lambda - \lambda^{\prime} )$, then the $\lambda$-integration gives $$\langle q - \frac{\alpha}{2} | H |q + \frac{\alpha}{2} \rangle = \frac{1}{2 \pi \hbar} \int dp \ h(q,p) \ e^{- \frac{i}{\hbar} \alpha p} .$$ Now, you have it: define $$x = q - \frac{\alpha}{2} , \ \ y = q + \frac{\alpha}{2},$$ and you find $$\langle x | H | y \rangle = \frac{1}{2 \pi \hbar} \int dp \ h \left( \frac{x + y}{2} , p \right) \ e^{\frac{i}{\hbar} p ( x - y ) } . \ \ \ (4)$$ Conversely, you can start from (4) and arrive at the Weyl formula (2).
Mathematicians generalized the above Weyl’s work on the abelian group $\mathbb{R}^{2n}$ to an arbitrary locally compact Lie group $G$ and developed what we now know as “abstract harmonic analysis” or “functional analysis on locally compact groups”: Given a unitary representation $\pi$ of a locally compact group $G$ on some Hilbert space $\mathcal{H}$: $\pi : G \to \mbox{U}(\mathcal{H})$, $g \mapsto \pi (g)$, and a “nice” function $a : G \to \mathbb{R}$ on $G$, then for all $a \in L^{1}(G)$, the operator $$\hat{\pi}(a) = \int_{G} d \mu (g) \ a(g) \ \pi (g) ,$$ is bounded $$\lVert \hat{\pi} \rVert \leq \int_{G} d \mu (g) \ \lvert a (g) \rvert .$$ Indeed, it is easy to show that $\hat{\pi} : L^{1}(G) \to \mathcal{L}(\mathcal{H})$ is a non-degenerate $\ast$-homomorphism from the group algebra $\left( L^{1}(G , d \mu ) , \ast \right)$ into the algebra of bounded operators on the same Hilbert space. That is, $$\hat{\pi} (a) \ \hat{\pi} (b) = \hat{\pi} ( a \ast b ),$$$$\hat{\pi}^{\ast}(a) = \hat{\pi} ( a^{\ast} ),$$ where $$\left( a \ast b \right) ( g ) = \int_{G} d \mu (h) \ a (h) \ b (h^{-1}g),$$ is the convolution product on $L^{1}(G)$, and $a \mapsto a^{\ast}$ is an involution on $L^{1}(G)$ defined by $$a^{\ast} (g) = \Delta (g^{-1}) \bar{a} (g^{-1}),$$ where $\Delta : G \to \mathbb{R}$ is the (continuous) modular function on $G$, a group homomorphism $\Delta (g) \Delta (h) = \Delta (gh)$. Moreover, we can show that there exists a bijective correspondence between the continuous unitary representations $\pi (G)$ in $\mathcal{H}$ and the non-degenerate $\ast$-representations $\hat{\pi}\left( L^{1}(G) \right)$ in the same Hilbert space.
So, Eq(3) of #4 is in fact the reason why we now do abstract harmonic analysis on locally compact Lie groups.

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#### vanhees71

Gold Member
Interesting, I think, it's just that I don't understand the notation. Particularly if you write a general operator valued function $H(\hat{x},\hat{p})$ it's not so clear how you define the matrix elements. At least you need some operator-ordering description to clearly define them. If I remember right, there's something called "Weyl ordering", where everything becomes well defined. My example was just for the most simple non-relativistic Lagrangian,
$$\hat{H}=\frac{1}{2m} \hat{p}^2 + V(\hat{x}),$$
where no operator-ordering problems occur.

This is already the case for the slightly more general form taking into account an external magnetic field in terms of its vector potential,
$$\hat{H}=\frac{1}{2m} [\hat{\vec{p}}-q \vec{A}(\hat{\vec{x}})]^2 + V(\vec{x}).$$
Of course, also here it's easy to calculate the matrix elements, and it finally boils down to
$$\hat{\vec{p}} \rightarrow \; -\mathrm{i} \vec{\nabla}$$
and
$$\hat{\vec{x}} \Rightarrow \; \vec{x},$$
in the position representation.

Of course, what the physists call Heisenberg algebra has the profound group-theoretical background of the theory of projective representations of the Galilei group you've indicated, but that's indeed not common knowledge in the physics community.

#### samalkhaiat

If I remember right, there's something called "Weyl ordering", where everything becomes well defined. My example was just for the most simple non-
Indeed, “Weyl symmetric ordering” or “Weyl quantization procedure” is exactly what Eq(3) of #4 does for you. Let me use your notations and consider the operator $H( \hat{x} , \hat{p})$. Let us order all the $\hat{x}$-operators to the left of the $\hat{p}$-operators in $H( \hat{x} , \hat{p})$. Then we can unambiguously write $$\langle x |H ( \hat{x} , \hat{p} ) | p \rangle = H (x , p) \ \langle x | p \rangle = \frac{1}{\sqrt{2 \pi \hbar}} H(x,p) \ e^{\frac{i}{\hbar}px} , \ \ \ \ (1)$$ where $H(x,p)$ is a $C^{\infty}$-function on the phase space $\mbox{T}^{\ast}(\mathbb{R}) \cong \mathbb{R}^{2}$, the ordinary continuous function $H: \mathbb{R}^{2} \to \mathbb{R}$.
But, we can also write (by inserting the identity $1 = \int dp |p \rangle \langle p |$) $$\langle x |H ( \hat{x} , \hat{p} ) | y \rangle = \int dp \ \langle x | H ( \hat{x} , \hat{p} ) | p \rangle \langle p | y \rangle = \frac{1}{\sqrt{2 \pi \hbar}} \int dp \ \langle x | H( \hat{x} , \hat{p}) |p \rangle \ e^{- \frac{i}{\hbar}py} . \ \ \ (2)$$ Putting (1) in (2), we get $$\langle x |H ( \hat{x} , \hat{p} ) | y \rangle = \frac{1}{2 \pi \hbar} \int dp \ H(x,p) \ e^{\frac{i}{\hbar} p (x-y)} .$$ Next, you do the same thing for the opposite ordering and then by lengthy induction process you find that symmetric ordering requires you to evaluate the phase space function $H(x,p)$ at the mid-point $\frac{x + y}{2}$. That is, you arrive at Eq(3) of #4: $$\langle x |H ( \hat{x} , \hat{p} ) | y \rangle = \frac{1}{2 \pi \hbar} \int dp \ H\left( \frac{x + y}{2},p \right) \ e^{\frac{i}{\hbar} p (x-y)} .$$

#### dextercioby

Homework Helper
Do you know any good book on the Weyl quantization? What is it any different to the Wigner phase space formulation of QM and the Moyal quantization?

#### samalkhaiat

Do you know any good book on the Weyl quantization?
1) Dubin, Hennings & Smith “Mathematical Aspects of Weyl Quantization and Phase” World Scientific (2000).
2) G. B. Folland “A Course in Abstract Harmonic Analysis” CRC Press (1995).
And of course Weyl own great book “The Theory of Groups and Quantum Mechanics”, Dover 1931(the original German edition: Leipzig 1928).

What is it any different to the Wigner phase space formulation of QM and the Moyal quantization?
Don’t get me started, unfortunately Weyl was the unlucky man in all of this:
1) He did all the work, and mathematicians invented the name “The Heisenberg Group $H_{(2n + 1)}$”, the universal covering group of the group of translations in phase space $\mathbb{R}^{2n}$. That is, the space $\mathbb{R}^{n} \oplus \mathbb{R}^{n} \oplus \mathbb{R}$ with multiplication $$(x_{1} , x_{2} , a) (y_{1} , y_{2} , b ) = ( x_{1} + y_{1}, x_{2} + y_{2} , a + b + x_{1}y_{2})$$ Weyl obtained his projective representation of $\mathbb{R}^{2n}$ from the unitary representation of $H_{(2n + 1)}$. Heisenberg was not an expert on group theory at that time.
2) von Neumann [*] was the first to point out that, in effect, Weyl quantization induces a new non-commutative product on (the space of) functions: Let $a , b \in C^{\infty}(\mathbb{R}^{2n})$, Weyl map then gives $\hat{\pi}(a) \hat{\pi}(b) = \hat{\pi}(c)$, where the phase space function $c(q,p)$ is determined by $\tilde{c} = \tilde{a} \ast_{\omega} \tilde{b}$, the twisted (i.e., non-commutative) convolution of (the Fourier transforms) $\tilde{a}$ and $\tilde{b}$ by the symplectic form (2-cocycle) $\omega$ on phase space $\mathbb{R}^{2n}$ defined by $\omega \left( (\sigma , \tau) ; (q , p)\right) = \tau \cdot q - \sigma \cdot p$, for $(\sigma , \tau) \in \hat{\mathbb{R}}^{2n} \cong \mathbb{R}^{2n}$ and $(q,p) \in T^{*}\mathbb{R}^{n} \cong \mathbb{R}^{2n}$ with $$\tilde{a}(\sigma , \tau) = \left(\mathscr{F}a\right)(\sigma , \tau) = \frac{1}{(2 \pi \hbar)^{2n}} \int_{S} d^{n}q d^{n}p \ e^{- \frac{i}{\hbar}(\tau \cdot q - \sigma \cdot p)} \ a (q , p) \ .$$ So, Weyl quantization is in fact a deformation quantization leading to the so-called Moyal star-product.
[*] J. von Neumann, Math. Anal. 104 (1931), 570-578.

3) Groenewold [**] was the first to introduce the $\star$-product (which defines the algebraic structure of deformation quantization) often, unjustly, associated with Moyal’s name even though it was not introduced in his paper [***]. And I believe that the $\star$-product was not fully understood by Moyal at that time.
[**] H.G. Groenewold, Physica 12 (1946), 405-460.
[***] J.E. Moyal, Proc. Cambridge Phil. Soc. 45 (1949), 99-124.
[****] E.P. Wigner, Phys. Rev. 40 (1932), 749-759.
In this paper, Wigner introduced his famous phase-space function $W(q,p)$ which is essentially the (phase-space) Weyl symbol of the density operator $|\psi \rangle \langle \psi |$. Mathematically speaking, Wigner map is nothing but the inverse of Weyl map, yet [****] makes no reference to Weyl work. So, was Wigner “unaware” of Weyl paper [5*], or he was just “naughty”? Well, Wigner was certainly aware of von Neumann’s work.
[5*] H. Weyl, Zeits. Phys. 46 (1927)
This classic paper of Weyl marks the start of group theoretical study of QM. In it weyl introduced two new aspects to the mathematical formulation of QM:
1) He showed how atomic spectra can be determined from the representation theory of compact groups on Hilbert space, which he worked out during 1924-25.
2) He introduced his elegant quantization method which associates a quantum mechanical operator $\hat{\pi}(a)$ to an arbitrary classical observable $a(q,p)$. In fact, Weyl’s letters to Born and Jordan indicates that he developed his quantization procedure back in 1925. He also showed how to obtain the Schrodinger representation from irreducible projective unitary representation of the Abelian group $\mathbb{R}^{2n}$ of translations in phase space, and came (infinitesimally) close of proving the so-called Stone-von Neumann theorem.

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