Understanding Quantum Mechanics: Wave Functions, Schrödinger Equation, & Uncertainty, Study notes of Quantum Physics

Download Understanding Quantum Mechanics: Wave Functions, Schrödinger Equation, & Uncertainty and more Study notes Quantum Physics in PDF only on Docsity! Physics 4617/5617: Quantum Physics Course Lecture Notes Dr. Donald G. Luttermoser East Tennessee State University Edition 5.1 Abstract These class notes are designed for use of the instructor and students of the course Physics 4617/5617: Quantum Physics. This edition was last modified for the Fall 2006 semester. 3. From the concept of the wave function, it becomes easier to see how the Heisenberg Uncertainty Principle arises in nature. The wave function will not allow you to predict with certainty the outcome of a simple experiment to measure a particle’s position — all quantum mechanics has to offer is statistical information about the possible results. B. Philosophical Interpretations. 1. The Realist Position: a) We view the microscopic world as probabilistic due to the fact that quantum mechanics is an incomplete theory. b) The particle really was at a specific position (say point C in Figure II-1), yet quantum mechanics was unable to tell us so. c) To the realist, indeterminacy is not a fact of nature, but a reflection of our ignorance. d) If this scenario is, in fact, the correct one, then Ψ is not the whole story — some additional information (known as a hidden variable) is needed to provide a complete description of the particle. 2. The orthodox position =⇒ the Copenhagen interpretation: a) The particle isn’t really anywhere in space. The act of the measurement forces the particle to take a stand — though how and why we dare not ask! b) Observations not only disturb what is to be measured, they produce it. II–3 c) Bohr and his followers put forward this interpretation of quantum mechanics. d) It is the most widely accepted position of the interpreta- tion of quantum mechanics in physics. 3. The agnostic position: a) Refuse to answer! What sense can there be in making assertions about the status of a particle before a measure- ment, when the only way of knowing whether you were right is precisely to conduct the measurement, in which case what you get is no longer before the measurement. b) This has been used as a fall-back position used by many physicists if one is unable to convince another of the or- thodox position. 4. In 1964, John Bell astonished the physics community by showing that it makes an observable difference if the particle had a precise (although unknown) position prior to its measurement. a) This discovery effectively eliminated the realist position. b) Bell’s Theorem showed that the orthodox position is the correct interpretation of quantum mechanics by proving that any local hidden variable theory is incompatible with quantum mechanics (see Bell, J.S. 1964, Physics, 1, 195). c) We won’t get into the details of Bell’s Theorem at this point in time. Suffice it to say that a particle does not have a precise position prior to the measurement, any more than ripples in a pond do =⇒ it is the measurement pro- cess that insists upon one particular number, and thereby in a sense creates the specific result. II–4 5. The act of the measurement collapses the wave function to a delta function (e.g., a sharp peak) at some position — Ψ soon spreads out again after the measurement in accordance to the Schrödinger equation. C. Probability and Normalization. 1. Because of the statistical interpretation, probability, P , plays a central role in quantum mechanics. a) A probability value is the likelihood of a sample point oc- curring in a given distribution of points, where a sample point is defined here as a possible outcome of an experi- ment. b) A distribution of points can either be a set of discrete values or a continuous set of values. 2. Discrete Measurements. Below are a few definitions concern- ing discrete measurements. a) The total number of particles (or measurements) in a sys- tem is N = ∞∑ j=1 N(j) , (II-4) where N(j) is the number of particles (or measurements) in state j. b) The probability of a particle being in state j is P (j) = N(j) N , (II-5) whereas the sum of all the probabilities is ∞∑ j=1 P (j) = 1 . (II-6) II–5 4. A uniform distribution (as shown in Figure II-2) is used when there is an equal probability that all of the possible measured values will occur. If we set the probability density to a constant value of ρ(x) = A between the limits of −a/2 to a/2 (hence the total width of the distribution is a), the probability integral can then be used to find the amplitude A with respect to the width a: P = ∫ a/2 −a/2 A dx = Aa = 1 , (II-20) hence the amplitude must be A = 1 a (II-21) for this probability function to be normalizable. 5. Meanwhile, a normal distribution (i.e., a Gaussian distribu- tion) can be used to describe the distribution of random events or observations. This distribution function is shown in Figure (II-3) and described by the equation ρ(x) = 1√ 2πσ e−(x−µ) 2/2σ2 . (II-22) a) This distribution is centered around the mean, µ. b) Here σ is the standard deviation of the distribution. c) The full-width-at-half-maximum (FWHM), Γ, is related to the standard deviation by Γ = 2.354 σ . (II-23) d) The probable error (P.E.) of a normalized distribution is defined to be the absolute value of the deviation |x − µ| such that the probability for the deviation of any random observation |xi − µ| to be less is equal to 1/2 =⇒ that is, II–8 -6 σ -4 σ -2 σ 0 +2 σ +4 σ +6 σ x - µ 0.00 0.10 0.20 0.30 0.40 ρ( x ) Γ µ σ P.E. Figure II–3: Probability density function for a normal distribution. half of the observations of an event are expected to fall within the boundaries denoted by µ ± P.E. P.E. = 0.6745 σ = 0.2865 Γ . (II-24) 6. The statistical interpretation of the wave function (e.g., Eq. II- 3), says that |Ψ(x, t)|2 is the probability density for finding the particle at point x, at time t. This dictates the following normal- ization: ∫ +∞ −∞ |Ψ(x, t)|2 dx = 1 . (II-25) Without this, the statistical interpretation would be nonsense. a) But is this normalization consistent with Schrödinger’s equation (i.e., Eq. II-1)? That is, is it really valid for all time? b) Let us take the time derivative of the LHS of Eq. (II-25), then d dt ∫ +∞ −∞ |Ψ(x, t)|2 dx = ∫ +∞ −∞ ∂ ∂t |Ψ(x, t)|2 dx . (II-26) II–9 c) Note that the integral is a function only of t, since the x term(s) will disappear when the limits are applied. As such, a total derivative (d/dt) is taken for the solution to the integral on the LHS of Eq. (II-26), but the integrand (i.e., the function inside the integral) is a function of x as well as t, so a partial derivative (∂/∂t) must be used when the derivative is taken inside the integral on the RHS of Eq. (II-26). d) By the product rule, we get ∂ ∂t |Ψ|2 = ∂ ∂t (Ψ∗Ψ) = Ψ∗ ∂Ψ ∂t + ∂Ψ∗ ∂t Ψ . (II-27) e) Schrödinger’s equation says that ∂Ψ ∂t = ih̄ 2m ∂2Ψ ∂x2 − i h̄ V Ψ , (II-28) and hence also using Schrödinger’s equation for the com- plex conjugate of Eq. (II-27) gives ∂Ψ∗ ∂t = − ih̄ 2m ∂2Ψ∗ ∂x2 + i h̄ V Ψ∗ . (II-29) f) Using these equations in Eq. (II-27) gives ∂ ∂t |Ψ|2 = ih̄ 2m  Ψ∗ ∂2Ψ ∂x2 − ∂ 2Ψ∗ ∂x2 Ψ   = ∂ ∂x [ ih̄ 2m ( Ψ∗ ∂Ψ ∂x − ∂Ψ∗ ∂x Ψ )] . (II-30) g) The integral in Eq. (II-25) can now be evaluated explicitly: d dt ∫ +∞ −∞ |Ψ(x, t)|2 dx = ih̄ 2m ( Ψ∗ ∂Ψ ∂x − ∂Ψ ∗ ∂x Ψ )∣∣∣∣∣ +∞ −∞ . (II-31) h) Ψ(x, t) must go to zero as x → ±∞, otherwise the wave function would not be normalizable. It follows that d dt ∫ +∞ −∞ |Ψ(x, t)|2 dx = 0 , (II-32) II–10 d) Since Ψ must be in the form of a group of limited spatial extent in order that the uncertainty in the x coordinate be relatively small, both the wave function and its derivatives must go to zero faster than x → ±∞. Consequently the integrand term is equal to zero, and we have ∫ ∞ −∞ xΨ ∂2Ψ∗ ∂x2 dx = − ∫ +∞ −∞ ∂(x Ψ) ∂x︸ ︷︷ ︸ u ∂Ψ∗ ∂x dx ︸ ︷︷ ︸ dv . (II-38) e) Integrating by parts again, ∫ +∞ −∞ ∂(x Ψ) ∂x ∂Ψ∗ ∂x dx = −  ∂(x Ψ) ∂x Ψ∗   +∞ −∞ + ∫ +∞ −∞ Ψ∗ ∂2(xΨ) ∂x2 dx . (II-39) Reducing this again gives the following result for Eq. (II- 38): ∫ ∞ −∞ xΨ ∂2Ψ∗ ∂x2 dx = ∫ +∞ −∞ Ψ∗ ∂2(xΨ) ∂x2 dx . (II-40) f) Putting this back into Eq. (II-36), we have d〈x〉 dt = ih̄ 2m ∫ +∞ −∞ Ψ∗  x ∂2Ψ ∂x2 − ∂ 2(xΨ) ∂x2   dx . (II-41) Consider the bracket in the integrand, it can be written x ∂2Ψ ∂x2 − ∂2(x Ψ) ∂x2 = x ∂2Ψ ∂x2 − ∂ ∂x ( x ∂Ψ ∂x + Ψ ) = x ∂2Ψ ∂x2 − x ∂2Ψ ∂x2 − ∂Ψ ∂x − ∂Ψ ∂x = −2∂Ψ ∂x . (II-42) g) Consequently, d〈x〉 dt = −ih̄ m ∫ +∞ −∞ ( Ψ∗ ∂Ψ ∂x ) dx . (II-43) II–13 h) At this point, we will postulate that the expectation value of the velocity of the particle is equal to the time derivative of the expectation value of the position of the particle : 〈v〉 = d〈x〉 dt . (II-44) i) Eq. (II-43) tells us, then, how to calculate 〈v〉 directly from Ψ. 3. Actually, it is customary to work with momentum (p = mv), rather than velocity: 〈p〉 ≡ m d〈x〉 dt = −ih̄ ∫ +∞ −∞ ( Ψ∗ ∂Ψ ∂x ) dx . (II-45) a) Let’s write expressions for 〈x〉 and 〈p〉 in a more suggestive way 〈x〉 = ∫ +∞ −∞ Ψ∗(x)Ψ dx (II-46) 〈p〉 = ∫ +∞ −∞ Ψ∗ ( h̄ i ∂ ∂x ) Ψ dx . (II-47) b) We say that the operator x represents position, and the operator (h̄/i)(∂/∂x) represents momentum, in quantum mechanics =⇒ to calculate expectation values, we sand- wich the appropriate operator between Ψ∗ and Ψ and in- tegrate. 4. All such dynamic variables can be written in terms of position and momentum. Kinetic energy is 〈T 〉 = 〈p2〉 2m = −h̄2 2m ∫ +∞ −∞ Ψ∗   ∂ 2 ∂x2  Ψ dx . (II-48) E. The Uncertainty Principle 1. The uncertainty in the measurement of an event is nothing more than the standard deviation σ = √ σ2 of the measurement (e.g., Eq. II-19). II–14 2. Hence, the Heisenberg Uncertainty Principle can be rewrit- ten in the form σx σp ≥ h̄ 2 , (II-49) where σx is the standard deviation in the x-position and σp is the standard deviation in the corresponding momentum. 3. We shall see from this point forward, that wave functions that describe real particles always obey Eq. (II-49). We shall prove this relation in §IV of the notes. Example II–2. Consider the wave function of Example II-1: Ψ(x, t) =  β 2 π   1/4 e−(β 2x2/2+iEt/h̄) . Show that this wave function satisfies the Heisenberg Uncertainty Relation- ship. Solution: First, calculate the various expectation values. 〈x〉 = ∫ +∞ −∞ √√√√β2 π e−(β 2x2/2−iEt/h̄) x e−(β 2x2/2+iEt/h̄) dx = √√√√β2 π ∫ +∞ −∞ x e−β 2x2 dx = 0 , since we are integrating an odd function over an even interval. 〈x2〉 = ∫ +∞ −∞ √√√√β2 π e−(β 2x2/2−iEt/h̄) x2 e−(β 2x2/2+iEt/h̄) dx = √√√√β2 π ∫ +∞ −∞ x2 e−β 2x2 dx = 2 √√√√β2 π ∫ +∞ 0 x2 e−β 2x2 dx = 2 √√√√β2 π 1 4β2 √√√√ π β2 = 1 2β2 . II–15