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hokhani
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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?
Thanks. However it seems to be five-diagonal (in one dimension) due to the second derivative in the Hamiltonian operator.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.
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##.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?
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".hokhani said:Thanks. However it seems to be five-diagonal (in one dimension) due to the second derivative in the Hamiltonian operator.
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".
Who said it is? [itex]H(x , \partial)[/itex] is not a function of two variables.Jazzdude said: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.
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.
samalkhaiat said:Who said it is? [itex]H(x , \partial)[/itex] is not a function of two variables.
Exactly, [itex]H(x , \partial)[/itex] is an 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].
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.samalkhaiat said:What is "five-diagonal"?
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.
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.
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.
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.
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.
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.