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Electron possessing zero K.E

  1. Jan 26, 2013 #1
    hi guys, is it possible for electron to have zero K.E in a confined region in space?

    My guessed is, due to the wave nature of electron, if electron were to possess zero KE, its wavelength will have to approach infinity since p = h / wavelength.
    therefore its not possible for electron to have zero KE in a confined space.
    But it can possess zero K.E in space that is not confined.
     
    Last edited: Jan 26, 2013
  2. jcsd
  3. Jan 26, 2013 #2

    vanhees71

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    An electron never can possess zero k.e. Let's argue within non-relativistic quantum theory. There its state is described by a square-integrable Schrödinger wave function, and for any state, Heisenberg's uncertainty relation tells you that
    [tex]\Delta p_x \Delta x \geq \hbar/2.[/tex]
    This means that the mean value of [itex]\vec{p}^2[/itex], and thus its kinetic energy [itex]T=\frac{\vec{p}^2}{2m}[/itex]
    has never expectation value 0.

    You can make this value arbitrarily small, but you can never make it 0. One should note that the plane-wave solution of Schrödinger's equation are not representing a quantum state. They are generalized functions in the sense of distribution theory. They are not square integrable but used to express the state in terms of the spectral representation of the momentum operator. The corresponding generalized eigenstates are normalized to a [itex]\delta[/itex] distribution,
    [tex]\langle \vec{p}|\vec{p}' \rangle = \delta^{(3)}(\vec{p}-\vec{p}'),[/tex]
    which again shows that these generalized eigenstates cannot represent a state of an electron that can be realized in nature.
     
  4. Jan 26, 2013 #3
    Sir, Heisenberg's uncertainty applies for electron in atom. but the case i stated is an isolated electron confined to space
     
    Last edited: Jan 26, 2013
  5. Jan 26, 2013 #4
    Heisenberg's uncertainty principle is valid for every quantum system. In your case, if the electron is space-confined this means that [itex]\Delta x<\infty[/itex]. So, from the uncertainty principle, you can never have [itex]\Delta p_x=0[/itex] as it would imply a completely delocalized electron.
     
  6. Jan 26, 2013 #5
    ok, thank you. Is my explanation in term of wave correct?
     
  7. Jan 26, 2013 #6
    Yes, it is. In fact, if p=0 then you can both say that the wavelenght is infinity or that the electron in completely deconfined, they mean the same thing.
     
  8. Jan 26, 2013 #7

    vanhees71

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    In fact in terms of wave mechanics, which is only one particular representation of non-relativistic quantum theory with a fixed number of particles, the uncertainty principle can be understood intuitively a bit better than in terms of the more abstract representation independent formulation by Dirac. The relation between both concepts is the following:

    In the Dirac formulation (using the Schrödinger picture of time evolution), the (pure) state of a particle at time, [itex]t[/itex], is represented by a normalized vector [itex]|\psi(t) \rangle[/itex] on an appropriate Hilbert space.

    The observables are represented by self-adjoint operators. Quantizing a classical point particle, the fundamental observables are position and momentum, and all other observables can be built as functions of them (like, orbital angular momentum and, particularly important, the Hamiltonian=energy operator). The observables can take values given by the spectral values (i.e., generalized eigenvalues) of the operators. The eigenvectors give generalized bases of the Hilbert space. A complete set of basis vectors is determined by "diagonalizing" a complete set of compatible observables, whose operators all commute.

    One complete set for a quantized classical point particle are the three components of its position vector [itex]\vec{x}[/itex]. In the here considered Schrödinger picture of the time evolution the operators representing observables are time independent, and that's then of course also true for its generalized eigenvectors. The spectrum is the entire [itex]\mathbb{R}^3[/itex], and the wave function in position representation then is given by
    [tex]\psi(t,\vec{x})=\langle \vec{x}|\psi(t) \rangle.[/tex]
    The physical meaning is as follows: If the particle is prepared to be described by the state, represented by this wave function, then the probability density to measure the particle at position [itex]\vec{x}[/itex] is given by Born's rule:
    [tex]P_{\psi}(t,\vec{x})=|\psi(t,\vec{x})|^2.[/tex]
    Now we only need the momentum operator in position representation. It can be derived from the Heisenberg commutation relation
    [tex][\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk}.[/tex]
    After some analysis it turns out that
    [tex]\hat{p}_j \psi(t,\vec{x}):=\langle \vec{x}|\hat{p}_j \psi(t) \rangle=-\mathrm{i} \hbar \frac{\partial}{\partial x_j} \psi(t,\vec{x}).[/tex]
    Now we can evaluate the generalized eigenvectors of [itex]\vec{p}[/itex] in the position representation. Setting
    [tex]u_{\vec{p}}(\vec{x})=\langle \vec{x}|\vec{p} \rangle,[/tex]
    we find the generalized eigenfunction by applying the above operator to the eigenvalue equation:
    [tex]\hat{p}_j u_{\vec{p}}(\vec{x})=-\mathrm{i} \hbar \partial_j u_{\vec{p}}(\vec{x}) \stackrel{!}{=} p_j u_{\vec{p}}(\vec{x}).[/tex]
    The solution is
    [tex]u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi \hbar)^{3/2}} \exp \left (\frac{\mathrm{i} \vec{p} \cdot \vec{x}}{\hbar} \right).[/tex]
    The normalization factor has been chosen such that
    [tex]\langle \vec{p}_1 | \vec{p}_2 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}_1}^*(\vec{x}) u_{\vec{p}_2}(\vec{x})=\delta^{(3)}(\vec{p}_1-\vec{p}_2).[/tex]
    Then you can express any state as well wrt. the generalized momentum-eigenbasis, defining the momentum-wave function by just inserting a completeness relation for the position eigenbasis,
    [tex]\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\vec{x} \rangle \langle \vec{x} |=1[/tex]
    into the definition of the momentum-wave function:
    [tex]\tilde{\psi}(t,\vec{p})=\langle \vec{p}|\psi(t) \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \langle \vec{p}|\vec{x} \rangle \langle \vec{x}|\psi(t) \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}}^*(\vec{x}) \psi(\vec{x}).[/tex]
    Plugging in the momentum eigenfunction, you see that the momentum-wave function is just the Fourier transform of the position wave function. Of course the opposite is true too, because the inverse of a Fourier transformation is again a Fourier transform, only with a flipped sign in the exponential. Indeed in an analogous way, by inserting a completeness relation for the momentum eigenbasis into the defintion of the position-wave function, you get
    [tex]\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}}(\vec{x}) \tilde{\psi}(t,\vec{p}).[/tex]
    Now the origin of the uncertainty relation becomes pretty clear: To prepare a particle such that it has a pretty certain position, means that its position wave function is peaked sharply around a certain value [itex]\vec{x}_0[/itex], i.e., the probability to find the particle is pretty close to one in a rather small neighborhood of [itex]\vec{x}_0[/itex].

    Now you ask, what does this preparation mean for the particle's momentum. Of course, I can as well measure momentum and ask for the probability to find a value for the momentum. The corresponding probability is
    [tex]P_{\psi}(t,\vec{p})=|\tilde{\psi}(t,\vec{p})|^2.[/tex]
    Now, as we have just derived, [itex]\tilde{\psi}(t,\vec{p})[/itex] is (up to some factors, keeping the normalization of the state correct) the Fourier transform of [itex]\psi(t,\vec{x})[/itex].

    Now the Fourier transform has the property, that a narrow function in [itex]\vec{x}[/itex]-space gives a wide-spread function in [itex]\vec{p}[/itex]-space. This means the probability density of the momentum is a pretty spread-out function, which means that the standard deviation of the momentum around its mean value is pretty large, and it must be the larger the narrower the position representation of the state, i.e., the smaller the standard deviation of the particle's position around the mean value [itex]\vec{x}_0[/itex] is.

    The opposite is of course true for a pretty sharply determined momentum of the particle. Then [itex]\tilde{\psi}(t,\vec{p})[/itex] is rather sharply peaked around a certain value [itex]\vec{p}_0[/itex], and the position wave function [itex]\psi(t,\vec{x})[/itex], which is the Fourier transform of [itex]\tilde{\psi}(t,\vec{p}[/itex] becomes a pretty wide-spread function.

    Quantitatively this mutual sharpness-wide-spreadedness of the Fourier transforms of functions is reflected in the Heisenberg-Robertson uncertainty relation between position and momentum. For any state (not only for the very special stationary states, which are the eigenstates of the Hamiltonian!) this uncertainty relation reads
    [tex]\Delta x \Delta p \geq \frac{\hbar}{2}.[/tex]
    The equality sign is reached for a Gaussian wave packet.

    The uncertainty relation can be generalized to any pair of observables [itex]A[/itex] and [itex]B[/itex]:
    [tex]\Delta A \Delta B \geq \frac{1}{2} |\langle \psi|[\hat{A},\hat{B}]|\psi|.[/tex]
    Here [itex]\Delta A[/itex] and [itex]\Delta B[/itex] are the standard deviations of the observables [itex]A[/itex] and [itex]B[/itex] caclulated with the probability distributions of these observables given according to the prepraration of the particle in the state [itex]|\psi \rangle[/itex] according to Born's rule.

    It is important to note that it reflects the statistical outcome of independent measurents of the two observables on ensembles of independently from each other but equally prepared quantum systems. The uncertainty relation has nothing to do with the disturbance of the value of the observables [itex]A[/itex] from measuring the other observable [itex]B[/itex].
     
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