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== 7.3#h(0.6em)The Schrӧdinger Equation


  In the preceding two sections, we described how to use a quantum mechanical wave function and discussed Heisenberg’s uncertainty principle. In this section, we present a complete and formal theory of quantum mechanics that can be used to make predictions. In developing this theory, it is helpful to review the wave theory of light. For a light wave, the electric field #emph[E] (#emph[x],#emph[t]) obeys the relation


  #math.equation(block: true, alt: "the fraction ∂ squared E over ∂ x squared equals the fraction 1 over c squared the fraction ∂ squared E over ∂ t squared ,")[$frac(∂^(2) E, ∂ x^(2)) = frac(1, c^(2)) #h(0.2em) frac(∂^(2) E, ∂ t^(2)) ,$]
  where #emph[c] is the speed of light and the symbol #math.equation(block: false, alt: "∂")[$∂$] represents a #emph[partial derivative]. (Recall from Oscillations that a partial derivative is closely related to an ordinary derivative, but involves functions of more than one variable. When taking the partial derivative of a function by a certain variable, all other variables are held constant.) A light wave consists of a very large number of photons, so the quantity #math.equation(block: false, alt: "| E open parenthesis x , t close parenthesis | squared")[$attach(| E ( x , t ) |, t: 2)$] can interpreted as a probability density of finding a single photon at a particular point in space (for example, on a viewing screen).


  There are many solutions to this equation. One solution of particular importance is


  #math.equation(block: true, alt: "E open parenthesis x , t close parenthesis equals A sin open parenthesis k x minus ω t close parenthesis ,")[$E ( x , t ) = A #h(0.2em) "sin" ( k x − ω t ) ,$]
  where #emph[A] is the amplitude of the electric field, #emph[k] is the wave number, and #math.equation(block: false, alt: "ω")[$ω$] is the angular frequency. Combing this equation with  gives


  #math.equation(block: true, alt: "k squared equals the fraction ω squared over c squared .")[$k^(2) = frac(ω^(2), c^(2)) .$]
  According to de Broglie’s equations, we have #math.equation(block: false, alt: "p equals ℏ k")[$p = ℏ k$] and #math.equation(block: false, alt: "E equals ℏ ω")[$E = ℏ ω$]. Substituting these equations in  gives


  #math.equation(block: true, alt: "p equals the fraction E over c ,")[$p = frac(E, c) ,$]
  or


  #math.equation(block: true, alt: "E equals p c .")[$E = p c .$]
  Therefore, according to Einstein’s general energy-momentum equation ,  describes a particle with a zero rest mass. This is consistent with our knowledge of a photon.


  This process can be reversed. We can begin with the energy-momentum equation of a particle and then ask what wave equation corresponds to it. The energy-momentum equation of a nonrelativistic particle in one dimension is


  #math.equation(block: true, alt: "E equals the fraction p squared over 2 m plus U open parenthesis x , t close parenthesis ,")[$E = frac(p^(2), 2 m) + U ( x , t ) ,$]
  where #emph[p] is the momentum, #emph[m] is the mass, and #emph[U] is the potential energy of the particle. The wave equation that goes with it turns out to be a key equation in quantum mechanics, called #strong[Schrӧdinger’s time-dependent equation].


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[The Schrӧdinger Time-Dependent Equation]


    The equation describing the energy and momentum of a wave function is known as the Schrӧdinger equation:


    #math.equation(block: true, alt: "minus the fraction ℏ squared over 2 m the fraction ∂ squared Ψ open parenthesis x , t close parenthesis over ∂ x squared plus U open parenthesis x , t close parenthesis Ψ open parenthesis x , t close parenthesis equals i ℏ the fraction ∂ Ψ open parenthesis x , t close parenthesis over ∂ t .")[$− frac(ℏ^(2), 2 m) #h(0.2em) frac(∂^(2) "Ψ" ( x , t ), ∂ x^(2)) + U ( x , t ) "Ψ" ( x , t ) = i ℏ frac(∂ "Ψ" ( x , t ), ∂ t) .$]
]


  As described in Potential Energy and Conservation of Energy, the force on the particle described by this equation is given by


  #math.equation(block: true, alt: "F equals minus the fraction ∂ U open parenthesis x , t close parenthesis over ∂ x .")[$F = − frac(∂ U ( x , t ), ∂ x) .$]
  This equation plays a role in quantum mechanics similar to Newton’s second law in classical mechanics. Once the potential energy of a particle is specified—or, equivalently, once the force on the particle is specified—we can solve this differential equation for the wave function. The solution to Newton’s second law equation (also a differential equation) in one dimension is a function #emph[x] (#emph[t]) that specifies where an object is at any time #emph[t]. The solution to Schrӧdinger’s time-dependent equation provides a tool—the wave function—that can be used to determine where the particle is #emph[likely] to be. This equation can be also written in two or three dimensions. Solving Schrӧdinger’s time-dependent equation often requires the aid of a computer.


  Consider the special case of a free particle. A free particle experiences no force #math.equation(block: false, alt: "open parenthesis F equals 0 close parenthesis .")[$( F = 0 ) .$] Based on , this requires only that


  #math.equation(block: true, alt: "U open parenthesis x , t close parenthesis equals U sub 0 equals constant .")[$U ( x , t ) = U_(0) = "constant" .$]
  For simplicity, we set #math.equation(block: false, alt: "U sub 0 equals 0")[$U_(0) = 0$]. Schrӧdinger’s equation then reduces to


  #math.equation(block: true, alt: "minus the fraction ℏ squared over 2 m the fraction ∂ squared Ψ open parenthesis x , t close parenthesis over ∂ x squared equals i ℏ the fraction ∂ Ψ open parenthesis x , t close parenthesis over ∂ t .")[$− frac(ℏ^(2), 2 m) #h(0.2em) frac(∂^(2) "Ψ" ( x , t ), ∂ x^(2)) = i ℏ frac(∂ "Ψ" ( x , t ), ∂ t) .$]
  A valid solution to this equation is


  #math.equation(block: true, alt: "Ψ open parenthesis x , t close parenthesis equals A e to the power i open parenthesis k x minus ω t close parenthesis .")[$"Ψ" ( x , t ) = A e^(i ( k x − ω t )) .$]
  Not surprisingly, this solution contains an #strong[imaginary number] #math.equation(block: false, alt: "open parenthesis i equals the square root of −1 close parenthesis")[$( i = sqrt(−1) )$] because the differential equation itself contains an imaginary number. As stressed before, however, quantum-mechanical predictions depend only on #math.equation(block: false, alt: "| Ψ open parenthesis x , t close parenthesis | squared")[$attach(| "Ψ" ( x , t ) |, t: 2)$], which yields completely real values. Notice that the real plane-wave solutions, #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis equals A sin open parenthesis k x minus ω t close parenthesis")[$"Ψ" ( x , t ) = A #h(0.2em) "sin" ( k x − ω t )$] and #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis equals A cos open parenthesis k x minus ω t close parenthesis ,")[$"Ψ" ( x , t ) = A #h(0.2em) "cos" ( k x − ω t ) ,$] do not obey Schrödinger’s equation. The temptation to think that a wave function can be seen, touched, and felt in nature is eliminated by the appearance of an imaginary number. In Schrӧdinger’s theory of quantum mechanics, the wave function is merely a tool for calculating things.


  If the potential energy function (#emph[U]) does not depend on time, it is possible to show that


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "Ψ open parenthesis x , t close parenthesis equals ψ open parenthesis x close parenthesis e to the power − i ω t")[$"Ψ" ( x , t ) = ψ ( x ) e^("−" i ω t)$]
]


  satisfies Schrӧdinger’s time-dependent equation, where #math.equation(block: false, alt: "ψ open parenthesis x close parenthesis")[$ψ ( x )$] is a #emph[time]-independent function and #math.equation(block: false, alt: "e to the power − i ω t")[$e^("−" i ω t)$] is a #emph[space]-independent function. In other words, the wave function is #emph[separable] into two parts: a space-only part and a time-only part. The factor #math.equation(block: false, alt: "e to the power − i ω t")[$e^("−" i ω t)$] is sometimes referred to as a #strong[time-modulation factor] since it modifies the space-only function. According to de Broglie, the energy of a matter wave is given by #math.equation(block: false, alt: "E equals ℏ ω")[$E = ℏ ω$], where #emph[E] is its total energy. Thus, the above equation can also be written as


  #math.equation(block: true, alt: "Ψ open parenthesis x , t close parenthesis equals ψ open parenthesis x close parenthesis e to the power − i E t / ℏ .")[$"Ψ" ( x , t ) = ψ ( x ) e^("−" i E t "/" ℏ) .$]
  Any linear combination of such states (mixed state of energy or momentum) is also valid solution to this equation. Such states can, for example, describe a localized particle


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A particle with mass #emph[m] is moving along the #emph[x]-axis in a potential given by the potential energy function #math.equation(block: false, alt: "U open parenthesis x close parenthesis equals 0.5 m ω squared x squared")[$U ( x ) = 0.5 m ω^(2) x^(2)$]. Compute the product #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis to the power * U open parenthesis x close parenthesis Ψ open parenthesis x , t close parenthesis .")[$"Ψ" attach(( x , t ), t: *) U ( x ) "Ψ" ( x , t ) .$] Express your answer in terms of the time-independent wave function, #math.equation(block: false, alt: "ψ open parenthesis x close parenthesis .")[$ψ ( x ) .$]


    
    #solutionbox[
#math.equation(block: true, alt: "0.5 m ω squared x squared ψ open parenthesis x close parenthesis to the power * ψ open parenthesis x close parenthesis")[$0.5 m ω^(2) x^(2) ψ attach(( x ), t: *) ψ ( x )$]
]
]


  Combining  and , Schrödinger’s time-dependent equation reduces to


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "minus the fraction ℏ squared over 2 m the fraction d squared ψ open parenthesis x close parenthesis over d x squared plus U open parenthesis x close parenthesis ψ open parenthesis x close parenthesis equals E ψ open parenthesis x close parenthesis ,")[$− frac(ℏ^(2), 2 m) #h(0.2em) frac(d^(2) ψ ( x ), d x^(2)) + U ( x ) ψ ( x ) = E ψ ( x ) ,$]
]


  where #emph[E] is the total energy of the particle (a real number). This equation is called #strong[Schrӧdinger’s time-independent equation]. Notice that we use “big psi” #math.equation(block: false, alt: "open parenthesis Ψ close parenthesis")[$( "Ψ" )$] for the time-dependent wave function and “little psi” #math.equation(block: false, alt: "open parenthesis ψ close parenthesis")[$( ψ )$] for the time-independent wave function. The wave-function solution to this equation must be multiplied by the time-modulation factor to obtain the time-dependent wave function.


  In the next sections, we solve Schrӧdinger’s time-independent equation for three cases: a quantum particle in a box, a simple harmonic oscillator, and a quantum barrier. These cases provide important lessons that can be used to solve more complicated systems. The time-independent wave function #math.equation(block: false, alt: "ψ open parenthesis x close parenthesis")[$ψ ( x )$] solutions must satisfy three conditions:


  - #math.equation(block: false, alt: "ψ open parenthesis x close parenthesis")[$ψ ( x )$] must be a continuous function.
- The first derivative of #math.equation(block: false, alt: "ψ open parenthesis x close parenthesis")[$ψ ( x )$] with respect to space, #math.equation(block: false, alt: "d ψ open parenthesis x close parenthesis / d x")[$d ψ ( x ) "/" d x$], must be continuous, unless #math.equation(block: false, alt: "V open parenthesis x close parenthesis equals ∞")[$V ( x ) = ∞$].
- #math.equation(block: false, alt: "ψ open parenthesis x close parenthesis")[$ψ ( x )$] must not diverge (“blow up”) at #math.equation(block: false, alt: "x equals ± ∞ .")[$x = "±" ∞ .$]


  The first condition avoids sudden jumps or gaps in the wave function. The second condition requires the wave function to be smooth at all points, except in special cases. (In a more advanced course on quantum mechanics, for example, potential spikes of infinite depth and height are used to model solids). The third condition requires the wave function be normalizable. This third condition follows from Born’s interpretation of quantum mechanics. It ensures that #math.equation(block: false, alt: "| ψ open parenthesis x close parenthesis | squared")[$attach(| ψ ( x ) |, t: 2)$] is a finite number so we can use it to calculate probabilities.


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Which of the following wave functions is a valid wave-function solution for Schrӧdinger’s equation?


      #figure(figph[Three graphs of Psi of x versus x are shown. The first rises then drops discontinuously to a lower value, rises again and then has a constant value. The second function looks like a breaking wave, with a crest overtaking the base. The third increases exponentially to infinity.], alt: "Three graphs of Psi of x versus x are shown. The first rises then drops discontinuously to a lower value, rises again and then has a constant value. The second function looks like a breaking wave, with a crest overtaking the base. The third increases exponentially to infinity.", caption: none)


    
    #solutionbox[
None. The first function has a discontinuity; the second function is double-valued; and the third function diverges so is not normalizable.
]
]


  
    
=== Summary


    - The Schrӧdinger equation is the fundamental equation of wave quantum mechanics. It allows us to make predictions about wave functions.
- When a particle moves in a time-independent potential, a solution of the time-dependent Schrӧdinger equation is a product of a time-independent wave function and a time-modulation factor.
- The Schrӧdinger equation can be applied to many physical situations.


  
  
    
=== Conceptual Questions


    
      What is the difference between a wave function #math.equation(block: false, alt: "ψ open parenthesis x , y , z close parenthesis")[$ψ ( x , y , z )$] and a wave function #math.equation(block: false, alt: "Ψ open parenthesis x , y , z , t close parenthesis")[$"Ψ" ( x , y , z , t )$] for the same particle?


    
    
      If a quantum particle is in a stationary state, does it mean that it does not move?


    
    #solutionbox[
No, it means that predictions about the particle (expressed in terms of probabilities) are time-independent.
]


    
      Explain the difference between time-dependent and -independent Schrӧdinger’s equations.


    
    
      Suppose a wave function is discontinuous at some point. Can this function represent a quantum state of some physical particle? Why? Why not?


    
    #solutionbox[
No, because the probability of the particle existing in a narrow (infinitesimally small) interval at the discontinuity is undefined.
]


  
  
    
=== Problems


    
      Combine  and  to show #math.equation(block: false, alt: "k squared equals the fraction ω squared over c squared .")[$k^(2) = frac(ω^(2), c^(2)) .$]


    
    #solutionbox[
Carrying out the derivatives yields #math.equation(block: false, alt: "k squared equals the fraction ω squared over c squared .")[$k^(2) = frac(ω^(2), c^(2)) .$]
]


    
      Show that #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis equals A e to the power i open parenthesis k x minus ω t close parenthesis")[$"Ψ" ( x , t ) = A e^(i ( k x − ω t ))$] is a valid solution to Schrӧdinger’s time-dependent equation.


    
    
      Show that #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis equals A sin open parenthesis k x minus ω t close parenthesis")[$"Ψ" ( x , t ) = A #h(0.2em) "sin" ( k x − ω t )$] and #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis equals A cos open parenthesis k x minus ω t close parenthesis")[$"Ψ" ( x , t ) = A #h(0.2em) "cos" ( k x − ω t )$] do not obey Schrӧdinger’s time-dependent equation.


    
    #solutionbox[
Carrying out the derivatives (as above) for the sine function gives a cosine on the right side the equation, so the equality fails. The same occurs for the cosine solution.
]


    
      Show that when #math.equation(block: false, alt: "Ψ sub 1 open parenthesis x , t close parenthesis")[$"Ψ"_(1) ( x , t )$] and #math.equation(block: false, alt: "Ψ sub 2 open parenthesis x , t close parenthesis")[$"Ψ"_(2) ( x , t )$] are solutions to the time-dependent Schrӧdinger equation and #emph[A],#emph[B] are numbers, then a function #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis")[$"Ψ" ( x , t )$] that is a superposition of these functions is also a solution: #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis equals A Ψ sub 1 open parenthesis x , t close parenthesis plus B Ψ sub 2 open parenthesis x , t close parenthesis")[$"Ψ" ( x , t ) = A "Ψ"_(1) ( x , t ) + B "Ψ"_(2) ( x , t )$].


    
    
      A particle with mass #emph[m] is described by the following wave function: #math.equation(block: false, alt: "ψ open parenthesis x close parenthesis equals A cos k x plus B sin k x")[$ψ ( x ) = A #h(0.2em) "cos" #h(0.2em) k x + B #h(0.2em) "sin" #h(0.2em) k x$], where #emph[A], #emph[B], and #emph[k] are constants. Assuming that the particle is free, show that this function is the solution of the stationary Schrӧdinger equation for this particle and find the energy that the particle has in this state.


    
    #solutionbox[
#math.equation(block: true, alt: "E equals ℏ squared k squared / 2 m")[$E = ℏ^(" " 2) k^(2) "/" 2 m$]
]


    
      Find the expectation value of the kinetic energy for the particle in the state, #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis equals A e to the power i open parenthesis k x minus ω t close parenthesis")[$"Ψ" ( x , t ) = A e^(i ( k x − ω t ))$]. What conclusion can you draw from your solution?


    
    
      Find the expectation value of the square of the momentum squared for the particle in the state, #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis equals A e to the power i open parenthesis k x minus ω t close parenthesis")[$"Ψ" ( x , t ) = A e^(i ( k x − ω t ))$]. What conclusion can you draw from your solution?


    
    #solutionbox[
#math.equation(block: false, alt: "ℏ squared k squared")[$ℏ^(" " 2) k^(" " 2)$]; The particle has definite momentum and therefore definite momentum squared.
]


    
      A free proton has a wave function given by #math.equation(block: false, alt: "Ψ open parenthesis x , t close parenthesis equals A e to the power i open parenthesis 5.02 times 10 to the power 11 x minus 8.00 times 10 to the power 15 t close parenthesis")[$"Ψ" ( x , t ) = A e^(" " i " " ( 5.02 #h(0.2em) × #h(0.2em) 10^(11) x − 8.00 #h(0.2em) × #h(0.2em) 10^(15) t ))$].


      The coefficient of #emph[x] is inverse meters #math.equation(block: false, alt: "open parenthesis m to the power −1 close parenthesis")[$( "m"^(−1) )$] and the coefficient on #emph[t] is inverse seconds #math.equation(block: false, alt: "open parenthesis s to the power −1 close parenthesis .")[$( "s"^(−1) ) .$] Find its momentum and energy.