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== 7.2#h(0.6em)The Heisenberg Uncertainty Principle


  #strong[Heisenberg’s uncertainty principle] is a key principle in quantum mechanics. Very roughly, it states that if we know #emph[everything] about where a particle is located (the uncertainty of position is small), we know #emph[nothing] about its momentum (the uncertainty of momentum is large), and vice versa. Versions of the uncertainty principle also exist for other quantities as well, such as energy and time. We discuss the momentum-position and energy-time uncertainty principles separately.


  
    
=== Momentum and Position


    To illustrate the momentum-position uncertainty principle, consider a free particle that moves along the #emph[x]-direction. The particle moves with a constant velocity #emph[u] and momentum #math.equation(block: false, alt: "p equals m u")[$p = m u$]. According to de Broglie’s relations, #math.equation(block: false, alt: "p equals ℏ k")[$p = ℏ k$] and #math.equation(block: false, alt: "E equals ℏ ω")[$E = ℏ ω$]. As discussed in the previous section, the wave function for this particle is given by


    #math.equation(block: true, alt: "ψ sub k open parenthesis x , t close parenthesis equals A [ cos open parenthesis ω t minus k x close parenthesis minus i sin open parenthesis ω t minus k x close parenthesis ] equals A e to the power − i open parenthesis ω t minus k x close parenthesis equals A e to the power − i ω t e to the power i k x")[$ψ_(k) ( x , t ) = A [ "cos" ( ω " " t − k " " x ) − i #h(0.2em) "sin" ( ω " " t − k " " x ) ] = A e^("−" i ( ω " " t − k " " x )) = A e^("−" i " " ω " " t) e^(i " " k " " x)$]
    and the probability density #math.equation(block: false, alt: "| ψ sub k open parenthesis x , t close parenthesis | squared equals A squared")[$| ψ_(k) ( x , t ) |^(" " 2) = A^(2)$] is #emph[uniform] and independent of time. The particle is equally likely to be found anywhere along the #emph[x]-axis but has definite values of wavelength and wave number, and therefore momentum. The uncertainty of position is infinite (we are completely uncertain about position) and the uncertainty of the momentum is zero (we are completely certain about momentum). This account of a free particle is consistent with Heisenberg’s uncertainty principle.


    Similar statements can be made of localized particles. In quantum theory, a localized particle is modeled by a linear superposition of free-particle (or plane-wave) states called a #strong[wave packet]. An example of a wave packet is shown. A wave packet contains many wavelengths and therefore by de Broglie’s relations many momenta—possible in quantum mechanics! This particle also has many values of position, although the particle is confined mostly to the interval #math.equation(block: false, alt: "Δ x")[$"Δ" x$]. The particle can be better localized #math.equation(block: false, alt: "open parenthesis Δ x")[$\( "Δ" x$] can be decreased) if more plane-wave states of different wavelengths or momenta are added together in the right way #math.equation(block: false, alt: "open parenthesis Δ p")[$\( "Δ" p$] is increased). According to Heisenberg, these uncertainties obey the following relation.


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[The Heisenberg Uncertainty Principle]


      The product of the uncertainty in position of a particle and the uncertainty in its momentum can never be less than one-half of the reduced Planck constant:


      #math.equation(block: true, alt: "Δ x Δ p greater than or equal to ℏ / 2 .")[$"Δ" x " " "Δ" p ≥ ℏ "/" 2 .$]
]


    This relation expresses Heisenberg’s uncertainty principle. It places limits on what we can know about a particle from simultaneous measurements of position and momentum. If #math.equation(block: false, alt: "Δ x")[$"Δ" x$] is large, #math.equation(block: false, alt: "Δ p")[$#h(0.2em) "Δ" p$] is small, and vice versa.  can be derived in a more advanced course in modern physics. Reflecting on this relation in his work #emph[The Physical Principles of the Quantum Theory], Heisenberg wrote “Any use of the words ‘position’ and ‘velocity’ with accuracy exceeding that given by \[the relation\] is just as meaningless as the use of words whose sense is not defined.”


    #figure(figph[Several waves are shown, all with equal amplitude but different. The result of adding these to form a wave packet is also shown. The wave packet is an oscillating wave whose amplitude increases to a maximum then decreases, so that its envelope is a pulse of width Delta x.], alt: "Several waves are shown, all with equal amplitude but different. The result of adding these to form a wave packet is also shown. The wave packet is an oscillating wave whose amplitude increases to a maximum then decreases, so that its envelope is a pulse of width Delta x.", caption: [Adding together several plane waves of different wavelengths can produce a wave that is relatively localized.])


    Note that the uncertainty principle has nothing to do with the precision of an experimental apparatus. Even for perfect measuring devices, these uncertainties would remain because they originate in the wave-like nature of matter. The precise value of the product #math.equation(block: false, alt: "Δ x Δ p")[$"Δ" x "Δ" p$] depends on the specific form of the wave function. Interestingly, the Gaussian function (or bell-curve distribution) gives the minimum value of the uncertainty product: #math.equation(block: false, alt: "Δ x Δ p equals ℏ / 2 .")[$"Δ" x " " "Δ" p = ℏ "/" 2 .$]


    #examplebox("Example 1")[The Uncertainty Principle Large and Small][
Determine the minimum uncertainties in the positions of the following objects if their speeds are known with a precision of #math.equation(block: false, alt: "1.0 times 10 to the power −3 m/s")[$1.0 #h(0.2em) × #h(0.2em) 10^(−3) "m/s"$]: (a) an electron and (b) a bowling ball of mass 6.0 kg.


      Strategy
Given the uncertainty in speed #math.equation(block: false, alt: "Δ u equals 1.0 times 10 to the power −3 m/s")[$"Δ" u = 1.0 #h(0.2em) × #h(0.2em) 10^(−3) "m/s"$], we have to first determine the uncertainty in momentum #math.equation(block: false, alt: "Δ p equals m Δ u")[$"Δ" p = m " " "Δ" u$] and then invert  to find the uncertainty in position #math.equation(block: false, alt: "Δ x equals ℏ / open parenthesis 2 Δ p close parenthesis")[$"Δ" x = ℏ "/" ( 2 "Δ" p )$].


      Solution


      + For the electron:
#math.equation(block: true, alt: "Δ p, equals, m Δ u equals open parenthesis 9.1 times 10 to the power −31 kg close parenthesis open parenthesis 1.0 times 10 to the power −3 m/s close parenthesis equals 9.1 times 10 to the power −34 kg times m/s ,; Δ x, equals, the fraction ℏ over 2 Δ p equals 5.8 cm .")[$"Δ" p & = & m "Δ" u = ( 9.1 #h(0.2em) × #h(0.2em) 10^(−31) #h(0.2em) "kg" ) ( 1.0 #h(0.2em) × #h(0.2em) 10^(−3) "m/s" ) = 9.1 #h(0.2em) × #h(0.2em) 10^(−34) #h(0.2em) "kg" · "m/s" " " , \ "Δ" x & = & frac(ℏ, 2 " " "Δ" p) = 5.8 #h(0.2em) "cm" "."$]
+ For the bowling ball:
#math.equation(block: true, alt: "Δ p, equals, m Δ u equals open parenthesis 6.0 kg close parenthesis open parenthesis 1.0 times 10 to the power −3 m/s close parenthesis equals 6.0 times 10 to the power −3 kg times m/s ,; Δ x, equals, the fraction ℏ over 2 Δ p equals 8.8 times 10 to the power −33 m .")[$"Δ" p & = & m "Δ" u = ( 6.0 #h(0.2em) "kg" ) ( 1.0 #h(0.2em) × #h(0.2em) 10^(−3) "m/s" ) = 6.0 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "kg" · "m/s" " " , \ "Δ" x & = & frac(ℏ, 2 " " "Δ" p) = 8.8 #h(0.2em) × #h(0.2em) 10^(−33) "m" #h(0.2em) "."$]


      Significance
Unlike the position uncertainty for the electron, the position uncertainty for the bowling ball is immeasurably small. Planck’s constant is very small, so the limitations imposed by the uncertainty principle are not noticeable in macroscopic systems such as a bowling ball.
]


    #examplebox("Example 2")[Uncertainty and the Hydrogen Atom][
Estimate the ground-state energy of a hydrogen atom using Heisenberg’s uncertainty principle. (#emph[Hint]: According to early experiments, the size of a hydrogen atom is approximately 0.1 nm.)


      Strategy
An electron bound to a hydrogen atom can be modeled by a particle bound to a one-dimensional box of length #math.equation(block: false, alt: "L equals 0.1 nm .")[$L = 0.1 #h(0.2em) "nm" "."$] The ground-state wave function of this system is a half wave, like that given in . This is the largest wavelength that can “fit” in the box, so the wave function corresponds to the lowest energy state. Note that this function is very similar in shape to a Gaussian (bell curve) function. We can take the average energy of a particle described by this function (#emph[E]) as a good estimate of the ground state energy #math.equation(block: false, alt: "open parenthesis E sub 0 close parenthesis")[$( E_(0) )$]. This average energy of a particle is related to its average of the momentum squared, which is related to its momentum uncertainty.


      Solution
To solve this problem, we must be specific about what is meant by “uncertainty of position” and “uncertainty of momentum.” We identify the uncertainty of position #math.equation(block: false, alt: "open parenthesis Δ x close parenthesis")[$( "Δ" x )$] with the standard deviation of position #math.equation(block: false, alt: "open parenthesis σ sub x close parenthesis")[$( σ_(x) )$], and the uncertainty of momentum #math.equation(block: false, alt: "open parenthesis Δ p close parenthesis")[$( "Δ" p )$] with the standard deviation of momentum #math.equation(block: false, alt: "open parenthesis σ sub p close parenthesis")[$( σ_(p) )$]. For the Gaussian function, the uncertainty product is


      #math.equation(block: true, alt: "σ sub x σ sub p equals the fraction ℏ over 2 ,")[$σ_(x) σ_(p) = frac(ℏ, 2) ,$]
      where


      #math.equation(block: true, alt: "σ x 2 equals x squared minus minus x minus squared and σ p 2 equals p squared minus minus p minus squared .")[$σ_(x)^(2) = accent(x^(2), −) − attach(accent(x, −), t: 2) #h(0.5em) "and" #h(0.5em) σ_(p)^(2) = accent(p^(2), −) − attach(accent(p, −), t: 2) .$]
      The bar represents an average, so #math.equation(block: false, alt: "x squared minus")[$accent(x^(2), −)$] is the average of #math.equation(block: false, alt: "x squared")[$x^(2)$], while #math.equation(block: false, alt: "x minus squared")[$attach(accent(x, −), t: 2)$] is the square of the average of #math.equation(block: false, alt: "x")[$x$], and similarly for #math.equation(block: false, alt: "p")[$p$].


      The particle is equally likely to be moving left as moving right, so #math.equation(block: false, alt: "p minus equals 0")[$accent(p, −) = 0$]. Also, the uncertainty of position is comparable to the size of the box, so #math.equation(block: false, alt: "σ sub x equals L .")[$σ_(x) = L .$] The estimated ground state energy is therefore


      #math.equation(block: true, alt: "E sub 0 equals E sub Gaussian equals the fraction p squared minus over m equals the fraction σ p 2 over 2 m equals the fraction 1 over 2 m open parenthesis the fraction ℏ over 2 σ sub x close parenthesis squared equals the fraction 1 over 2 m open parenthesis the fraction ℏ over 2 L close parenthesis squared equals the fraction ℏ squared over 8 m L squared .")[$E_(0) = E_("Gaussian") = frac(accent(p^(2), −), m) = frac(σ_(p)^(2), 2 m) = frac(1, 2 m) attach(( frac(ℏ, 2 σ_(x)) ), t: 2) = frac(1, 2 m) attach(( frac(ℏ, 2 L) ), t: 2) = frac(ℏ^(2), 8 m L^(2)) .$]
      Multiplying numerator and denominator by #math.equation(block: false, alt: "c squared")[$c^(2)$] gives


      #math.equation(block: true, alt: "E sub 0 equals the fraction open parenthesis ℏ c close parenthesis squared over 8 open parenthesis m c squared close parenthesis L squared equals the fraction open parenthesis 197.3 eV times nm close parenthesis squared over 8 open parenthesis 0.511 times 10 to the power 6 eV close parenthesis open parenthesis 0.1 nm close parenthesis squared equals 0.952 eV approximately equals 1 eV .")[$E_(0) = frac(attach(( ℏ "c" ), t: 2), 8 ( m c^(2) ) L^(2)) = frac(attach(( 197.3 #h(0.2em) "eV" · "nm" ), t: 2), 8 ( 0.511 · 10^(6) "eV" ) attach(( 0.1 #h(0.2em) "nm" ), t: 2)) = 0.952 #h(0.2em) "eV" ≈ 1 #h(0.2em) "eV" .$]
      Significance
Based on early estimates of the size of a hydrogen atom and the uncertainty principle, the ground-state energy of a hydrogen atom is in the eV range. The ionization energy of an electron in the ground-state energy is approximately 10 eV, so this prediction is roughly confirmed. (#emph[Note:] The product #math.equation(block: false, alt: "ℏ c")[$ℏ c$] is often a useful value in performing calculations in quantum mechanics.)
]


  
  
    
=== Energy and Time


    Another kind of uncertainty principle concerns uncertainties in simultaneous measurements of the energy of a quantum state and its lifetime,


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "Δ E Δ t greater than or equal to the fraction ℏ over 2 ,")[$"Δ" E "Δ" t ≥ frac(ℏ, 2) ,$]
]


    where #math.equation(block: false, alt: "Δ E")[$"Δ" E$] is the uncertainty in the energy measurement and #math.equation(block: false, alt: "Δ t")[$"Δ" t$] is the uncertainty in the lifetime measurement. The #strong[energy-time uncertainty principle] does not result from a relation of the type expressed by  for technical reasons beyond this discussion. Nevertheless, the general meaning of the energy-time principle is that a quantum state that exists for only a short time cannot have a definite energy. The reason is that the frequency of a state is inversely proportional to time and the frequency connects with the energy of the state, so to measure the energy with good precision, the state must be observed for many cycles.


    To illustrate, consider the excited states of an atom. The finite lifetimes of these states can be deduced from the shapes of spectral lines observed in atomic emission spectra. Each time an excited state decays, the emitted energy is slightly different and, therefore, the emission line is characterized by a #emph[distribution] of spectral frequencies (or wavelengths) of the emitted photons. As a result, all spectral lines are characterized by spectral widths. The average energy of the emitted photon corresponds to the theoretical energy of the excited state and gives the spectral location of the peak of the emission line. Short-lived states have broad spectral widths and long-lived states have narrow spectral widths.


    #examplebox("Example 3")[Atomic Transitions][
An atom typically exists in an excited state for about #math.equation(block: false, alt: "Δ t equals 10 to the power −8 s")[$"Δ" t = 10^(−8) "s"$]. Estimate the uncertainty #math.equation(block: false, alt: "Δ f")[$"Δ" " " f$] in the frequency of emitted photons when an atom makes a transition from an excited state with the simultaneous emission of a photon with an average frequency of #math.equation(block: false, alt: "f equals 7.1 times 10 to the power 14 Hz")[$f = 7.1 #h(0.2em) × #h(0.2em) 10^(14) #h(0.2em) "Hz"$]. Is the emitted radiation monochromatic?


      Strategy
We invert  to obtain the energy uncertainty #math.equation(block: false, alt: "Δ E approximately equals ℏ / 2 Δ t")[$"Δ" E ≈ ℏ "/" 2 "Δ" t$] and combine it with the photon energy #math.equation(block: false, alt: "E equals h f")[$E = h " " f$] to obtain #math.equation(block: false, alt: "Δ f")[$"Δ" " " f$]. To estimate whether or not the emission is monochromatic, we evaluate #math.equation(block: false, alt: "Δ f / f")[$"Δ" f "/" f$].


      Solution
The spread in photon energies is #math.equation(block: false, alt: "Δ E equals h Δ f")[$"Δ" " " E = h "Δ" " " f$]. Therefore,


      #math.equation(block: true, alt: "Δ E, approximately equals, the fraction ℏ over 2 Δ t ⇒ h Δ f approximately equals the fraction ℏ over 2 Δ t ⇒ Δ f approximately equals the fraction 1 over 4 π Δ t equals the fraction 1 over 4 π open parenthesis 10 to the power −8 s close parenthesis equals 8.0 times 10 to the power 6 Hz ,; the fraction Δ f over f, equals, the fraction 8.0 times 10 to the power 6 Hz over 7.1 times 10 to the power 14 Hz equals 1.1 times 10 to the power −8 .")[$"Δ" E & ≈ & frac(ℏ, 2 "Δ" t) ⇒ h "Δ" f ≈ frac(ℏ, 2 "Δ" t) ⇒ "Δ" f ≈ frac(1, 4 "π" "Δ" t) = frac(1, 4 "π" ( 10^(−8) "s" )) = 8.0 #h(0.2em) × #h(0.2em) 10^(6) #h(0.2em) "Hz" "," \ frac("Δ" f, f) & = & frac(8.0 #h(0.2em) × #h(0.2em) 10^(6) #h(0.2em) "Hz", 7.1 #h(0.2em) × #h(0.2em) 10^(14) #h(0.2em) "Hz") = 1.1 #h(0.2em) × #h(0.2em) 10^(−8) .$]
      Significance
Because the emitted photons have their frequencies within #math.equation(block: false, alt: "1.1 times 10 to the power −6")[$1.1 #h(0.2em) × #h(0.2em) 10^(−6)$] percent of the average frequency, the emitted radiation can be considered monochromatic.
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A sodium atom makes a transition from the first excited state to the ground state, emitting a 589.0-nm photon with energy 2.105 eV. If the lifetime of this excited state is #math.equation(block: false, alt: "1.6 times 10 to the power −8 s")[$1.6 #h(0.2em) × #h(0.2em) 10^(−8) "s"$], what is the uncertainty in energy of this excited state? What is the width of the corresponding spectral line?


      
      #solutionbox[
#math.equation(block: true, alt: "4.1 times 10 to the power −8 eV")[$4.1 #h(0.2em) × #h(0.2em) 10^(−8) "eV"$]; #math.equation(block: true, alt: "1.1 times 10 to the power −5 nm")[$1.1 #h(0.2em) × #h(0.2em) 10^(−5) "nm"$]
]
]


  
  
    
=== Summary


    - The Heisenberg uncertainty principle states that it is impossible to simultaneously measure the #emph[x]-components of position and of momentum of a particle with an arbitrarily high precision. The product of experimental uncertainties is always larger than or equal to #math.equation(block: false, alt: "ℏ / 2 .")[$ℏ "/" 2 .$]
- The limitations of this principle have nothing to do with the quality of the experimental apparatus but originate in the wave-like nature of matter.
- The energy-time uncertainty principle expresses the experimental observation that a quantum state that exists only for a short time cannot have a definite energy.


  
  
    
=== Conceptual Questions


    
      If the formalism of quantum mechanics is ‘more exact’ than that of classical mechanics, why don’t we use quantum mechanics to describe the motion of a leaping frog? Explain.


    
    
      Can the de Broglie wavelength of a particle be known precisely? Can the position of a particle be known precisely?


    
    #solutionbox[
Yes, if its position is completely unknown. Yes, if its momentum is completely unknown.
]


    
      Can we measure the energy of a free localized particle with complete precision?


    
    
      Can we measure both the position and momentum of a particle with complete precision?


    
    #solutionbox[
No. According to the uncertainty principle, if the uncertainty on the particle’s position is small, the uncertainty on its momentum is large. Similarly, if the uncertainty on the particle’s position is large, the uncertainty on its momentum is small.
]


  
  
    
=== Problems


    
      A velocity measurement of an #math.equation(block: false, alt: "α")[$α$]-particle has been performed with a precision of 0.02 mm/s. What is the minimum uncertainty in its position?


    
    
      A gas of helium atoms at 273 K is in a cubical container with 25.0 cm on a side. (a) What is the minimum uncertainty in momentum components of helium atoms? (b) What is the minimum uncertainty in velocity components? (c) Find the ratio of the uncertainties in (b) to the mean speed of an atom in each direction.


    
    #solutionbox[
a. #math.equation(block: false, alt: "Δ p greater than or equal to 2.11 times 10 to the power −34 N times s")[$"Δ" p ≥ 2.11 #h(0.2em) × #h(0.2em) 10^(−34) "N" · "s"$]; b. #math.equation(block: false, alt: "Δ v greater than or equal to 6.31 times 10 to the power −8 m")[$"Δ" v ≥ 6.31 #h(0.2em) × #h(0.2em) 10^(−8) "m"$]; c. #math.equation(block: false, alt: "Δ v / the square root of k sub B T / m sub α equals 5.94 times 10 to the power −11")[$"Δ" v "/" sqrt(k_("B") T "/" m_(α)) = 5.94 #h(0.2em) × #h(0.2em) 10^(−11)$]
]


    
      If the uncertainty in the #math.equation(block: false, alt: "y")[$y$]-component of a proton’s position is 2.0 pm, find the minimum uncertainty in the simultaneous measurement of the proton’s #math.equation(block: false, alt: "y")[$y$]-component of velocity. What is the minimum uncertainty in the simultaneous measurement of the proton’s #math.equation(block: false, alt: "x")[$x$]-component of velocity?


    
    
      Some unstable elementary particle has a rest energy of 80.41 GeV and an uncertainty in rest energy of 2.06 GeV. Estimate the lifetime of this particle.


    
    #solutionbox[
#math.equation(block: true, alt: "Δ τ greater than or equal to 1.6 times 10 to the power −25 s")[$"Δ" τ ≥ 1.6 #h(0.2em) × #h(0.2em) 10^(−25) "s"$]
]


    
      An atom in a metastable state has a lifetime of 5.2 ms. Find the minimum uncertainty in the measurement of energy of the excited state.


    
    
      Measurements indicate that an atom remains in an excited state for an average time of 50.0 ns before making a transition to the ground state with the simultaneous emission of a 2.1-eV photon. (a) Estimate the uncertainty in the frequency of the photon. (b) What fraction of the photon’s average frequency is this?


    
    #solutionbox[
a. #math.equation(block: false, alt: "Δ f greater than or equal to 1.59 MHz")[$"Δ" f ≥ 1.59 #h(0.2em) "MHz"$]; b. #math.equation(block: false, alt: "Δ ω / ω sub 0 equals 3.135 times 10 to the power −9")[$"Δ" ω "/" ω_(0) = 3.135 #h(0.2em) × #h(0.2em) 10^(−9)$]
]


    
      Suppose an electron is confined to a region of length 0.1 nm (of the order of the size of a hydrogen atom). (a) What is the minimum uncertainty of its momentum? (b) What would the uncertainty in momentum be if the confined length region doubled to 0.2 nm?