Hamiltonian in the position basis

In summary: I call it...I don't know, preference for the main diagonal - for me diagonal means main diagonal by default, but I realize that you may have another convention and I should have mentioned it explicitly.Cheers, JazzIn summary, the conversation discussed the expression <x|H|x\prime>=(-\hbar ^2 /2m \frac{\partial^2 }{\partial x^2}+v(x)) \delta (x-x\prime) and whether it implies that the Hamiltonian is always diagonal in the position basis. It was concluded that the Hamiltonian is not diagonal in the usual sense because |x\rangle is not an eigenstate of H, but it can be considered "diagonal" if
  • #1
hokhani
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8
According to [tex]<x|H|x\prime>=(-\hbar ^2 /2m \frac{\partial^2 }{\partial x^2}+v(x)) \delta (x-x\prime)[/tex] can one draw the conclusion that the Hamiltonian is always diagonal in the position basis?
 
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  • #2
Yes, the expression you wrote means that the Hamiltonian is diagonal in position space. Otherwise, the Schrodinger equation would not be PDE for the wavefunction.
 
  • #3
samalkhaiat said:
Yes, the expression you wrote means that the Hamiltonian is diagonal in position space. Otherwise, the Schrodinger equation would not be PDE for the wavefunction.
Thanks. However it seems to be five-diagonal (in one dimension) due to the second derivative in the Hamiltonian operator.
 
  • #4
hokhani said:
According to [tex]<x|H|x\prime>=(-\hbar ^2 /2m \frac{\partial^2 }{\partial x^2}+v(x)) \delta (x-x\prime)[/tex] can one draw the conclusion that the Hamiltonian is always diagonal in the position basis?

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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  • #5
hokhani said:
According to [tex]<x|H|x\prime>=(-\hbar ^2 /2m \frac{\partial^2 }{\partial x^2}+v(x)) \delta (x-x\prime)[/tex] can one draw the conclusion that the Hamiltonian is always diagonal in the position basis?
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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  • #6
hokhani said:
Thanks. However it seems to be five-diagonal (in one dimension) due to the second derivative in the Hamiltonian operator.
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".
 
  • #7
samalkhaiat said:
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".

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.

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
 
  • #8
Jazzdude said:
H is not a scalar factor depending on x and y,
Who said it is? [itex]H(x , \partial)[/itex] is not a function of two variables.
it's a differential operator.
Exactly, [itex]H(x , \partial)[/itex] is an 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.

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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  • #9
samalkhaiat said:
Who said it is? [itex]H(x , \partial)[/itex] is not a function of two variables.

Nobody said it was, but it would be the only way to make your argument work.

Exactly, [itex]H(x , \partial)[/itex] is an operator.

Yet you don't seem to understand what this implies.

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].

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).

[Mentor's note: edited to remove some personal argumentation]
 
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  • #10
samalkhaiat said:
What is "five-diagonal"?
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.
 
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  • #11
hokhani said:
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.

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:
[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
 

Related to Hamiltonian in the position basis

1. What is a Hamiltonian in the position basis?

A Hamiltonian in the position basis is a mathematical operator that represents the total energy of a system in terms of the position coordinates of its constituent particles. It is an important concept in quantum mechanics and is used to describe the dynamics of a system over time.

2. How is the Hamiltonian in the position basis different from the Hamiltonian in the momentum basis?

The Hamiltonian in the position basis is defined in terms of the position coordinates of a system, while the Hamiltonian in the momentum basis is defined in terms of the momentum coordinates. While they both represent the total energy of a system, they are mathematically different operators and can yield different results when used in calculations.

3. What is the significance of the Hamiltonian in the position basis in quantum mechanics?

The Hamiltonian in the position basis is a fundamental concept in quantum mechanics and is used to describe the evolution of a system in time. It is used to calculate important properties of a system, such as energy levels and transition probabilities, and is essential in understanding the behavior of quantum systems.

4. How is the Hamiltonian in the position basis related to the Schrödinger equation?

The Schrödinger equation is a fundamental equation in quantum mechanics that describes the time evolution of a quantum system. The Hamiltonian in the position basis is one of the terms in the Schrödinger equation and is used to determine the energy of a system at a given time.

5. Can the Hamiltonian in the position basis be used to calculate the position of a particle in a system?

No, the Hamiltonian in the position basis is not used to directly calculate the position of a particle in a system. It is used to calculate the total energy of a system and the probability of finding a particle at a certain position. The position of a particle is determined by the wave function, which is also described by the Schrödinger equation.

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