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samalkhaiat

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Thanks. However it seems to be five-diagonal (in one dimension) due to the second derivative in the Hamiltonian operator.

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No, this Hamiltonian is not diagonal for meaningful definitions of "diagonal" for linear operators on ##L^2(\mathbb{R})##. If it were, the position basis vectors would be eigenvectors of the Hamiltonian and the derivative operator would have to commute with the potential.

Expand the derivative over the Dirac distribution using the chain rule and the weak derivative of the distribution to see that you get terms that are not a function of x multiplied with the Dirac distribution.

Cheers,

Jazz

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No, because ##\partial/\partial x## is not diagonal in the position basis. ##|x>## is not an eigenstate of ##\partial/\partial x##.

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samalkhaiat

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What is "five-diagonal"? Look, the Hamiltonian is not diagonal in the usual sense because [itex]|x\rangle[/itex] is not an eigenstate of [itex]H[/itex]. However, if you have no problem calling [itex]\langle x | y \rangle = \delta (x-y)[/itex] an orthognality relation, and interpret [itex]\delta(x,y)[/itex] as an infinite-dimensional "unit matrix", then you can call [itex]\langle x | H ( \hat{x} , \hat{p})| y \rangle = H(x , \partial_{x}) \delta(x , y)[/itex] "diagonal".Thanks. However it seems to be five-diagonal (in one dimension) due to the second derivative in the Hamiltonian operator.

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No, you can't. H is not a scalar factor depending on x and y, it's a differential operator. This is analog to the fact that the product of a non-diagonal matrix and a diagonal matrix is in general not a diagonal matrix.What is "five-diagonal"? Look, the Hamiltonian is not diagonal in the usual sense because [itex]|x\rangle[/itex] is not an eigenstate of [itex]H[/itex]. However, if you have no problem calling [itex]\langle x | y \rangle = \delta (x-y)[/itex] an orthognality relation, and interpret [itex]\delta(x,y)[/itex] as an infinite-dimensional "unit matrix", then you can call [itex]\langle x | H ( \hat{x} , \hat{p})| y \rangle = H(x , \partial_{x}) \delta(x , y)[/itex] "diagonal".

Also, n-diagonal is defined for discrete bases where you can have sparse matrices that are populated on the minor diagonals too. And the typical discretisation of the Hamiltonian in position basis would lead to a tridiagonal matrix.

Cheers,

Jazz

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samalkhaiat

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Who said it is? [itex]H(x , \partial)[/itex] is not a function of two variables.H is not a scalar factor depending on x and y,

Exactly, [itex]H(x , \partial)[/itex] is an operator.it's a differential operator.

No, [itex]H(x , \partial_{x}) = V(x) + F(\partial_{x})[/itex] is not indexed by two continuous labels. So, it is not an analogue of infinite-dimensional “matrix”, i.e. not a function of two variables [itex]f(x,y)[/itex], rather it is the sum of two columns with infinite entries, i.e [itex]\infty \times 1[/itex] matrix. But, if we define the matrix [itex]H(x,y) \equiv \langle x | H (\hat{x} , \hat{p}) | y \rangle[/itex], then we have the following "matrix equation" [tex]H(x,y) = H(x , \partial_{x}) \delta (x,y) ,[/tex] with [itex]H(x,y) = 0[/itex] when [itex]x \neq y[/itex].This is analog to the fact that the product of a non-diagonal matrix and a diagonal matrix is in general not a diagonal matrix.

[Mentor's note: edited to remove some personal argumentation]

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Nobody said it was, but it would be the only way to make your argument work.Who said it is? [itex]H(x , \partial)[/itex] is not a function of two variables.

Yet you don't seem to understand what this implies.Exactly, [itex]H(x , \partial)[/itex] is an operator.

That's not how operators work. The derivative operator acts to the right. You can't plug in values for x and y before you evaluate the derivative. Doing that would be just as wrong as commuting the two parts. Arguing about tempered distributions is always a bit tricky, but let's just discuss the differential operator. It acts on functions at a single point, but needs an open neighbourhood of that point. That means you can't regard it as a diagonal operator, even though it's distributional representation ##\delta'## only has a single point as support (the requirement for the open set is in the integral used for the evaluation in that case).No, [itex]H(x , \partial_{x}) = V(x) + F(\partial_{x})[/itex] is not indexed by two continuous labels. So, it is not an analogue of infinite-dimensional “matrix”, i.e. not a function of two variables [itex]f(x,y)[/itex], rather it is the sum of two columns with infinite entries, i.e [itex]\infty \times 1[/itex] matrix. But, if we define the matrix [itex]H(x,y) \equiv \langle x | H (\hat{x} , \hat{p}) | y \rangle[/itex], then we have the following "matrix equation" [tex]H(x,y) = H(x , \partial_{x}) \delta (x,y) ,[/tex] with [itex]H(x,y) = 0[/itex] when [itex]x \neq y[/itex].

[Mentor's note: edited to remove some personal argumentation]

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The second derivative in the Hamiltonian is [tex]\frac{\partial ^2 f}{\partial x^2}=\frac{f(i+1)-2f(i)+f(i-1)}{\delta x^2}[/tex] and so relates the main diagonal (corresponding to x(i)) to the other two lines- above (corresponding to x(i+1) ) and below (corresponding to x(i-1)) the main diagonal. Therefore the matrix is three-diagonal as mentioned by Jazz in the post 7. Excuse me for my mistake.What is "five-diagonal"?

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samalkhaiat

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You are expecting too much from formal expression. Distribution in general and the Delta “function” in particular is determined by means of continuous functions as a linear (continuous) functional on those functions, i.e., the value of the functional [itex]\delta[/itex] on the function [itex]\varphi[/itex] is the number [itex]\varphi (0)[/itex] and it is denoted by [itex]( \delta , \varphi ) = \varphi (0)[/itex]. By analogy with ordinary functions, we formally write [itex]\delta (x)[/itex] instead of [itex]\delta[/itex], and regard [itex]x[/itex] as the argument of the (good) functions on which the functional [itex]\delta[/itex] operates: the role of the integral [itex]\int dx \delta(x) \varphi(x) = \varphi(0)[/itex] is played here by the quantity [itex](\delta , \varphi)[/itex], which is the value of the functional [itex]\delta[/itex] on the function [itex]\varphi[/itex]. We never need to carry out any differentiation on the delta “function”, thanks to the relation [tex](D^{a} \delta , \varphi ) = (-1)^{|a|} ( \delta , D^{a}\varphi) = (-1)^{|a|} \varphi^{(a)}(0) .[/tex] So, the formal expression [itex]\langle x|H(\hat{x},\hat{p})|y\rangle = H(x,\partial_{x}) \delta(x,y)[/itex] is useful only when you need to convert the representation-free Schrodinger equation, [itex]i \partial_{t} |\Psi (t) \rangle = H(\hat{x},\hat{p}) | \Psi(t) \rangle[/itex], into a partial differential equation in the position-space:The second derivative in the Hamiltonian is [tex]\frac{\partial ^2 f}{\partial x^2}=\frac{f(i+1)-2f(i)+f(i-1)}{\delta x^2}[/tex] and so relates the main diagonal (corresponding to x(i)) to the other two lines- above (corresponding to x(i+1) ) and below (corresponding to x(i-1)) the main diagonal. Therefore the matrix is three-diagonal as mentioned by Jazz in the post 7. Excuse me for my mistake.

[tex]

i \partial_{t} \langle x | \Psi (t) \rangle = \int dy \ \langle x | H ( \hat{x} , \hat{p}) | y \rangle \langle y | \Psi (t) \rangle .[/tex]

By the above differentiation role, we obtain the PDE of Schrodinger [tex]i \partial_{t} \Psi (x,t) = \int dy \ \delta (x,y) \ H(y,\partial_{y}) \Psi ( y,t ) = H(x,\partial_{x}) \Psi (x,t) .[/tex] In momentum space, we introduce the matrix [itex]\langle p | H(\hat{x},\hat{p}) | \bar{p}\rangle[/itex] into the representation-free Shrodinger equation and transform it into an integral equation [tex]i \partial_{t} \Psi (p,t) = \frac{p^{2}}{2m} \Psi (p,t) + \int d \bar{p} \ V(p - \bar{p}) \Psi ( \bar{p} , t) .[/tex]

Sam

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