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Time independent schrödinger equation query

  1. Feb 15, 2015 #1

    Maylis

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    Hello, the TISE can be simplified
    $$H \psi = E \psi$$
    Where ##H## is the Hamiltonian, and ##E## is the eigenvalue, but why don't the ##\psi## terms cancel, leaving ##H = E##?

    Also, what the heck does the eigenvalue ##E## have to do with the eigenvalue that I have previously encountered in mathematics (linear algebra)

    ##A \textbf {v} = \lambda \textbf {v}##
    where ##A## is a matrix, ##\textbf {v}## is a vector, and ##\lambda## is the eigenvalue. Where was linear algebra ever invoked when doing TISE to use such a name for E, Eigenvalue?
     
  2. jcsd
  3. Feb 15, 2015 #2
    The Hamiltonian is a differential operator, not a mumber, you cant cancel the wave functions on both sides for this reason.
     
  4. Feb 15, 2015 #3

    Maylis

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    That was my suspicion
     
  5. Feb 15, 2015 #4

    Maylis

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    Well, I have some more issues with this. I don't understand why there are two ways of defining and then calling them equivalent, ##\hat {H} \psi = E \psi##

    So I think the quantum mechanical Hamiltonian is defined

    $$\hat {H} = - \frac {\hbar^{2}}{2m} \frac {\partial^{2}}{\partial x^{2}} + V $$

    So I will just take this for granted as a letter that will signify the sum of the kinetic and potential energy, with the kinetic energy having a fancy operator attached to it. Now if I multiply through by ##\psi##, I will get

    $$ \hat {H} \psi = - \frac {\hbar^{2}}{2m} \frac {\partial^{2} \psi}{\partial x^{2}} + V \psi $$

    But for some reason, in deriving the TISE, we set
    $$ i \hbar \frac {1}{f} \frac {\partial f}{\partial t} = - \frac {\hbar^{2}}{2m} \frac {1}{\psi} \frac {d^{2} \psi}{dx^{2}} + V = E $$

    Where ##f## is a function of time, taken from separating variables of the wave function ##\Psi (x,t) = \psi (x) f(t) ##

    Why create this constant, ##E##? At the end of it all, it becomes ##\hat {H} \psi = E \psi##. I don't understand why this is necessary to even have a letter ##E##.

    Is it perhaps meaningless to write ##\Psi (x,t) = \psi(x) e^{\frac {-i \hat {H} t}{\hbar}}## since ##\hat {H}## contains the second derivative operator, and for that reason ##E## is useful?
     
    Last edited: Feb 15, 2015
  6. Feb 15, 2015 #5

    vela

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    E is a separation constant. The term involving ##f## is a function of ##t## only while the stuff in the middle is a function of ##x## only. The only way this equality can hold is if they're equal to a constant.
     
  7. Feb 15, 2015 #6

    Maylis

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    What the heck is the difference between E and the Hamiltonian?
     
  8. Feb 15, 2015 #7

    Nugatory

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    The potential energy ##V## is also an operator and it may be a function of ##x##. For example, if we're looking for the wave function of a particle in a harmonic oscillator, ##V=\frac{1}{2}kx^2## and the time-independent Schrodinger equation becomes
    $$\frac {\hbar^{2}}{2m} \frac {\partial^{2} \psi}{\partial x^{2}}+\frac{1}{2}kx^2\psi=E\psi$$

    Solutions (other than the trivial ##\psi=0##) to this differential equation only exist for certain values of ##E##.

    This is the same concept of eigenvalue and eigenvector that you saw in linear algebra. All that's going on now is that the vector space on which the operator H is acting is the space of functions ##\psi##. H maps the function ##\psi## onto the function ##H\psi##, and the TISE is looking for functions that are mapped to scalar multiples of themselves by the operator H.
     
    Last edited: Feb 15, 2015
  9. Feb 15, 2015 #8

    vanhees71

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    The fundamental equation of non-relativistic quantum theory in its formulation as "wave mechanics" (discovered by Erwin Schrödinger in 1926 when on skiing vacation) is the time-dependent Schrödinger equation
    $$\mathrm{i} \hbar \partial_t \psi(t,x)=\hat{H} \psi(t,x),$$
    where the Hamiltonian in the most simple case of a point particle moving in a potential is given by
    $$\hat{H}=-\frac{\hbar^2}{2m} \partial_x^2 + V(x).$$
    Given the wavefunction at ##t=0##, you can calculate the wave function at any later time by solving this partial differential equation (with some boundary conditions, ensuring that ##|\psi(t,x)|^2## can be integrated over ##x \in \mathbb{R}##.

    One method to solve linear partial differential equations is a separation ansatz. This lets you look for solutions of the equation which factor in a product of functions of ##t## and ##x##:
    $$\psi(t,x)=f(t) \Psi(x).$$
    Plugging this into the Schrödinger equation, you get
    $$\mathrm{i} \hbar \Psi(x) \frac{\mathrm{d}}{\mathrm{d} t} f(t) =f(t) \left [-\frac{\hbar^2 \Psi''(x)}{2m} + V(x) \Psi(x) \right ].$$
    Now you divide this equation by ##f(t) \Psi(x)##. Then you get
    $$\mathrm{i} \hbar \frac{1}{f(t)} \frac{\mathrm{d}}{\mathrm{d} t} f(t) =\frac{1}{\Psi(x)} \left [-\frac{\hbar^2 \Psi''(x)}{2m} + V(x) \Psi(x) \right ].$$
    Now the left-hand side does not depend on ##x## and the right-hand side does not depend on ##t##. This implies that it cannot depend on both ##x## and ##t##, i.e., it's a constant, which we suggestively call ##E##.

    From the left-hand side this means
    $$\mathrm{i} \hbar \frac{\dot{f}}{f}=E.$$
    This you can immediately integrate to get
    $$f(t)=f_0 \exp \left (-\frac{\mathrm{i} E t}{\hbar} \right )=u_E(t).$$
    For the right-hand side you get
    $$\hat{H} \Psi(x)=E \Psi(x).$$
    Together with the approriate boundary conditions you get the possible set of eigenvalues ##E## of the Hamiltonian and the corresponding eigensolutions, ##\Psi_E(x)##. Usually the possible eigensolutions involve a set of discrete energy eigenvalues ##E_j \in \mathbb{R}## ("bound states") and a continuous set ##S(\hat{H}) \subseteq \mathbb{R}##. Then you can write any wavefunction in terms of these eigensolutions:
    $$\psi(t,x)=\sum_n \exp \left(-\frac{\mathrm{i} E t}{\hbar} \right) A_n \Psi_{E_n}(x) + \int_{S(\hat{H})} \mathrm{d} E A(E) \Psi_{E}(x).$$
     
  10. Feb 15, 2015 #9

    Maylis

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    I know ##\Psi (x,t)## is the wave function, what does ##\psi(x)## by itself represent or mean?
     
  11. Feb 15, 2015 #10

    bhobba

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    Its the expansion of a pure state in terms of the position eigenfunctions.

    Here are the two axioms of QM:

    Axiom 1
    Associated with each measurement we can find a Hermitian operator O, called the observations observable, such that the possible outcomes of the observation are its eigenvalues yi.

    Axiom 2 - called the Born Rule
    Associated with any system is a positive operator of unit trace, P, called the state of the system, such that expected value of of the outcomes of the observation is Trace (PO).

    Axiom 2 is to some extent, without going into the details, implied by axiom 1 via Gleason's theorem - but no need to go into that here - we will simply focus on axiom 1.

    Given a Hermitian operator O by the spectral theorem it can be expanded in terms of its eigenvalues and eigenvectors O = ∑yi |yi><yi|. Since the |yi> form an orthonormal basis any vector |v> can be expanded in terms of that basis |v> = ∑vi |yi>. The vi is called the representation of the vector v in terms of the observable O. Ignoring the subtlety of a continuous basis whose rigorous resolution involves what's called a Rigged Hilbert Space the wave-function is the representation of the systems state in terms of the position observable.

    Without proving it, it turns out, from the Born rule, given a wave-function ψ(x), the probability of finding a particle at position x if you observe it is |ψ(x)|^2.

    To really understand this stuff get a copy of Ballentine - Quantum Mechanics - A Modern Development:
    https://www.amazon.com/Quantum-Mechanics-Modern-Development-Edition/dp/9814578584

    Read the first three chapters,

    Thanks
    Bill
     
    Last edited: Feb 15, 2015
  12. Feb 15, 2015 #11

    Nugatory

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    If ##H## is not a function of ##t## (which is equivalent to saying that the potential is constant over time), then we can solve the time-dependent Schrodinger equation using separation of variables. The solutions will be ##\Psi(x,t)=\psi(x)e^{-iEt / \hbar}## where ##\psi(x)## is the solution of the time-independent Schrodinger equation ##H\psi=E\psi## with ##E## a scalar.

    Because ##H## is Hermitian we know that ##E## is real. Therefore the magnitude of the exponential is equal to one, and ##|\Psi(x,t)|^2##, the probability distribution of finding a particle with energy ##E## at position ##x## at time ##t##, is equal to ##|\psi(x)|^2##. More complicated states with more complicated and interesting probability distributions will be the square of linear combinations of different solutions of the TISE with different values of ##E##.
     
    Last edited: Feb 15, 2015
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