rtal said:If the energy Eigen values for hydrogen atom can be precisely determined, does that not violate uncertainty?
Since DeltaE is 0 in this case as energy (and therefore momentum) is known exactly?
eaglelake said:The energy uncertainty is zero, but the momentum uncertainty is infinite. The uncertainty principle is not violated.
sweet springs said:Hi.
If momentum uncertainty is infinite, uncertainty of kinetic energy becomes infinite! Electron in hydrogen atom has finite momentum uncertainty and position uncertainty satisfying uncertainty relation.
Regards.
eaglelake said:The position-momentum uncertainty relation is not violated in an energy eigenstate. In fact it is always valid because position and momentum never commute.
SpectraCat said:That is correct, even in the case of the particle in an infinite square well, where the momentum and energy operators *do* commute.
sweet springs said:Just a short calculation.
In the case of square well potential, both finite and infinite depth, the momentum and energy operators do not commute.
H=P^2/2m + V(x), V(x)=-V for -a<x<a, 0 for otherwise. Then [P,H]=[P,V(x)]=-ih'[d/dx,V(x)]=ih' V{δ(x+a)-δ(x-a)}, not zero. Applet http://www.falstad.com/qm1d/index.html shows us distribution of momentum. Particle is not in simple to and fro motion of the constant speed.
Regards.
sweet springs said:Just a short calculation.
In the case of square well potential, both finite and infinite depth, the momentum and energy operators do not commute.
H=P^2/2m + V(x), V(x)=-V for -a<x<a, 0 for otherwise. Then [P,H]=[P,V(x)]=-ih'[d/dx,V(x)]=ih' V{δ(x+a)-δ(x-a)}, not zero. Applet http://www.falstad.com/qm1d/index.html shows us distribution of momentum. Particle is not in simple to and fro motion of the constant speed.
Regards.
eaglelake said:We know that observables must be Hermitian operators. It turns out that the momentum operator is not Hermitian for the infinite square well! Any wave function in the infinite well must vanish at the boundaries. But the momentum eigenfunctions exp(ikx) do not vanish anywhere.
I am not sure that we can even discuss momentum in such a case.
sweet springs said:Hi.
Take it easy. Let ψ(x) and its Fourier transform ψ(p) be wave functions of a energy eigenstate of a finite square well potential of depth V. We get their formula easily. Then we enlarge V in formula of ψ(x) and ψ(p) to infinity and get the wave functions for infinite square well potential.
Regards.
eaglelake said:If we want to measure the momentum of a particle known to be in the energy eigenstate \psi _n (x) of the infinitely deep square well, then we must write \psi _n (x) in the momentum representation. Here, the basis functions are the momentum eigenfunctions \varphi (x) which satisfy the momentum eigenvalue equation
<br /> \eqalign{<br /> & - i\hbar {{d\varphi (x)} \over {dx}} = p\varphi (x)
subject to the boundary conditions \cr <br /> & \varphi ( - a) = \varphi (a) = 0 \cr} <br />
But the solutions of the differential equation e^{ipx/\hbar } cannot satisfy the boundary conditions! The only solution of the momentum eigenvalue equation is \varphi (x) = 0 everywhere. The Fourier integral doesn't exist. The probability of getting a value, any value, of momentum is zero! Does this mean that we cannot measure the momentum? Is this result simply because the infinite square well is unrealistic? I welcome everyone’s response.
It is true that the momentum eigenvalue equation does have solutions for the finite well, but when we let V \to \infty, I think we still have the difficulty with the boundary conditions.
However, there is a way to give meaning to the Fourier transform: Given the particle in energy state \psi _n (x), imagine that we suddenly remove the walls of the box containing the particle. Now we have a free particle in the energy state \psi _n (x). The momentum eigenvalue problem has non-zero solutions for a free particle and we can now proceed in the usual way.
Best wishes
eaglelake said:It is true that the momentum eigenvalue equation does have solutions for the finite well, but when we let V \to \infty, I think we still have the difficulty with the boundary conditions.
SpectraCat said:your analysis is rather flawed ... the ground state of a particle in an infinite square well can be written as a linear combination of momentum eigenstates. What's hard about that? Then the probability of finding the system in a particular eigenstate when a measurement is taken is given by the square modulus of the expansion coefficient for that state. Now, since the momentum eigenstates are continuous, we represent the "linear combination" as an integral rather than a sum, but the meaning is analogous. So, with this continuous distribution the probability of measuring a momentum between p and dp is given by:
\tilde{\psi}\left(p\right)^{*}\tilde{\psi}\left(p\right)dp
where \tilde{\psi}\left(p\right) is the Fourier transform of the position representation of the wavefunction. Now, the fact that there are discontinuities in the slope of the wavefunction at the boundaries of the well means that the Fourier transform can have non-zero components across the entire range of possible values (i.e. from \infty to -\infty). However, this is not a problem, since we know from Parseval's theorem that the momentum space and position space representations of the wavefunction must have the same norm. The position space eigenstate is certainly normalizable, so the FT in momentum space must also be.
I don't have time to derive the precise form of the momentum space wavefunction right now, but I know how to do it ... you just create a step function that is 1 inside the well and zero outside it by adding together two Heaviside functions, and take the product with the ground-state wavefunction (just a sine function). You then take the Fourier transform of this product, using the convolution theorem. The FT's of the step functions and the sine function are well-known, so it should be simple enough to write the solution ...
eaglelake said:You misunderstood what is bothering me! I apologize for the confusion. Let me try again.
We want to determine the possible results of a measurement and the probability distribution of those results. These are the fundamental calculations that we make in quantum mechanics. The postulates tell us how to do this. Among other things, we are told to solve the eigenvalue equation for the observable to be measured.
For a particle in an infinitely deep square well, the energy eigenvalue equation (Schrodinger’s time independent equation) is
- {{\hbar ^2 } \over {2m}}{{d^2 \psi (x)} \over {dx^2 }} = E\psi (x)
where \psi ( - a) = \psi (a) = 0
This is a boundary value problem; the boundary conditions cannot be ignored. We solve for \psi (x) without difficulty. Since we have written \psi (x) in position space, \psi (x) = \left\langle {x}\mathrel{\left | {\vphantom {x \psi }}\right. \kern-\nulldelimiterspace}{\psi } \right\rangle is the amplitude for finding the particle, known to be in state \psi (x), at position x.
Now we want to measure the momentum for a particle known to be in energy eigenstate \psi _n (x). We proceed in the same way. We need to write \psi _n (x) in momentum space, where the basis functions are the eigenfunctions of the momentum operator. You state, “the ground state of a particle in an infinite square well can be written as a linear combination of momentum eigenstates. What's hard about that? “ It is not “hard”, rather I think, it’s impossible! Let me try to explain------
By definition, the eigenfunctions are solutions of the eigenvalue equation. (I apologize for lecturing, but I want to be clear on this.) We must solve the momentum eigenvalue equation for a particle in the infinitely deep well, <br /> - i\hbar {{d\varphi } \over {dx}} = p\varphi <br />
where <br /> \varphi ( - a) = \varphi (a) = 0<br />
There is no solution because the boundary conditions cannot be satisfied! You assume incorrectly that <br /> e^{ikx} <br /> is the momentum eigenfunction. You cannot write the Fourier transform <br /> \left\langle {k}<br /> \mathrel{\left | {\vphantom {k {\psi _n }}}<br /> \right. \kern-\nulldelimiterspace}<br /> {{\psi _n }} \right\rangle = \int_{ - a}^a {\varphi ^* (k,x)\psi _n (x)dx} <br /> because <br /> \varphi (k,x) = 0<br /> everywhere inside the well. What am I missing?
eaglelake said:You misunderstood what is bothering me! I apologize for the confusion. Let me try again.
We want to determine the possible results of a measurement and the probability distribution of those results. These are the fundamental calculations that we make in quantum mechanics. The postulates tell us how to do this. Among other things, we are told to solve the eigenvalue equation for the observable to be measured.
For a particle in an infinitely deep square well, the energy eigenvalue equation (Schrodinger’s time independent equation) is
- {{\hbar ^2 } \over {2m}}{{d^2 \psi (x)} \over {dx^2 }} = E\psi (x)
where \psi ( - a) = \psi (a) = 0
This is a boundary value problem; the boundary conditions cannot be ignored. We solve for \psi (x) without difficulty. Since we have written \psi (x) in position space, \psi (x) = \left\langle {x}\mathrel{\left | {\vphantom {x \psi }}\right. \kern-\nulldelimiterspace}{\psi } \right\rangle is the amplitude for finding the particle, known to be in state \psi (x), at position x.Now we want to measure the momentum for a particle known to be in energy eigenstate \psi _n (x). We proceed in the same way. We need to write \psi _n (x) in momentum space, where the basis functions are the eigenfunctions of the momentum operator. You state, “the ground state of a particle in an infinite square well can be written as a linear combination of momentum eigenstates. What's hard about that? “ It is not “hard”, rather I think, it’s impossible! Let me try to explain------
By definition, the eigenfunctions are solutions of the eigenvalue equation. (I apologize for lecturing, but I want to be clear on this.) We must solve the momentum eigenvalue equation for a particle in the infinitely deep well, <br /> - i\hbar {{d\varphi } \over {dx}} = p\varphi <br />
where <br /> \varphi ( - a) = \varphi (a) = 0<br />
There is no solution because the boundary conditions cannot be satisfied! You assume incorrectly that <br /> e^{ikx} <br /> is the momentum eigenfunction. You cannot write the Fourier transform <br /> \left\langle {k}<br /> \mathrel{\left | {\vphantom {k {\psi _n }}}<br /> \right. \kern-\nulldelimiterspace}<br /> {{\psi _n }} \right\rangle = \int_{ - a}^a {\varphi ^* (k,x)\psi _n (x)dx} <br /> because <br /> \varphi (k,x) = 0<br /> everywhere inside the well. What am I missing?
eaglelake said:For a particle in an infinitely deep square well, the energy eigenvalue equation (Schrodinger’s time independent equation) is
- {{\hbar ^2 } \over {2m}}{{d^2 \psi (x)} \over {dx^2 }} = E\psi (x)
where \psi ( - a) = \psi (a) = 0
eaglelake said:By definition, the eigenfunctions are solutions of the eigenvalue equation. (I apologize for lecturing, but I want to be clear on this.) We must solve the momentum eigenvalue equation for a particle in the infinitely deep well, <br /> - i\hbar {{d\varphi } \over {dx}} = p\varphi <br />
where <br /> \varphi ( - a) = \varphi (a) = 0<br />
There is no solution because the boundary conditions cannot be satisfied!
