PUEC 2018 Quantum Mechanics Section
Team: ShutUpAndLovePhysics Sabina Drăgoi Ioana Stelea Daniel Sîreţanu Felix Puşcaşu
November 2018
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ShutUpAndLovePhysics
PUEC 2018
Contents 1
Wave Functions, Observables, and the Schrödinger Equation 1.1 The Wave Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Schrödinger’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Mathematical Formalism and Quantum Quirks 2.1 Solving Schrödinger’s Equation . . . . . . . . 2.2 The Hamiltonian and Hermitian Operators . . 2.3 The Harmonic Oscillator . . . . . . . . . . . 2.4 The Commutator . . . . . . . . . . . . . . .
3
Experiment
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4 4 5 9 9 11 22 28 31
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1 1.1
PUEC 2018
Wave Functions, Observables, and the Schrödinger Equation The Wave Function
Answer The statistical interpretation of Ψ(𝑥, 𝑡) is that |Ψ(𝑥, 𝑡)|2 𝑑𝑥 represents the probability of finding the particle between 𝑥 and 𝑥 + 𝑑𝑥, at time 𝑡. +∞
Therefore, when we do the integral of ∫−∞ |Ψ(𝑥, 𝑡)|2 𝑑𝑥 from −∞ to ∞ (covering all the space) we must get 1 because the particle has to be there. So the probability of finding the particle is equal to unity. Exercise We will first prove that the normalization condition for Ψ(𝑥, 0) is respected: +∞
∫−∞ +∞ [(
∫−∞
𝑚𝜔 𝜋ℏ
)2
𝑒
|Ψ(𝑥, 𝑡)|2 𝑑𝑥 = 1
] 𝑑𝑥 =
𝑚𝜔𝑥2 2ℏ
+∞ (
∫−∞ √
𝑚𝜔 𝜋ℏ
)1
2
𝑒
𝑚𝜔𝑥2 2ℏ
𝑑𝑥 = 1
√ 𝑥 for which we have that 𝑑𝑧 = 𝑚𝜔 𝑑𝑥. We denote 𝑧 = 𝑚𝜔 ℏ ℏ This means we are left to prove that: +∞ 1 −𝑧2 √ 𝑒 𝑑𝑧 = 1 ∫−∞ 𝜋 We will now prove the result from the Gaussian integral obtained above. 𝐼2 =
(
∞
∫−∞
2
𝑒−𝑥 𝑑𝑥
)2
∞
=
∫−∞
∞
2
𝑒−𝑥 𝑑𝑥
∫−∞
∞
∞
2
𝑒−𝑦 𝑑𝑦 =
∫−∞ ∫−∞
2 +𝑦2 )
𝑒−(𝑥
𝑑𝑥𝑑𝑦
Moving into polar coordinates, we would get: 2𝜋
𝐼2 =
+∞
𝑑𝜃
∫0
∫0
+∞
2
𝑒−𝑟 𝑟𝑑𝑟 = 2𝜋
∫0
2
𝑒−𝑟 𝑟𝑑𝑟
Changing the variable to 𝑎 = 𝑟2 , we get 𝑑𝑎 = 2𝑟𝑑𝑟, and +∞
𝐼 2 = 2𝜋
∫0
1 −𝑎 𝑒 𝑑𝑎 = 𝜋(𝑒0 − 𝑒−∞ ) = 𝜋 2 +∞
⇒𝐼 =
∫−∞
2
𝑒−𝑥 𝑑𝑥 =
Finally, this means our normalization condition becomes: √ +∞ 𝜋 1 −𝑧2 √ 𝑒 𝑑𝑧 = √ = 1 ∫−∞ 𝜋 𝜋
√ 𝜋
True
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Next, we move on to compute the expectation value for x: +∞ (
+∞
⟨𝑥⟩(𝑡=0) =
2
∫−∞
|Ψ(𝑥, 0)| 𝑑𝑥 =
∫−∞
𝑚𝜔 𝜋ℏ
)1 2
𝑒
𝑚𝜔𝑥2 ℏ
𝑥𝑑𝑥
𝑚𝜔𝑥2
Because 𝑒 ℏ 𝑥 is an odd function, which means that is symmetric to the origin, it results that the integral from −𝑎 to 𝑎 is 0, and therefore that ⟨𝑥⟩ = 0.
Figure 1: Plot depicting an odd function √ √ 𝑚𝜔 Doing the math in which we used again the notation 𝑧 = 𝑚𝜔 𝑥 for which we have that 𝑑𝑧 = 𝑑𝑥, ℏ ℏ we get: √ √ )1 +∞ +∞ +∞ ( 2 𝑚𝜔 2 | 1 1 ℏ ℏ 𝑚𝜔 2 𝑚𝜔𝑥 | − 𝑥 −𝑧 ⇒ 𝑒 𝑑𝑧 = − 𝑒 ℏ | 𝑒 ℏ 𝑥𝑑𝑥 = | ∫−∞ 𝜋ℏ 2 ∫−∞ 𝑚𝜔𝜋 𝑚𝜔𝜋 2 | −∞
⟨𝑥⟩ = 0
1.2
Schrödinger’s Equation
Answer In order to provide a comparison with classical mechanics, we will try to deduce a correction to Newton’s 2𝑛𝑑 Law. We start by looking at the way we write in classical mechanics the momentum in terms of the potential: d𝑝𝑥 𝜕𝑉 (𝑥) =− d𝑡 𝜕𝑥 And now in quantum theory:
d ⟨𝑝𝑥 ⟩ 𝜕𝑉 (𝑥) = −⟨ ⟩ d𝑡 𝜕𝑥 In words, it can be seen that the main difference is that in quantum mechanics we have ⟨𝑥⟩, ⟨𝑝⟩, which are the expectation values, instead of simply x and p. For a complete analogy to be made, instead of ⟨ 𝜕𝑉𝜕𝑥(𝑥) ⟩ we Page 5
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𝜕𝑉 (⟨𝑥⟩) ⟩. 𝜕𝑥
should be looking for − ⟨ Writing the Taylor Series: 𝑉 (𝑥)for 𝑎=⟨𝑥⟩ =
∞ ∑ 𝑉 𝑖 (⟨𝑥⟩) 𝑖=0
𝑖!
= 𝑉 (⟨𝑥⟩) +
(𝑥 − ⟨𝑥⟩)𝑖
𝜕𝑉 (⟨𝑥⟩) 1 𝜕 3 𝑉 (⟨𝑥⟩) 1 𝜕 2 𝑉 (⟨𝑥⟩) (𝑥 − ⟨𝑥⟩)2 + (𝑥 − ⟨𝑥⟩)3 + … (𝑥 − ⟨𝑥⟩) + 𝜕𝑥 2 𝜕𝑥2 6 𝜕𝑥3
Taking the derivative with respect to x and keeping in mind that ⟨𝑥⟩ is constant : 𝜕𝑉 (⟨𝑥⟩) 𝜕 2 𝑉 (⟨𝑥⟩) 𝜕𝑉 (𝑥) 1 𝜕 3 𝑉 (⟨𝑥⟩) (𝑥 − ⟨𝑥⟩) + ⟨ (𝑥 − ⟨𝑥⟩)2 ⟩ + … =0+ + 𝜕𝑥 𝜕𝑥 2 𝜕𝑥3 𝜕𝑥2 Taking the expectation values for the whole equation: ⟨
𝜕𝑉 (⟨𝑥⟩) 𝜕 2 𝑉 (⟨𝑥⟩) 𝜕𝑉 (𝑥) 1 𝜕 3 𝑉 (⟨𝑥⟩) ⟩=⟨ ⟩+⟨ (𝑥 − ⟨𝑥⟩)⟩ + ⟨ (𝑥 − ⟨𝑥⟩)2 ⟩ + … 𝜕𝑥 𝜕𝑥 2 𝜕𝑥3 𝜕𝑥2
We also see that: ⟨𝑥 − ⟨𝑥⟩⟩ = ⟨𝑥⟩ − ⟨𝑥⟩ = 0 From all these, it follows that: ⟨
𝜕𝑉 (⟨𝑥⟩) 𝜕𝑉 (𝑥) 1 𝜕 3 𝑉 (⟨𝑥⟩) ⟩=⟨ ⟩+0+⟨ (𝑥 − ⟨𝑥⟩)2 ⟩ 𝜕𝑥 𝜕𝑥 2 𝜕𝑥3
And therefore,
𝑑 ⟨𝑝⟩ 𝜕𝑉 (𝑥) 1 𝜕 3 𝑉 (⟨𝑥⟩) = −⟨ ⟩+⟨ (𝑥 − ⟨𝑥⟩)2 ⟩ 𝑑𝑡 𝜕𝑥 2 𝜕𝑥3
If 𝑉 (𝑥) has a form similar to 𝑉 (𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐, we would have an exact analogy of Newton’s 2𝑛𝑑 law for quantum mechanics. However, if 𝑉 (𝑥) has a higher polynomial degree, it will pick up other terms, which will introduce corrections to Newton’s 2𝑛𝑑 law. To answer the question, Newton’s second law, in the form without expectation values, is a good approximation for day-to-day experiences, but needs corrections for other set-ups, such as the physics of particles and probabilities. Exercise 1 We start by writing what we know: 𝑑 ⟨𝑥⟩ ⟨𝑝⟩ = 𝑚 𝑑𝑡 𝜕Ψ ℏ2 𝜕 2 Ψ =− + 𝑉 Ψ (Schrödinger’s equation) 𝜕𝑡 2𝑚 𝜕𝑥2 Next, we take a closer look at the expectation value of x: 𝑖ℏ
+∞
⟨𝑥⟩ =
∫−∞
+∞
𝑥|Ψ(𝑥, 𝑡)|2 𝑑𝑥 =
∫−∞
𝑥Ψ(𝑥, 𝑡) ⋅ Ψ∗ (𝑥, 𝑡)𝑑𝑥
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Taking the derivative with respect to time: +∞ 𝑑 ⟨𝑥⟩ 𝜕 = 𝑥 (Ψ ⋅ Ψ∗ )𝑑𝑥 ∫ 𝑑𝑡 𝜕𝑡 −∞ ) +∞ ( 𝜕Ψ ∗ 𝜕Ψ∗ = 𝑥 Ψ + Ψ 𝑑𝑥 ∫−∞ 𝜕𝑡 𝜕𝑡
Keeping in mind the result from above, we wil now move our attention toward expressing using Schördinger’s equation: 𝑖ℏ 𝜕 2 Ψ 𝑖 𝜕Ψ = − 𝑉Ψ 𝜕𝑡 2𝑚 𝜕𝑥2 ℏ
𝜕Ψ 𝜕𝑡
and
𝜕Ψ∗ 𝜕𝑡
by
𝜕Ψ∗ 𝑖ℏ 𝜕 2 Ψ∗ 𝑖 =− + 𝑉 Ψ∗ 𝜕𝑡 2𝑚 𝜕𝑥2 ℏ Substituting this in
𝑑⟨𝑥⟩ : 𝑑𝑡
( ) ( ) ] +∞ [ 𝑑 ⟨𝑥⟩ 𝑖ℏ 𝜕 2 Ψ 𝑖 𝑖ℏ 𝜕 2 Ψ∗ 𝑖 = − 𝑉 Ψ Ψ∗ + − + 𝑉 Ψ∗ Ψ 𝑥𝑑𝑥 ∫−∞ 2𝑚 𝜕𝑥2 𝑑𝑡 ℏ 2𝑚 𝜕𝑥2 ℏ ( ) +∞ 𝑑 ⟨𝑥⟩ 𝑖ℏ 𝜕 𝜕Ψ 𝜕Ψ∗ = 𝑥 𝜓∗ − Ψ 𝑑𝑥 ∫−∞ 2𝑚 𝜕𝑥 𝑑𝑡 𝜕𝑥 𝜕𝑥 Re ordering the terms in the equation: 𝑚
( ) +∞ 𝑑 ⟨𝑥⟩ 𝑖ℏ 𝜕 𝜕Ψ 𝜕Ψ∗ = 𝑥 Ψ∗ − Ψ 𝑑𝑥 ∫−∞ 2 𝜕𝑥 𝑑𝑡 𝜕𝑥 𝜕𝑥
And therefore:
+∞
⟨𝑝⟩ =
∫−∞
( ) 𝑖ℏ 𝜕 𝜕Ψ 𝜕Ψ∗ 𝑥 Ψ∗ − Ψ 𝑑𝑥 2 𝜕𝑥 𝜕𝑥 𝜕𝑥
We will now integrate by parts. In formulas, we use the general form: ∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢, where
𝜕Ψ 𝜕Ψ∗ − Ψ 𝜕𝑥 𝜕𝑥 ( ) 𝜕 𝜕Ψ∗ ∗ 𝜕Ψ 𝑑𝑣 = Ψ − Ψ 𝑑𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑣 = Ψ∗
𝑢1 = 𝑥 𝑑𝑢1 = 𝑑𝑥
This means: ( )|+∞ ) +∞ ( 𝑖ℏ 𝜕Ψ∗ 𝑖ℏ 𝜕Ψ∗ | ∗ 𝜕Ψ ∗ 𝜕Ψ ⟨𝑝⟩ = ⋅𝑥⋅ Ψ − Ψ | − Ψ − Ψ 𝑑𝑥 | 2 𝜕𝑥 𝜕𝑥 2 ∫−∞ 𝜕𝑥 𝜕𝑥 | −∞
We see that when 𝑥 ⟶ ±∞, then both Ψ and
𝜕Ψ 𝜕𝑥
→ 0. This leads to:
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+∞ (
) 𝜕Ψ∗ Ψ − Ψ 𝑑𝑥 𝜕𝑥 𝜕𝑥 ) ( ∗ +∞ We use again the method of integrating by parts, this time for ∫−∞ − 𝜕Ψ Ψ 𝑑𝑥. 𝜕𝑥 𝑖ℏ ⟨𝑝⟩ = − 2 ∫−∞
∗ 𝜕Ψ
𝑢2 = Ψ∗ 𝜕Ψ∗ 𝑑𝑢2 = 𝑑𝑥 𝜕𝑥 Then, +∞
∫−∞
𝑣=Ψ 𝜕Ψ 𝑑𝑣 = 𝑑𝑥 𝜕𝑥
( ∗ ) ) ) +∞ ( +∞ ( |+∞ 𝜕Ψ 𝜕Ψ ∗ 𝜕Ψ∗ | − Ψ 𝑑𝑥 = −Ψ ⋅ Ψ∗ | − Ψ 𝑑𝑥 = Ψ 𝑑𝑥 − | ∫−∞ ∫−∞ 𝜕𝑥 𝜕𝑥 𝜕𝑥 | −∞
From all of the above, we finally obtain∶ +∞
⟨𝑝⟩ = −
𝜕Ψ 𝑖ℏ 2Ψ∗ 𝑑𝑥 ∫ 2 −∞ 𝜕𝑥 +∞
⟨𝑝⟩ =
ℏ 𝜕Ψ Ψ∗ 𝑑𝑥 ∫ 𝑖 −∞ 𝜕𝑥
Exercise 2 Using the above result, we re-write: ⟨𝑝⟩ =
∫
Ψ∗
( − 𝑖ℏ
𝑝̂𝑥 = −𝑖ℏ ⟨𝑝⟩ = ∫ Ψ∗ 𝑝̂𝑥 Ψ𝑑𝑥 ⟨𝑥⟩ = ∫ Ψ∗ 𝑥Ψ𝑑𝑥 Therefore, we have for any dynamical variables: ⟨𝐹 (𝑥, 𝑝)⟩ =
∫
) 𝜕 Ψ𝑑𝑥 𝜕𝑥
𝜕 𝜕𝑥
momentum operator
}
position operator
Ψ∗ 𝐹 (𝑥, −𝑖ℏ
𝜕 )Ψ𝑑𝑥 = ⟨Ψ|𝐹̂ Ψ⟩ 𝜕𝑥
⇒
(1)
Although we also used the hat notation, we will only make use of it only later. For instance, we will look at one of the most basic dynamical variables: the kinetic energy 𝑇 , for which we have: +∞ ⟨𝑝⟩2 1 ∗ 𝜕 = 𝜓 𝑇 (𝑥, −𝑖ℏ )𝜓𝑑𝑥 ∫−∞ 2𝑚 2𝑚 𝜕𝑥 +∞ 𝜕2𝜓 1 ⟨𝑇 ⟩ = 𝜓 ∗ ⋅ (−𝑖ℏ)2 ⋅ 𝑑𝑥 2𝑚 ∫−∞ 𝜕𝑥2
⟨𝑇 ⟩ =
⟨𝑇 ⟩ = −
𝜕2𝜓 ℏ2 𝜓∗ 𝑑𝑥 2𝑚 ∫ 𝜕𝑥2
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Exercise 3 If the normalization condition has to be satisfied for all 𝑡, in formulas this is equivalent to: +∞
+∞
𝑑 𝜕 |𝜓(𝑥, 𝑡)|2 𝑑𝑥 = |𝜓(𝑥, 𝑡)|2 𝑑𝑥 = 0 ∫ ∫ 𝑑𝑡 −∞ −∞ 𝜕𝑡 Looking more closely at the time derivative: 𝜕𝜓 𝜕𝜓 ∗ 𝜕 𝜕 |𝜓|2 = |𝜓 ∗ 𝜓| = 𝜓 ∗ + 𝜓 𝜕𝑡 𝜕𝑡 𝜕𝑡 𝜕𝑡 We now remember the Schrödinger’s equation for Ψ and Ψ∗ 2 ⎧ 𝜕𝜓 = 𝑖ℏ 𝜕 𝜓 − 𝑖𝑉 𝜓 ⎪ 𝜕𝑡 2𝑚 𝜕𝑥2 ℏ ⎨ ∗ 2 ∗ ∗ ⎪ 𝜕𝜓 = − 𝑖ℏ 𝜕 𝜓 + 𝑖𝑉 𝜓 ⎩ 𝜕ℏ 2𝑚 𝜕𝑥2 ℏ
We implement this in the time derivative and obtain: 𝜕 || 2 || 𝑖ℏ || ∗ 𝜕 2 𝜓 ∗ 𝜕 2 𝜓 ∗ || 𝑖ℏ 𝜕 | ∗ 𝜕𝜓 𝜕𝜓 ∗ | 𝜓 = 𝜓 − 𝜓 = − 𝜓| |𝜓 𝜕𝑡 || || 2𝑚 || 𝜕𝑥 | 𝜕𝑥2 𝜕𝑥2 || 2𝑚 𝜕𝑥 | 𝜕𝑥 This gives us:
∗ ||+∞ |2 | |𝜓(𝑥, 𝑡)| 𝑑𝑥 = 𝑖ℏ |𝜓 ∗ 𝜕𝜓 − 𝜕𝜓 𝜓 ||| | | 2𝑚 || 𝜕𝑥 𝜕𝑥 |||| | | −∞
+∞ |
𝑑 𝑑𝑡 ∫−∞ As 𝑥 goes to ±∞, we note that
𝜓(𝑥, 𝑡) → 0 𝜕 𝜓(𝑥, 𝑡) 𝜕𝑡
And therefore,
}
→0
⇒
+∞
𝜕 |𝜓(𝑥, 𝑡)|2 𝑑𝑥 = 0 ∫−∞ 𝜕𝑡 This means that 𝜓 stays normalized all the time.
2 2.1
Mathematical Formalism and Quantum Quirks Solving Schrödinger’s Equation
Exercise 1 We start by writing Ψ(𝑥, 𝑡) = 𝜙(𝑡)𝜓(𝑥). This means
And that
𝑑𝜙 𝜕Ψ =𝜓 𝜕𝑡 𝑑𝑡 𝑑2𝜓 𝜕2Ψ =𝜙 𝜕𝑥2 𝑑𝑥2 Page 9
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Put into Schrödinger’s equation, it gives us: 𝑖ℏ
𝜕Ψ ℏ2 𝜕 2 Ψ + 𝑉 (𝑥)Ψ =− 𝜕𝑡 2𝑚 𝜕𝑥2
𝑑𝜙 ℏ2 𝑑 2 𝜓 =− 𝜙 + 𝑉 𝜙𝜓 𝑑𝑡 2𝑚 𝑑𝑥2 We then divide by 𝜓𝜙, noting that they cannot be zero for all t, or else there would be no problem at all. ⇒
𝑖ℏ𝜓
𝑖ℏ
1 𝑑𝜙 ℏ 1 𝑑2𝜓 = − +𝑉 𝜙 𝑑𝑡 2𝑚 𝜓 𝑑𝑥2
} Right side is a function of 𝑡 Left side is a function of 𝑥
⇒
This is true as long as both sides are constant
We then write: 𝑖ℏ
1 𝑑𝜙 =𝐸 𝜓 𝑑𝑡
Or in another form: 𝑑𝜙 𝑖𝐸 =− 𝜙 𝑑𝑡 ℏ For later use, we will also write it as: 𝜙 ∼ 𝑒−
𝑖𝐸𝑡 ℏ
Substituting, we have: −
ℏ2 𝑑 2 𝜓 + 𝑉 (𝑥)𝜓 = 𝐸𝜓 2𝑚 𝑑𝑥2
(2)
Some applications 1. Stationary state Using the exponential form for 𝜙 and that Ψ(𝑥, 𝑡) = 𝜙(𝑡)𝜓(𝑥), it follows that: Ψ(𝑥, 𝑡) = 𝜓(𝑥)𝑒−
𝑖𝐸𝑡 ℏ
Then, |Ψ(𝑥, 𝑡)|2 = 𝜓 ∗ 𝑒
𝑖𝐸𝑡 ℏ
𝜓𝑒−
𝑖𝐸𝑡 ℏ
= |𝜓(𝑥)|2
In words, this means that every expected value is constant in time. ⟨𝑥⟩ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⟨𝑝⟩ = 𝑚
𝑑 ⟨𝑥⟩ =0 𝑑𝑡
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2. In classical mechanics, the total energy is called the Hamiltonian and is defined as: 𝐻(𝑥, 𝑝) =
𝑝2 + 𝑉 (𝑥) 2𝑚
The corresponding Hamiltonian operator is ℏ2 𝜕 2 𝐻̂ = − + 𝑉 (𝑥) 2𝑚 𝜕𝑥2 And has the property that: ̂ = 𝐸𝜓 𝐻𝜓 Computing its expectation value, we get: ⟨𝐻⟩ =
∫
̂ 𝜓 ∗ (𝐻𝜓)𝑑𝑥 =𝐸
∫
|𝜓|2 𝑑𝑥 = 𝐸 𝑡𝑜𝑡
Now computing the expectation value of 𝐻 2 : ̂ 𝐻𝜓| ̂ ̂ 𝐻̂ 2 𝜓 = 𝐻| = 𝐻|𝐸𝜓| = 𝐸2𝜓 Therefore, its ’deviation’ gives: 2 = ⟨𝐻⟩2 − ⟨𝐻⟩2 = 0 𝜎𝐻
In words, this can be interpreted that every measurement must share the same value of 𝐸. 3. Because of the raised difficulty of Schrödinger’s equation in its raw form, the general solution will be written as a linear combination of separable solutions. The same approach can be found, at high school level, in driven and damped harmonic motion: 𝑥̈ + 2𝛽 𝑥̇ + 𝜔20 𝑥 = 𝐹 (𝑥). So the solution will be the sum of the homogeneous and the particular solutions. Ψ1 (𝑥, 𝑡) = 𝜓(𝑥)𝑒− Ψ2 (𝑥, 𝑡) = 𝜓(𝑥)𝑒−
𝑖𝐸1 𝑡 ℏ 𝑖𝐸2 𝑡 ℏ
… Ψ(𝑥, 𝑡) =
∞ ∑
𝐶𝑛 𝜓𝑛 (𝑥)𝑒−
𝑖𝐸𝑛 𝑡 ℏ
𝑛=1
We will investigate the mathematics behind this writing in more detail in the next sections, but however considered it worth to mention it here as well.
2.2
The Hamiltonian and Hermitian Operators
Exercise 1 We will start our analysis with operator 𝑥: ̂ ⟨𝑥𝜓|𝜓⟩ ̂ =
∫
(𝑥𝜓) ̂ ∗ 𝜓𝑑𝑥 =
∫
(𝑥𝜓)∗ 𝜓𝑑𝑥 =
∫
(𝜓)∗ 𝑥𝜓𝑑𝑥 = ⟨𝜓|𝑥𝜓⟩ ̂ ⇒ Hermitian Page 11
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Next, we have operator 𝑝̂ =
PUEC 2018
ℏ 𝜕 𝑖 𝜕𝑥
⟨𝑝𝜓|𝜓⟩ ̂ = = 𝑖ℏ
∫
(𝑝𝜓) ̂ ∗ 𝜓𝑑𝑥 = (
∫
𝑑𝜓 𝑑𝑡
)∗
∫
(−𝑖ℏ
𝜓𝑑𝑥 = 𝑖ℏ
𝜕𝜓 ∗ ) 𝜓𝑑𝑥 𝜕𝑥
𝑑𝜓 ∗ 𝜓
∫
Using integrals by parts, this becomes: ⟨𝑝𝜓|𝜓⟩ ̂ = 𝑖ℏ𝜓 ∗ 𝜓 − 𝑖ℏ = −𝑖ℏ
∫
𝜓∗
∫
𝑑𝜓𝜓 ∗ =
𝑑𝜓 𝑑𝑥 = ⟨𝜓|𝑝𝜓⟩ ̂ ⇒ Hermitian 𝑑𝑥
̂ 𝑇̂ + 𝑉̂ (𝑥). ̂ We will first try to prove that 𝑇 is Hermitian. Lastly, we look at operator 𝐻= By definition, 𝑝̂2 𝑇̂ = 2𝑚 This means, 1 ⟨𝑇̂ 𝜓|𝜓⟩ = (𝑇̂ 𝜓)∗ 𝜓𝑑𝑥 = (𝑝̂2 𝜓)∗ 𝜓𝑑𝑥 = ∫ 2𝑚 ∫ =
1 (𝑝𝜓) ̂ ∗ (𝑝𝜓)𝑑𝑥 ̂ 2𝑚 ∫
From (𝑝̂ ⋅ 𝑝𝜓) ̂ ∗ = 𝑝̂∗ ⋅ 𝜓 ∗ ⋅ 𝑝̂ ⋅ 𝜓 = (𝑝𝜓) ̂ ∗ 𝑝𝜓 ̂ It follows that ⟨𝑇̂ 𝜓|𝜓⟩ =
1 𝜓 ∗ 𝑝̂2 𝜓𝑑𝑥 2𝑚 ∫
= 𝜓 ∗ 𝑇̂ 𝜓𝑑𝑥 = ⟨𝜓|𝑇̂ 𝜓⟩ ⇒ }
𝑇̂ is Hermitian 𝑉̂ is Hermitian by nature, because it is an operator of 𝑥̂
⇒
𝐻̂ is Hermitian because it is the sum of 2 Herminitans
Exercise 2 In formulas, proving that hermitian operators have real eigenvalues, translates as ̂ = 𝜆𝑥 ⇒ 𝜆 ∈ R. proving that if 𝐻𝑥 For starters we will note that: ̂ ̂ ⟨𝑥|𝐻|𝑥⟩ = ⟨𝑥|(𝐻𝑥⟩) = ⟨𝑥|𝜆|𝑥⟩ = 𝜆 ⟨𝑥|𝑥⟩ We also deduce that because 𝐻̂ is Hermitian → 𝐻̂ = 𝐻̂ ∗ ∗ ̂ Generally speaking, we know that for any 𝐴̂ and ∀ |𝑥⟩ , |𝑦⟩ ∈ V ⇒ ⟨𝑥|𝐴|𝑦⟩ = ⟨𝑦|𝐴̂ 𝑡 |𝑥⟩ . Therefore, that means for us that ∗ ̂ ̂ ⟨𝑥|𝐻|𝑥⟩ = ⟨𝑥|𝐻|𝑥⟩ ⇒ Page 12
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𝜆 ⟨𝑥|𝑥⟩ = (𝜆 ⟨𝑥|𝑥⟩)∗ = 𝜆∗ (⟨𝑥|𝑥⟩)∗ We also know, ⟨𝑥|𝑥⟩ = ⟨𝑥|𝑥⟩∗ for all 𝑥 ∈ R And then, ⇒𝜆=𝜆∗ Since this can only be true for real numbers, we conclude that 𝜆 ∈ R. Exercise 3 Let’s suppose we have 2 eigenfunctions 𝜓 and 𝜙. Then we write: 𝐴𝜓 = 𝜆1 𝜓 𝐴𝜙 = 𝜆2 𝜙 Making the product ⟨𝜙|𝐴|𝜓⟩ = ⟨𝜙|𝐴𝜓⟩ in 2 different ways, we get the following: ⟨𝜙|𝐴|𝜓⟩ = ⟨𝜙|𝐴𝜓⟩ = ⟨𝐴𝜙|𝜓⟩ = 𝜆2 ⟨𝜙|𝜓⟩ = ⟨𝜙𝜆1 𝜓⟩ = 𝜆1 ⟨𝜙|𝜓⟩ Because we started with the premises that the eigenvalues are different, that is 𝜆1 ≠ 𝜆2 , then the only way for the equality above to hold is if ⟨𝜙|𝜓⟩ = 0 Exercise 4 aka Question ∑ By writing 𝜓 = 𝑛 𝐶𝑛 𝜓𝑛 we assume that each one of the n separate solution has a contribution (probability) of 𝐶𝑛 to the final form of 𝜓. As the product ⟨𝜓𝑛 (𝑥)|𝜓(𝑥, 0)⟩, in its integral form can be interpreted just as that, it would seem intuitive to write 𝐶𝑛 = ⟨𝜓𝑛 (𝑥)|𝜓(𝑥, 0)⟩. Pushing our analogy between 𝐶𝑛 and probabilities even further, we could say that a ’normalization’ relation ∑ would be written as 𝑛 |𝐶𝑛 |2 = 1. However, we will also rigorously prove this in the later exercises. Exercise 5 For starters we have the linear combination: ∑ 𝐶𝑛 𝜓 𝑛 𝜓= 𝑛
We now look at the product: ⟨𝜓𝑚 (𝑥)|𝜓(𝑥, 0)⟩ = =
∫ ∑ 𝑛
=
∑
𝜓𝑚∗ (𝑥)𝜓(𝑥)𝑑𝑥 ( ) ∗ 𝐶𝑛 𝜓 (𝑥)𝜓𝑛 (𝑥)𝑑𝑥 ∫ 𝑚 𝐶𝑛 ⟨𝜓𝑚 |𝜓𝑛 ⟩
𝑛
Using the result from above regarding orthogonality of eigenfunctions with different eigenvalues, we can write that: { 0 if 𝑚 ≠ 𝑛 ⟨𝜓𝑚 |𝜓𝑛 ⟩ = 𝛿𝑚𝑛 (Kronecker delta) = 1 if 𝑚 = 𝑛 Page 13
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Substituted in the equation from above, it results that ⟨𝜓𝑚 (𝑥)|𝜓(𝑥, 0)⟩ = 𝐶𝑚 Which is equivalent to 𝐶𝑛 = ⟨𝜓𝑛 (𝑥)|𝜓(𝑥, 0)⟩ Exercise 6 We return to the way we wrote the general solution as a linear combination: Ψ(𝑥, 𝑡) =
∞ ∑
𝐶𝑛 𝑒−
𝑖𝐸𝑛 𝑡 ℏ
𝜓𝑛 (𝑥) =
𝑛=1
∞ ∑
|𝐶𝑛 |𝜓𝑛 (𝑥)
𝑛=1
Normalizing, | |2 |Ψ(𝑥, 𝑡)| = 1 | ∫ | ( ) ∑ | |2 |𝐶𝑛 |2 |Ψ(𝑥, 𝑡)| = Re 𝜓 2 (𝑥, 𝑡) = | | As ⟨𝜓𝑛 |𝜓𝑚 ⟩ = 0 for all 𝑚 ≠ 𝑛, this gives: ∑ | |2 ∑ ⇒ |Ψ(𝑥, 𝑡)| |𝐶𝑛 |2 |𝜓𝑛 |2 = |𝐶𝑛 |2 = 1 | | , which is what we wanted to prove. Exercise 7 The given function can also be written as: (
𝜋𝑖 2 𝜓 + 𝑒 2 𝜓2 𝑖 1 = 𝑁(−2𝑖𝜓1 − 𝑖𝜓2 )
𝜓(𝑥, 0) = 𝑁
)
The normalizing condition: |𝜓|2 = ⟨𝜓|𝜓⟩ = 1 This means, 𝑁 2 (4 + 1) = 1; Furthermore, we have:
{
1 𝑁=√ 5
𝐻𝜓1 = 𝐸1 𝜓1 𝐻𝜓2 = 𝐸2 𝜓2
Applying the result deduced in the past exercise, we know: 𝑃1 = |𝐶1 |2 =
4 5
𝑃2 = |𝐶2 |2 =
1 5 Page 14
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Therefore, ⟨𝐻⟩ = Exercise 8 We have
{ 𝑉 (𝑥) =
4 1 𝐸 + 𝐸 5 1 5 2
0 ∞
0<𝑥<𝐿 elsewhere
Re- writing equation (8) from the subject, we get: 𝐸𝜓 = 𝑉 (𝑥)𝜓 −
ℏ2 𝑑 2 𝜓 2𝑚 𝑑𝑥2
in the Ist case: for 𝑉 (𝑥) = ∞ ⇒ It results from Schrödinger’s equation that Ψ is constant with 𝑥, and therefore equal to the value it has at the boundaries: 𝜓(0) = 𝜓(𝐿) = 0 in the IInd case: for 𝑉 (𝑥) = 0 ⇒ 𝐸𝜓 = −
ℏ2 𝑑 2 𝜓 2𝑚 𝑑𝑥2
𝑑2𝜓 2𝑚𝐸 =− 𝜓 𝑑𝑥2 ℏ2 2𝑚𝐸 = 𝜔2 → simple harmonic oscillation equation ℏ2 The general solution for this equation: 𝜓(𝑥) = 𝐴 sin 𝜔𝑥 + 𝐵 cos 𝜔𝑥 From the continuity of 𝜓(𝑥) we get the following boundary conditions: 𝜓(0) = 𝜓(𝐿) = 0 𝜓(0) = 𝐴 sin 0 + 𝐵 cos 0 = 0 ⇒ 𝐵 = 0 𝜓(𝐿) = 𝐴 sin 𝜔𝐿 = 0
⇒ 𝜔𝐿 = 𝜋𝑛 ; 𝑛 ∈ Z Since for 𝑛 = 0 ⇒ 𝜓(𝑥) = 0, we will discard this solution. For 𝑛 < 0, we don’t get anything new since sin 𝜃 = − sin (−𝜃), and the sign is absorbed by 𝐴. ⇒ 𝜔𝑛 = 𝐸𝑛 =
ℏ2 𝜔2𝑛 2𝑚
=
𝑛𝜋 𝐿 𝑛2 𝜋 2 ℏ2 2𝑚𝐿2 Page 15
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By normalizing 𝜓(𝑥) ⇒ 𝐿
1=
|𝐴|2 sin2 𝑘𝑥𝑑𝑥 =
∫0 𝐿
1 − cos 2𝑘𝑥 𝑑𝑥 |𝐴|2 2 ( ) |𝐿 𝐿 2 𝐿 | = |𝐴| − sin 2𝑘𝑥| 2 4 |0 2𝐿 = |𝐴| 2 √ 2 𝐴= 𝐿 =
∫0
Finally, we can write:
√ 𝜓𝑛 (𝑥) =
( ) 2 𝑛𝜋 sin 𝑥 𝐿 𝐿
Exercise 9 For extra fun (just kidding, it was by mistake), we first considered Ψ(𝑥, 0) = ℵ sin 𝜋𝑥 . We eventually read 2 the question correctly, but we just didn’t want to cut an entire page. However, you can skip reading this part if not interested, although we have a nice parallel made at the end between the two forms of the function. We normalize the function Ψ(𝑥, 0) = ℵ sin 𝜋𝑥 as follows: 2 𝐿 |2 𝜋𝑥 |Ψ(𝑥, 0)| = ℵ2 sin2 𝑑𝑥 = 1 | | ∫ 𝐿 | | 0
+∞ |
∫−∞ This gives us:
√ ℵ=
2 𝐿
And therefore, Ψ(𝑥, 𝑡) = ℵ sin We remember that we once wrote 𝜙(𝑡) = 𝑒−
𝑖𝐸𝑡 ℏ
𝜋𝑥 𝜙(𝑡) 𝐿
. This further implies:
√ ⎧ 2 − 𝑖𝑡ℏ 𝐸 𝜋𝑥 ⎪ Ψ(𝑥, 𝑡) = ⋅𝑒 sin ⎪ 𝐿 2 ⎨ 2 2 ⎪ ℏ 𝜋 ⎪where𝐸 = 2 2𝑚𝐿 ⎩
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We now compute the expectation value of x: |2 𝜋𝑥 || 2 | 𝑑𝑥 sin2 𝜙(𝑡) | ∫0 𝐿 𝐿 || | [ ] 𝐿 2 2 2|𝜙| 𝐿 2𝜋𝑥 − 𝑥 cos 𝑑𝑥 = ∫0 𝐿 2 𝐿 𝐿
⟨𝑥⟩ =
𝑥
𝐿
=
2𝜋𝑥 𝐿 1 − 𝑥 cos 𝑑𝑥 2 2 ∫0 𝐿
Changing the variable to 𝑦=
2𝜋𝑥 𝐿
Our integral becomes 2𝜋 |2𝜋 | sin 𝑦𝑑𝑦 𝑦 cos 𝑦𝑑𝑦 = 𝑦 sin 𝑦| − = | ∫0 ∫0 |0 |2𝜋 | = − cos 𝑦| | |0 =0 2𝜋
Finally, this gives us ⇒ ⟨𝑥⟩ =
𝐿 2
Next, we want to compute the expectation value for the energy: ∑ |𝐶𝑛 |2 𝐸𝑛 ⟨𝐸⟩ = 𝑛
Where, 𝐸𝑛 =
𝜋 2 ℏ2 2 𝑛 2𝑚𝐿2
Therefore, ∑| |2 |𝜓𝑛 (𝑥, 𝑡)| 𝐸𝑛 | | | 𝑛 | ( ) ∑2 𝑛𝜋𝑥 𝜋 2 ℏ2 2 sin2 𝑛 = 𝐿 𝐿 2𝑚𝐿2 𝑛
⟨𝐸⟩ =
𝐿 |2 2 𝜋𝑥 𝜋 2 ℏ2 |𝜓𝑛 (𝑥, 𝑡)| 𝐸𝑑𝑥 = sin2 𝑑𝑥 | | ∫−∞ | ∫0 𝐿 𝐿 2𝑚𝐿2 | 2𝜋𝑥 𝐿 1 − cos 𝜋 2 ℏ2 𝐿 = 𝑑𝑥 2 𝑚𝐿3 ∫0
=
+∞ |
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⇒ ⟨𝐸⟩ = CORRECT VERSION WITH Ψ(𝑥, 0) = ℵ sin3 We normalize the function as follows:
𝜋 2 ℏ2 2𝑚𝐿2
𝜋𝑥 . 2
+∞ |
∫−∞ We make the notation 𝜃 =
𝜋𝑥 𝐿
𝐿 |2 𝜋𝑥 |Ψ(𝑥, 0)| 𝑑𝑥 = ℵ2 𝑑𝑥 = 1 sin6 | | ∫0 𝐿 | |
and inputting the integral in Wolfram Alpha we get,
( )𝜋 | 1 1=ℵ 60𝜃 − 45 sin 2𝜃 + 9 sin 4𝜃 − sin 6𝜃 || 𝜋 192 |0 2𝐿
1 = ℵ2
𝐿 5𝜋 𝜋 16
Taking the positive solution, 4 ⇒ ℵ= √ 5𝐿 Now let’s do the necessary trigonometry: sin 3𝑥 = sin 2𝑥 cos 𝑥 + sin 𝑥 cos 2𝑥 = 2 sin 𝑥 cos2 𝑥 + (1 − 2 sin2 𝑥) sin 𝑥 = 2 sin 𝑥(1 − sin2 𝑥) + sin 𝑥 − 2 sin3 𝑥 = 3 sin 𝑥 − 4 sin3 𝑥
Therefore, sin3 𝑥 =
3 1 sin 𝑥 − sin 3𝑥 4 4
Now we write the equivalent form ( ) 3 𝜋𝑥 1 𝜋𝑥 𝜋𝑥 Ψ(𝑥, 0) = ℵ sin =ℵ sin − sin 3 𝐿 4 𝐿 4 𝐿 3
, where we take the positive solution 4 ℵ= √ 5𝐿 However, we can also write Ψ(𝑥, 𝑡) =
∑
𝑐𝑛 𝜓𝑛 (𝑥)𝑒−𝑖
𝐸𝑛 𝑡 ℏ
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, where 𝜓𝑛 takes the form found earlier in the exercises 𝜓𝑛 = Ψ(𝑥, 0) =
√
2 𝐿
sin
(
𝑛𝜋𝑥 𝐿
)
and the energy is 𝐸𝑛 =
𝜋 2 ℏ2 2 𝑛 . 2𝑚𝐿2
∑
𝑐𝑛 𝜓𝑛 (𝑥) ( ) 4 3 𝜋𝑥 1 𝜋𝑥 =√ sin − sin 3 𝐿 4 𝐿 5𝐿 4
Making the analogies,
√ ⎧ 2 3 3 =√ ⇒ 𝐶1 = √ ⎪𝐶1 𝐿 ⎪ 5𝐿 10 ⎨ √ 2 1 1 ⎪ ⇒ 𝐶3 = − √ ⎪𝐶3 𝐿 = − √ ⎩ 5𝐿 10
We note that 𝐶𝑛 = 0 for all 𝑛 ≠ 1, 3. We conclude that we have, ( ) ( ) 9𝜋2 ℏ 𝜋2 ℏ 𝑡 𝑡 3 𝜋𝑥 −𝑖 2𝑚𝐿 1 3𝜋𝑥 −𝑖 2𝑚𝐿 2 − 2 Ψ(𝑥, 𝑡) = √ sin 𝑒 𝑒 √ sin 𝐿 𝐿 10 10 Now, let’s compute the expectation value for x. By definition, 𝐿
⟨𝑥⟩ = For simplicity, we will denote Then, our function becomes:
𝜋2 ℏ 2𝑚𝐿2
∫0
|2 | 𝑥||Ψ(𝑥, 𝑡)|| 𝑑𝑥 | |
= 𝜏1 .
3 𝜋𝑥 − 𝑖𝑡𝜏 1 3𝜋𝑥 − 9𝑖𝑡𝜏 Ψ(𝑥, 𝑡) = √ sin 𝑒 − √ sin 𝑒 𝐿 𝐿 10 10 𝑖𝑡 [ ] 𝜋𝑥 𝑒− 𝜏 3𝜋𝑥 − 8𝑖𝑡𝜏 3 sin =√ − sin ⋅𝑒 𝐿 𝐿 10
Let us now look at the following ( ) |2 | |2 | |Ψ(𝑥, 𝑡)| = 1 |3 sin 𝜋𝑥 − sin 3𝜋𝑥 cos 8𝑡 − 𝑖 sin 8𝑡 | | | | 10 | 𝐿 𝐿 𝜏 𝜏 || | | [( ) ( ) ] 1 𝜋𝑥 3𝜋𝑥 8𝑡 2 3𝜋𝑥 8𝑡 2 = 3 sin − sin cos + sin sin 10 𝐿 𝐿 𝜏 𝐿 𝜏 [ ] 1 𝜋𝑥 3𝜋𝑥 𝜋𝑥 3𝜋𝑥 8𝑡 = 9 sin2 + sin2 − 6 sin sin cos 10 𝐿 𝐿 𝐿 𝐿 𝜏 We substitute sin
[ ] 𝜋𝑥 3𝜋𝑥 1 2𝜋𝑥 2𝜋𝑥 sin = cos − cos 𝐿 𝐿 2 𝐿 𝐿
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| |2 We treat each term in ||Ψ(𝑥, 𝑡)|| separately. | | We integrate the first part:
∫0
1 − cos 𝜋𝑥 𝑑𝑥 = 𝑥 ∫0 𝐿 2 𝐿
𝐿
𝑥 sin2
2𝜋𝑥 𝐿
𝑑𝑥
𝐿
=
𝐿 1 2𝜋𝑥 − cos 2 2 ∫0 𝐿
Again, separately, 𝐿
∫0
cos
( ) 𝐿 2𝜋𝑥 𝐿 2𝜋𝑥 𝑑𝑥 = 𝑥𝑑 sin ∫0 𝐿 𝐿 2𝜋 [ ] 𝐿 𝐿 2𝜋𝑥 ||𝐿 2𝜋𝑥 = 𝑥 sin − sin 𝑑𝑥 2𝜋 𝐿 ||0 ∫0 𝐿 𝐿 2 𝐿 2𝜋𝑥 || = cos 2 𝐿 ||0 4𝜋 =0
Therefore, 𝐿
𝜋𝑥 9 𝐿 9 𝑥 sin2 𝑑𝑥 = 10 𝐿 10 2
𝐿
1 1 𝐿 3𝜋𝑥 sin2 𝑑𝑥 = 10 𝐿 10 2
∫0 Similarly, for ∫0 Our next integral is
𝐿
∫0
cos
2𝜋𝑥 𝑑𝑥 𝐿
cos
4𝜋𝑥 𝑑𝑥 𝐿
, which we proved above that is equal to 0. And the final integral is 𝐿
∫0 , which also gives 0. This leaves us with 𝐿
⟨𝑥⟩ =
9𝑥 2 𝜋𝑥 sin 𝑑𝑥 + ∫0 10 𝐿
∫0 𝐿 = (9 + 1) 20 𝐿 = 2
𝐿
𝑥 3𝜋𝑥 sin2 𝑑𝑥 10 𝐿
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We observe that it is independent of time and (fun fact) actually has the same value we obtained when we . assumed Ψ(𝑥, 0) = ℵ sin 𝜋𝑥 2 Now, let’s compute the expectation value for E. Also by definition, and using 𝐸𝑛 = ⟨𝐸⟩ =
∑ ∑
𝜋 2 ℏ2 2 𝑛 , 2𝑚𝐿2
we get
⟨Ψ𝑛 |𝐸|Ψ𝑛 ⟩
|𝐶𝑛 |2 𝐸𝑛 )2 ( 1 9ℏ2 𝜋 2 ℏ2 𝜋 2 3 + = √ 2𝑚𝐿2 10 2𝑚𝐿2 10 =
=
9 ℏ2 𝜋 2 10 𝑚𝐿2
We see that ⟨𝐸⟩ is independent of time. Exercise and Question 10 For 𝑉 = 0, Schrödinger’s equation gives us 𝑖ℏ
𝜕𝜓 ℏ2 𝜕 2 𝜓 =− 𝜕𝑡 2𝑚 𝜕𝑥2
For free particles, the energy is 𝐸 = ℏ𝜔. This means: 𝜙(𝑡) = 𝑒−
𝑖𝐸𝑡 ℏ
= 𝑒−𝑖𝜔𝑡
𝜕2𝜓 2𝑚𝐸 =− 𝜓 = −𝑘2 𝜓 𝜕𝑥2 ℏ2 It follows immediately that 𝜓(𝑥) = 𝑒𝑖𝑘𝑥 and because Ψ(𝑥, 𝑡) = 𝜓(𝑥)𝜙(𝑡), we have thus came to deduce the fundamental equation of a wave: Ψ(𝑥, 𝑡) = 𝑒𝑖(𝑘𝑥−𝜔𝑡) This function is obviously not square- integrable and therefore cannot be normalized in the sense of equation (2) from the subjects. This means we made a ’hidden assumption’ that in this case is not respected. We indirectly supposed that it is possible for a function 𝜓 to be square integrable but to be unbounded! However, this property can only be true if 𝑙𝑖𝑚𝑥→∞ 𝜓(𝑥) = 0, which in the case of a wave does not verify. If a function 𝜓 is square integrable but unbounded, one can think of the following example: Let’s say we have a rectangle with height 2𝑘 and base of width 213𝑘 .
1 1 Let f be a function whose graph is a constant function of height 2𝑘 for 𝑥 ∈ [𝑘− 23𝑘+1 , 𝑘+ 23𝑘+1 ] for all positive 2 2𝑘 integers k and zero elsewhere. Then 𝑓 has height 2 over all of those intervals which means that the area enclosed by each rectangle (tall, but thin rectangles) is 21𝑘 . Hence ∞ 1 1 1 1 𝑓 2 𝑑𝑥 = + + ... 𝑘 + .... = −1=1 ∫−∞ 2 4 2 1− 1 2
Therefore, f is certainly square integrable but is unbounded. However, at infinity it approaches 0. Page 21
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2.3
PUEC 2018
The Harmonic Oscillator
Just writing the main formulas we will use in our further problems: 𝑉 (𝑥) =
1 𝑚𝜔2 𝑥2 2
1 ℏ2 𝑑 2 𝜓 𝑚𝜔2 𝑥2 − 2 2𝑚 𝑑𝑥2 1 𝑎± = √ (∓𝑖𝑝 + 𝑚𝜔𝑥) 2ℏ𝑚𝜔 [ ] 1 𝑝2 + (𝑚𝜔𝑥)2 − 𝑖𝑚𝜔(𝑥𝑝 − 𝑝𝑥) 𝑎− 𝑎+ = 2ℏ𝑚𝜔 [ ] 𝑂̂ 1 , 𝑂̂ 2 = 𝑂̂ 1 𝑂̂ 2 − 𝑂̂ 2 𝑂̂ 1 𝐸𝜓 =
Exercise 1 Applying the definitions of operators straightforward: ( ) d𝑓 (𝑥) d (𝑥̂ 𝑝̂ − 𝑝̂𝑥)𝑓 ̂ (𝑥) = −𝑖ℏ 𝑥 − (𝑥𝑓 (𝑥)) d𝑥 d𝑥 = 𝑖ℏ ⋅ 𝑓 (𝑥) [ ] ⇒ 𝑥, 𝑝 = 𝑖ℏ Exercise 2 Substituting in 𝑎+ and 𝑎− : 𝑎+ 𝑎− = =
1 (−𝑖𝑝 + 𝑚𝜔𝑥)(𝑖𝑝 + 𝑚𝜔𝑥) 2ℏ𝑚𝜔
1 (𝑝2 + 𝑚2 𝜔2 𝑥2 + 𝑖𝑚𝜔(𝑥𝑝 − 𝑝𝑥)) 2ℏ𝑚𝜔
By defitinion of the commutator [𝑎− , 𝑎+ ] = 𝑎− 𝑎+ − 𝑎+ 𝑎− [ ] 1 𝑖 = 2(𝑥𝑝 − 𝑝𝑥) ⋅ 𝑖𝑚𝜔 = − ⋅ 𝑥, 𝑝 2ℏ𝑚𝜔 ℏ ℏ = =1 ℏ We can also write it as 𝑎− 𝑎+ = 𝑎+ 𝑎− + 1 Knowing 𝐻 = ℏ𝜔(𝑎− 𝑎+ ) −
1 2
𝐻 = ℏ𝜔(𝑎+ 𝑎− ) +
1 2
We get:
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Exercise 3 To start with, we have: 1 ℏ𝜔(𝑎± 𝑎∓ ± )𝜓 = 𝐸𝜓 2 We now apply the property of the hermitian with th eigenfunction 𝑎+ 𝜓 and do the computations to verify the validty of equation (27) from the subjects: 𝐻(𝑎+ 𝜓) = (𝐸 + ℏ𝜔)(𝑎+ 𝜓) 𝐻(𝑎+ 𝜓) − 𝐸(𝑎 + 𝜓) = ℏ𝜔(𝑎+ 𝜓) 𝐻𝑎+ 𝜓 − 𝑎+ (𝐸𝜓) = ℏ𝜔(𝑎+ 𝜓) Substituting H with both the forms found above, ) ( ) ( 1 1 𝑎 𝜓 − ℏ𝜔𝑎+ 𝑎− 𝑎+ − 𝜓 = ℏ𝜔(𝑎+ 𝜓) ℏ𝜔 𝑎+ 𝑎− + 2 + 2 ℏ𝜔𝑎+ 𝜓 = ℏ𝜔(𝑎+ 𝜓) TRUE ⇒ the equation verifies This means, we now know, 𝐻(𝑎+ 𝜓) = (𝐸 + ℏ𝜔)(𝑎+ 𝜓) Similarly, we also have 𝐻(𝑎− 𝜓) = (𝐸 − ℏ𝜔)(𝑎− 𝜓) Exercise 4 aka Question In classical mechanics, where energies occupy a continuous spectrum of values, the ground state, that is the state with the smallest energy, obviously has value 0. However, in quantum mechanics, energies can no longer take ( ) every real value, but instead occupy a discrete spectrum of values described by equation ⇒ 𝐸𝑛 =
𝑛+
1 2
ℏ𝜔, where n is a natural number. It is then easy
to see that the smallest possible value, ’the ground state’ energy, is 𝐸0 = 12 ℏ𝜔. Exercise 5 First, we prove that: 𝑖) [𝑎+ 𝑎− , 𝑎+ ] = 𝑎+ 𝑎− 𝑎+ − 𝑎2+ 𝑎− = 𝑎+ (𝑎− 𝑎+ − 𝑎+ 𝑎− ) = 𝑎+
𝑖𝑖) [𝑎+ 𝑎− , 𝑎− ] = 𝑎+ 𝑎2− − 𝑎− 𝑎+ 𝑎− = (𝑎+ 𝑎− − 𝑎− 𝑎+ )𝑎− = −𝑎−
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We also recall that [𝑎− , 𝑎+ ] = 1 Now the commutation properly yields: ( ) (𝑎+ 𝑎− )𝑎+ 𝜓𝑛 = [𝑎+ 𝑎− , 𝑎+ ] + 𝑎+ (𝑎+ 𝑎− ) 𝜓𝑛 = 𝑎+ (1 + 𝑎+ 𝑎− )𝜓𝑛 = 𝑎+ (1 + 𝑛)𝜓𝑛
Similarly, we have (𝑎+ 𝑎− )𝑎− 𝜓𝑛 = 𝑎− (𝑛 − 1)𝜓𝑛 Now, we recall the result from the previous exercise { (𝑎+ 𝑎− )𝜓𝑛 = 𝑛𝜓𝑛 (𝑎− 𝑎+ )𝜓𝑛 = (𝑛 + 1)𝜓𝑛 In vector form, ⟨𝜓𝑛 |𝑎+ 𝑎− |𝜓𝑛 ⟩ = 𝑛 (𝑎− 𝜓𝑛 )† 𝑎− |𝜓𝑛 ⟩ = 𝑛 This means that ⟨𝜓𝑛 |𝑎− 𝑎+ |𝜓𝑛 ⟩ = ⟨𝜓𝑛 |(1 + 𝑎+ 𝑎− )|𝜓𝑛 ⟩ = 𝑛 + 1 Making use of 𝑎± ’s hermitian nature and expanding in the right side of the equality, (𝑎+ 𝜓𝑛 )† 𝑎+ |𝜓𝑛 ⟩ = (𝑛 + 1) ⟨𝜓𝑛+1 |𝜓𝑛+1 ⟩ (𝑎+ |𝜓𝑛 ⟩)2 = (𝑛 + 1)(|𝜓𝑛+1 ⟩)2 √ 𝑎+ |𝜓𝑛 ⟩ = 𝑛 + 1 |𝜓𝑛+1 ⟩ By induction from 𝑛 = 0 to any 𝑛 ∈ N 𝑎𝑛+ |𝜓𝑛 ⟩ = √ |𝜓0 ⟩ 𝑛! Exercise 6 First, we write
+∞
⟨𝑥4 ⟩ = , where
∫−∞
𝑥4 |𝜓𝑛 |2 𝑑𝑥
𝑎𝑛+ ( 𝑚𝜔 ) 14 − 𝑚𝜔 𝑥2 𝜓𝑛 = √ 𝑒 2ℏ 𝑛! 𝜋ℏ
From the way we defined 𝑎± , it results that 1
𝑎+ + 𝑎− = √ 2𝑚𝜔𝑥 = 2ℏ𝑚𝜔
√
2𝑚𝜔 𝑥 ℏ Page 24
ShutUpAndLovePhysics
PUEC 2018 √
ℏ (𝑎 + 𝑎− ) 2𝑚𝜔 + ( ) ℏ 2 𝑥4 = (𝑎+ + 𝑎− )4 2𝑚𝜔 This means, we now look to solve what might seem easy at first, but it definitely isn’t. 𝑥=
⟨𝑥4 ⟩ =
(
ℏ 2𝑚𝜔
)2
+∞
∫−∞
𝜓𝑛∗ (𝑎+ + 𝑎− )4 𝜓𝑛 𝑑𝑥
We further use the following formula, in which 𝛿𝑚𝑛 is the Kronecker delta function defined at page 13: +∞
∫−∞
𝜓𝑛∗ (𝑥)𝜓𝑚 (𝑥)𝑑𝑥 = 𝛿𝑚𝑛
Just as an example: √
] [ +∞ +∞ ℏ ∗ ∗ = ⟨𝑥⟩ = 𝜓 𝑎 𝜓 𝑑𝑥 𝜓 𝑎 𝜓 𝑑𝑥 + ∫−∞ ∫−∞ 𝑛 − 𝑛 2𝑚𝜔 ∫−∞ 𝑛 + 𝑛 √ ] [ +∞ +∞ ℏ ∗ = 𝜓𝑛+1 𝜓𝑛 𝑑𝑥 𝜓𝑛∗ 𝜓𝑛+1 𝑑𝑥 + ∫−∞ 2𝑚𝜔 ∫−∞ +∞
𝜓𝑛∗ 𝑥𝜓𝑛 𝑑𝑥
Because of the orthogonality for different coefficients, we conclude that ⟨𝑥⟩ = 0 Now, let’s solve our exercise, (𝑎+ + 𝑎− )4 = (𝑎+ + 𝑎− )(𝑎+ + 𝑎− )(𝑎+ + 𝑎− )2 = (𝑎2+ + 𝑎2− + 𝑎+ 𝑎− + 𝑎2− 𝑎+ )(𝑎+ + 𝑎− )(𝑎+ + 𝑎− ) = (𝑎3+ + 𝑎2+ 𝑎− + 𝑎2− 𝑎+ + 𝑎3− + 𝑎+ 𝑎− 𝑎+ + 𝑎2+ 𝑎2− + 𝑎− 𝑎2+ + 𝑎− 𝑎+ 𝑎− )(𝑎+ + 𝑎− ) = 𝑎4+ + 𝑎2+ 𝑎− 𝑎+ + 𝑎2− 𝑎2+ + 𝑎3− 𝑎+ + 𝑎+ 𝑎− 𝑎2+ + 𝑎+ 𝑎2− 𝑎+ + 𝑎− 𝑎3+ + 𝑎− 𝑎+ 𝑎− 𝑎+ + 𝑎− 𝑎+ 𝑎2− + 𝑎− 𝑎2+ 𝑎− + 𝑎+ 𝑎3− + 𝑎4− + + 𝑎2− 𝑎+ 𝑎− + 𝑎2+ 𝑎2− + 𝑎3+ 𝑎−
Some simplifications can be done by grouping terms as follows and eliminating the terms that are 0 because of orthogonality: ⎧𝑎2 𝑎− 𝑎+ = 𝑎2 (1 + 𝑎+ 𝑎− ) + ⎪ + ⎪𝑎+ 𝑎− 𝑎+ = (𝑎− 𝑎+ − 1)𝑎2+ ⎪ ⎪𝑎+ 𝑎2− 𝑎+ = (𝑎− 𝑎+ − 1)𝑎− 𝑎+ ⎨ 2 ⎪𝑎− 𝑎+ 𝑎− = 𝑎− 𝑎+ (𝑎− 𝑎+ − 1) = (𝑎+ 𝑎− + 1)𝑎+ 𝑎− ⎪ 2 ⎪𝑎− 𝑎+ 𝑎− 2 = (𝑎+ 𝑎− + 1)𝑎− ⎪𝑎2 𝑎 𝑎 = 𝑎2 (𝑎 𝑎 − 1) ⎩ − + − − − + Page 25
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PUEC 2018
Introducing these in our integral, we get +∞
𝜓𝑛∗ (𝑎2− 𝑎2+ + 𝑎2+ 𝑎2− + 2𝑎− 𝑎+ 𝑎− 𝑎+ + 2𝑎+ 𝑎− 𝑎+ 𝑎− − 1)𝜓𝑛 𝑑𝑥
∫−∞ Some more simplifications,
⎧ +∞ +∞ 2 ⎪ 𝜓𝑛∗ 𝑎+ 𝑎− 𝜓𝑛 𝑑𝑥 = 𝑛𝜓𝑛+1 𝑑𝑥 = 𝑛 ∫−∞ ⎪∫−∞ ⎨ +∞ +∞ 2 ⎪ 𝜓𝑛∗ 𝑎− 𝑎+ 𝜓𝑛 𝑑𝑥 = (𝑛 + 1)𝜓𝑛+1 𝑑𝑥 = 𝑛 + 1 ⎪∫−∞ ∫−∞ ⎩ Looking at every term in our integral, we have, +∞
+∞
⎧ 2 𝜓𝑛∗ 𝑎2− 𝑎2+ 𝜓𝑛 𝑑𝑥 = (𝑛 + 1)(𝑛 + 2)𝜓𝑛+2 𝑑𝑥 = (𝑛 + 1)(𝑛 + 2) ⎪∫ ∫ −∞ −∞ ⎪ +∞ ⎪ 2 (𝑛 + 1)(𝑛 + 2)𝜓𝑛+2 𝑑𝑥 = (𝑛 + 1)(𝑛 + 2) ⎪∫ ⎪ −∞ +∞ ⎪ +∞ ∗ 2 2 𝑛(𝑛 − 1)𝜓𝑛−2 𝑑𝑥 = 𝑛(𝑛 − 1) 𝜓𝑛 𝑎+ 𝑎− 𝜓𝑛 𝑑𝑥 = ⎨∫ ∫−∞ ⎪ −∞ +∞ +∞ ⎪ +∞ ∗ ∗ 𝜓𝑛 𝑎− 𝑎+ 𝑎− 𝑎+ 𝜓𝑛 𝑑𝑥 = 𝜓𝑛+1 𝑎+ 𝑎− 𝜓𝑛+1 𝑑𝑥(𝑛 + 1) == 𝜓 2 (𝑛 + 1)2 𝑑𝑥 = (𝑛 + 1)2 ⎪ ∫−∞ ∫−∞ 𝑛 ⎪∫−∞ +∞ +∞ ⎪ +∞ ∗ ⎪ 𝜓𝑛∗ 𝑎+ 𝑎− 𝑎+ 𝑎− 𝜓𝑛 𝑑𝑥 = 𝜓𝑛+1 𝑎− 𝑎+ 𝜓𝑛−1 𝑑𝑥𝑛 = 𝑛2 𝜓 2 𝑑𝑥 = 𝑛2 ∫−∞ ∫−∞ 𝑛 ⎩∫−∞ The big moment has come for us to input all discussed above (
) ] ℏ 2[ (𝑛 + 1)(𝑛 + 2) + (𝑛 − 1)𝑛 + 2𝑛2 + 2(𝑛 + 1)2 − 1 2𝑚𝜔 ( ) ) ℏ 2( 2 𝑛 + 3𝑛 + 2 + 𝑛2 − 𝑛 + 2𝑛2 + 2𝑛2 + 4𝑛 + 2 − 1 = 2𝑚𝜔 ) ( ) ℏ 2( 2 6𝑛 + 6𝑛 + 3 = 2𝑚𝜔 ( 2 ) ( ) ℏ 2 3 2𝑛 + 2𝑛 + 1 = ⋅ 2𝑚𝜔 4
⟨𝑥4 ⟩ =
In a final form, ⟨𝑥4 ⟩ =
(
ℏ 2𝑚𝜔
)2
) 3( 2 𝑛 + (𝑛 + 1)2 4
In the particular case in which 𝑛 = 75, we compute that ⟨𝑥4 ⟩ =
(
ℏ 2𝑚𝜔
)2
( ) 34203 ℏ 2 ≃ 2137, 69 4 𝑚𝜔 Page 26
ShutUpAndLovePhysics
PUEC 2018
Exercise 7 We start with analysing the potential energy. +∞
⟨𝑈𝑛 ⟩ =
∫−∞
𝜓𝑛∗
(
) 𝑘𝑥2 𝜓𝑛 𝑑𝑥 2
+∞
=
𝑘 𝜓 ∗ (𝑎2 + 𝑎2− + 𝑎+ 𝑎− + 𝑎− 𝑎+ )𝜓𝑛 𝑑𝑥 2 ∫−∞ 𝑛 + +∞
𝑘 𝜓 ∗ (𝑎 𝑎 + 𝑎− 𝑎+ )𝜓𝑛 𝑑𝑥 2 ∫−∞ 𝑛 + − [ +∞ ] +∞ 𝑘 ℏ 2 2 = (𝑛 + 1)𝜓𝑛+1 𝑑𝑥 + 𝑛𝜓𝑛−1 𝑑𝑥 ∫ ∫ 2 −∞ 2𝑚𝜔 −∞
=
Just as for a classical oscillator, we have 𝑘 = 𝜔2 𝑚 This gives us, ⟨𝑈𝑛 ⟩ =
𝐸 ℏ𝜔 1 ℏ𝜔 (2𝑛 + 1) = (𝑛 + ) = 𝑛 4 2 2 2
Next, we look at the kinetic energy. +∞
⟨𝐸𝑘,𝑛 ⟩ = , in which
∫−∞ √
𝑝=𝑖
𝜓𝑛∗
( 𝑝2 ) 𝜓 𝑑𝑥 2𝑚 𝑛
ℏ𝑚𝜔 (𝑎+ − 𝑎− ) 2
Similarly, +∞
ℏ𝜔 𝜓 ∗ (𝑎2 + 𝑎2− − 𝑎+ 𝑎− − 𝑎− 𝑎+ )𝜓𝑛 𝑑𝑥 4 ∫−∞ 𝑛 + 𝐸 ℏ𝜔 = + (2𝑛 + 1) = 𝑛 4 2
⟨𝐸𝑘,𝑛 ⟩ = −
We can observe that in a system with a total energy of 𝐸𝑛 , the energy splits equally between the potential and kinetic energy. In formulas, 𝐸 ⟨𝑈𝑛 ⟩ = ⟨𝐸𝑘,𝑛 ⟩ = 𝑛 2 The virial theorem states that for a potential energy of the form 𝑉 (𝑥) = 𝐶 ⋅ 𝑥𝑛 , where 𝐶 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. ⟨𝐸𝑘𝑖𝑛 ⟩ =
𝑛 ⟨𝑉 ⟩ 2 𝑛
It is therefore clear that our problem is a particular case for which 𝑛 = 2. ⟨𝐸𝑘𝑖𝑛 ⟩ =
𝑛 ⟨𝑉 ⟩ = ⟨𝑉𝑛 ⟩ 2 𝑛 Page 27
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2.4
PUEC 2018
The Commutator
We write the general formula 𝑑 ⟨𝑂⟩ 𝜕Ψ ̂ 𝜕 𝑂̂ 𝜕Ψ = ⟨ |𝑂Ψ⟩ + ⟨Ψ| Ψ⟩ + ⟨Ψ|𝑂̂ ⟩ 𝑑𝑡 𝜕𝑡 𝜕𝑡 𝜕𝑡 Exercise 8 aka Question The second equality in formula (34) is justified because we have proved in section 2.2. that H is Hermitian, ̂ = ⟨𝐻𝛼|𝛽⟩, ̂ ̂ which means that ⟨𝛼|𝐻𝛽⟩ where, in our case 𝛼 = Ψ and 𝛽 = 𝑂Ψ. Exercise 9 𝑑 and that [𝑥, 𝑝] = 𝑖ℏ. Firstly, we know 𝑝̂ = −𝑖ℏ 𝑑𝑥 If we take 𝑂̂ = 𝑥, we have from equation (34) in the subjects that: 𝑑 ⟨𝑥⟩ 𝑖 𝜕 𝑥̂ ̂ 𝑥]⟩ = ⟨[𝐻, ̂ +⟨ ⟩ 𝑑𝑡 ℏ 𝜕𝑡
(3)
And if we take that 𝑂̂ = 𝑝, we similarly get: 𝑑 ⟨𝑝⟩ 𝜕 𝑝̂ 𝑖 ̂ 𝑝]⟩ = ⟨[𝐻, ̂ +⟨ ⟩ 𝑑𝑡 ℏ 𝜕𝑡
(4)
, where 𝑝̂2 + 𝑉̂ (𝑥) 𝐻̂ = 2𝑚 𝜔2 𝑚 2 𝑘𝑥̂ 2 = 𝑥̂ 𝑉̂ = 2 2 For the sake of simplicity, we will give up the hat notation, but keep in mind they are operators. [𝐻, 𝑥] = 𝐻𝑥 − 𝑥𝐻 ( 2 ) ( 2 ) 𝑝 𝑝 = +𝑉 𝑥−𝑥 +𝑉 2𝑚 2𝑚 =
𝑝2 𝑥 − 𝑥𝑝2 2𝑚
We know that [𝑥, 𝑝] = 𝑥𝑝 − 𝑝𝑥 = 𝑖ℏ. This means, 𝑝 ⋅ (𝑝𝑥) − 𝑥𝑝2 = 2𝑚 𝑝(𝑥𝑝 − 𝑖ℏ) − (𝑝𝑥 + 𝑖ℏ)𝑝 = 2𝑚 𝑖ℏ𝑝 =− 𝑚
⇒ [𝐻, 𝑥] =
Substituting in equation (3) from above, we have 𝑑 ⟨𝑥⟩ 1 = ⟨𝑝⟩ 𝑑𝑡 𝑚 Page 28
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𝜕 , we now focus on Keeping in mind that 𝑝̂ = −𝑖ℏ 𝜕𝑥
[𝐻, 𝑝] = 𝐻𝑝 − 𝑝𝐻 ( 2 ) ( 2 ) 𝑝 𝑝 = +𝑉 𝑝−𝑝 +𝑉 2𝑚 2𝑚 = 𝑉 𝑝 − 𝑝𝑉
𝜕𝑓 𝑑 + 𝑖ℏ (𝑉 𝑓 ) 𝜕𝑥 𝑑𝑥 ( ) 𝑑𝑓 𝑑𝑓 𝑑𝑉 = −𝑖ℏ𝑉 + 𝑖ℏ 𝑓 + 𝑖ℏ𝑉 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑉 𝑓 = 𝑖ℏ 𝑑𝑥
⇒ [𝐻, 𝑝]𝑓 (𝑥) = −𝑖ℏ𝑉
We obtained that [𝐻, 𝑝] = 𝑖ℏ
𝑑𝑉 𝑑𝑥
This means that equation (4) from above becomes 𝑑 ⟨𝑝⟩ 𝜕 𝑝̂ 𝑑𝑉 = −⟨ ⟩+ 𝑑𝑡 𝑑𝑥 𝜕𝑡 𝜕 𝑝̂ 𝜕𝑡
Since operator 𝑝̂ is not time dependent ⇔
= 0, then we get
𝑑 ⟨𝑝⟩ 𝑑𝑉 = −⟨ ⟩ = − ⟨𝑉 ′ (𝑥)⟩ 𝑑𝑡 𝑑𝑥 Exercise 10 - the final one ( )2 By definition 𝜎02 = ⟨ 𝑂̂ − ⟨𝑂⟩ ⟩. We will denote: 𝐴 = 𝑂̂ 1 − ⟨𝑂1 ⟩ 𝐵 = 𝑂̂ 2 − ⟨𝑂2 ⟩ This means we have: 𝜎𝑂2 = ⟨Ψ|𝐴2 |Ψ⟩ = ⟨Ψ|𝐴𝐴Ψ⟩ 1
𝜎𝑂2 2
= ⟨Ψ|𝐵 2 |Ψ⟩ = ⟨Ψ|𝐵𝐵Ψ⟩
From the postulate at the bottom of page 6 on the subjects, and knowing 𝑂̂ 1,2 are observables, this also means they are hermitian.
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As a consequence, ( )2 𝜎𝑂2 = ⟨Ψ| 𝑂̂ 1 − ⟨𝑂1 ⟩ Ψ⟩ = 1 ) ( = ⟨Ψ| 𝑂̂ 1 𝑂̂ 1 − 𝑂̂ 1 ⟨𝑂1 ⟩ − ⟨𝑂1 ⟩ 𝑂̂ 1 + ⟨𝑂1 ⟩2 Ψ⟩ = = ⟨Ψ|𝑂̂ 1 𝑂̂ 1 Ψ⟩ − ⟨Ψ|𝑂̂ 1 ⟨𝑂1 ⟩ Ψ⟩ − ⟨Ψ| ⟨𝑂1 ⟩ 𝑂̂ 1 Ψ⟩ + ⟨Ψ|𝑂̂ 12 Ψ⟩ = = ⟨𝑂̂ 1 Ψ|𝑂̂ 1 Ψ⟩ − ⟨𝑂̂ 1 Ψ| ⟨𝑂1 ⟩ Ψ⟩ − ⟨⟨𝑂1 ⟩ Ψ|𝑂̂ 1 Ψ⟩ + ⟨⟨𝑂1 ⟩ Ψ|Ψ ⟨𝑂1 ⟩⟩ = ( ) ( ) = ⟨𝑂̂ 1 Ψ| 𝑂̂ 1 − ⟨𝑂1 ⟩ Ψ⟩ − ⟨⟨𝑂1 ⟩ Ψ| 𝑂̂ 1 − ⟨𝑂1 ⟩ Ψ⟩ = ( ) ( ) = ⟨ 𝑂̂ 1 − ⟨𝑂1 ⟩ Ψ| 𝑂̂ 1 − ⟨𝑂1 ⟩ Ψ⟩ = = ⟨𝐴Ψ|𝐴Ψ⟩ ( ) This means that adding a constant ⟨𝑂1 ⟩ does not change its hermitian nature. ⇔ 𝐴,𝐵 are also hermitians. Similarly, we get 𝜎𝑂2 = ⟨𝐵Ψ|𝐵Ψ⟩. We will now use the Cauchy Schwartz inequality, which states that for (∀) 𝑢, 𝑣 ∈ V; we have: |⟨𝑢, 𝑣⟩|2 ≤ ⟨𝑢, 𝑢⟩ ⋅ ⟨𝑣, 𝑣⟩ It clearly follows that in order to use it, we replace: { 𝑢 = 𝐴Ψ 𝑣 = 𝐵Ψ 𝜎𝑂2 𝜎𝑂2 = ⟨𝐴Ψ, 𝐴Ψ⟩ ⋅ ⟨𝐵Ψ, 𝐵Ψ⟩ ≥ 2
1
≥ |⟨𝐴Ψ, 𝐵Ψ⟩|2
Using the triangle inequality |𝑧|2 ≥ [𝐼𝑚(𝑧)]2 =
[
1 (𝑧 − 𝑧∗ ) 2𝑖
𝜎𝑂2 𝜎𝑂2 ≥ 1
2
[
]2
, where 𝑧 = ⟨𝐴Ψ, 𝐵Ψ⟩, we get:
]2 1 (𝑧 − 𝑧∗ ) 2𝑖
Because 𝐴 = hermitian⇒ 𝑧 = ⟨Ψ|𝐴𝐵|Ψ⟩. We now use a property we have used before at the second exercise from 2.2., while trying to prove that hermitian operators have real eigenvalues, that states for any |𝑥⟩, |𝑦⟩, 𝐴̂ ∈ V, we have ∗ ̂ ⟨𝑥|𝐴|𝑦⟩ = ⟨𝑦|𝐴̂ 𝑡 |𝑥⟩
Applying this in our case, we would have ⟨𝑥|𝐴𝐵|𝑦⟩ = ⟨𝑥|𝐴𝐵𝑦⟩ = ⟨𝐴𝑥|𝐵𝑦⟩ ∗
= ⟨𝑦|𝐵 𝑡 |𝐴𝑥⟩ = ⟨𝑦|𝐵𝐴|𝑥⟩∗ Page 30
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Because 𝐴, 𝐵 = hermitian ⇒ 𝐴 = 𝐴𝑡 and 𝐵 = 𝐵 𝑡 . Using all of the above, we get 𝑧∗ = ⟨Ψ|𝐴𝐵|Ψ⟩∗ = ⟨Ψ|𝐵𝐴|Ψ⟩ and that
]2 [ ]2 1 1 ⟨{Ψ|(𝐴𝐵 − 𝐵𝐴)|Ψ⟩ = ⟨{Ψ|[𝐴, 𝐵]|Ψ⟩ 1 2 2𝑖 2𝑖 All that is left to prove is that the commutator of A and B is the same with the one of 𝑂1 and 𝑂2 . ( )( ) ( )( ) [𝐴, 𝐵] = 𝑂̂ 1 − ⟨𝑂1 ⟩ 𝑂̂ 2 − ⟨𝑂2 ⟩ − 𝑂̂ 2 − ⟨𝑂2 ⟩ 𝑂̂ 1 − ⟨𝑂1 ⟩ = 𝑂̂ 1 𝑂̂ 2 − 𝑂̂ 2 𝑂̂ 1 + 𝑂̂ 2 ⟨𝑂1 ⟩ − ⟨𝑂1 ⟩ 𝑂̂ 2 − 𝑂̂ 1 ⟨𝑂2 ⟩ 𝜎𝑂2 𝜎𝑂2 ≤
[
+ ⟨𝑂1 ⟩ ⟨𝑂2 ⟩ − ⟨𝑂1 ⟩ ⟨𝑂2 ⟩ + 𝑂̂ 1 ⟨𝑂2 ⟩ = 𝑂̂ 1 𝑂̂ 2 − 𝑂̂ 2 𝑂̂ 1 [ ] = 𝑂̂ 1 𝑂̂ 2 here ⟨𝑂1 ⟩ 𝑂̂ 2 = 𝑂̂ 2 ⟨𝑂1 ⟩ and ⟨𝑂2 ⟩ 𝑂̂ 2 = 𝑂̂ 1 ⟨𝑂2 ⟩ because ⟨𝑂1 ⟩ , ⟨𝑂2 ⟩ are constants Finally, this means that 𝜎𝑂2 𝜎𝑂2 ≥ 1
3
2
[
]2 1 ̂ ̂ ⟨[𝑂1 𝑂2 ]⟩ 2𝑖
Experiment
Some of the most well known applications of quantum theory refer to wave nature of light. Although there can be constructed experiments as laborious as one may want, we will look at one the ’basic’ ones, but which describes very easily how quantum theory applies in every day life - the famous, or perhaps infamous, Double slit experiment, also known as Young’s double-slit interferometer. Double slit experiment Along our discussion, we will often compare the observed results with the ones yielded by the classical approach, in order to show how the experiment supports the quantum theory. Experimental set-up As mentioned above, the experiment set up is pretty trivial, but can lead to important results if done properly. This also means that it can be done at home, as long the interfering waves are coherent and the slits small enough. For starters we will use a long ruler or a measuring tape and align on the same direction a laser, a screen with two small holes in them and a screen in this order.
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Seeing that the alignment is done correctly - even though there can always be corrections made, we can measure the distance between the the 2 screens, and will next denote it by D. Let’s say that the laser has wavelength 𝜆. When we turn it on, we observe an interference pattern - alternating fringes of light and dark on the second screen (as can be seen in figure 2)- and can measure the distance between them simply with a ruler. We will write the periodical spacing between 2 consecutive maximums (or minimums) of intensity with 𝑖.
Figure 2: Picture of the entire set-up In geometric optics we would simply have 2 dots of light. As this is not the case, we will further investigate the physics behind the actual results. For starters, we will imagine having only one slit. In this case, the particle’s wave function that hits the barrier would be simply blocked or absorbed, while downstream of the barrier, the wave function has a packet-shaped profile. The evolution of the packet-shape, with position, downstream of the slit, is up to a certain degree, the same as the evolution of one-dimensional wave packets. We know that wave packets will spread in the transverse direction as they propagate downstream. This phenomenon is known by the name of diffraction, illustrated in the figure below.
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Figure 3: An incident wave passes through a slit and diffracts We consider the slit to act as a barrier, whose fraction of incident wave function is transmits rather than being absorbed, equal to a Gaussian. Then, the transverse profile of the beam just downstream of the barrier is also a Gaussian. Now let’s consider two slits. The two “Gaussian slits” are centered, say, at 𝑥 = 𝑎 and 𝑥 = −𝑎 . Then, the transverse profile of the wave function just behind the barrier will be given by Ψ(𝑥, 0) = Ψ𝐺 (𝑥 − 𝑎, 0) + Ψ𝐺 (𝑥 + 𝑎, 0) Formally, we can say that since the Schrödinger equation is linear in the solution (𝑥, 𝑡) for this initial state – a superposition of two Gaussians – is just the corresponding superposition of the solutions for the two superposed terms individually. Thus, we could write: 2 2 ⎡ − (𝑥−𝑎)𝑖ℏ𝑦 − (𝑥+𝑎)𝑖ℏ𝑦 ⎤ 2+ 2+ 4(𝜎 ) 4(𝜎 ) ⎢𝑒 2𝑚𝑣 𝑒 2𝑚𝑣 ⎥ Ψ(𝑥, 𝑦) ∼ √ ⎢ ⎥ 𝑖ℏ𝑦 𝜎 2 2𝑚𝑣 ⎣ ⎦
1
This is slightly messy to work with, but the idea qualitatively is that, as the two individual Gaussians begin to spread, they start to overlap. There can be either constructive or destructive interference depending on the relative phases in the region of overlap. For example, along the symmetry line, 𝑥 = 0, the phases of the two terms will always match and we will therefore always have constructive interference, corresponding to a large value of |Ψ|2 , i.e., a high probability for the particle (if looked for) to be detected. But if we move a little bit to the side (say, in the positive x-direction) we are moving toward the central peak of one of the Gaussians and away from the central peak of the other, and so the phases of the two terms change at different rates, and eventually we find a spot where there is (at least for large y, nearly complete) destructive interference, corresponding to |Ψ|2 = 0. Page 33
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Moving even farther in the positive x-direction eventually yields another spot where there is constructive interference, and so on. The intensity pattern that results is shown in Fig. 3.
Figure 4: Intensity of light aka density plot of |Ψ2 | done in Matlab Finally, some approximate mathematics would yield the final formula: 𝑖 = 𝐷𝜆∕2𝑎 Comments There a few major factors that could drastically influence the interference patterns and another minor ones. We’ll start with the major ones: 1)Distance between the screens. If the two screens are close enough that the wavelength is not negligible, then Fresnel’s theory for ondulatory optics applies instead of the classic Frauenhoffen’s. However, the result can still be expressed using quantum theory through Schrodinger’s equation, but it would yield a much complicated mathematical pattern. 2)Coherence. Even if it might seem trivial, coherence between the 2 incident light waves must be respected in order to obtain an interference patters. Some additional comments:
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- Note how the expression ’interference pattern’ that we used throughout our analysis refers to the superposition of 2 waves, and leads one to think about Fourier transform of there were more slits. - If the alignment is not precise, then results may encounter some errors. - Although we used in our experiment the photons from a laser, we could theoretically use any high energy particles. Electrons would be also be easy to use since their mass is small. An example of such an experiment is Tonomura et al. for a two-slit experiment with individual electrons. Motivation and Goals Even though our goal was to prove the existence of quantum theory, the experiment itself is used for other purposes. Using the relation deduced above involving the wave length 𝜆, distance between the screen the barrier D, and the pattern sizing i, one can measure one of them in terms of the others. The idea which ultimately led to the discovery of gravitational waves was based on a similar approach. They measured the distance D and calculated the small changes that occurred because of the gravitational waves.
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References [1] Time Evolution in Quantum Mechanics, https://ocw.mit.edu/courses/nuclear-engineering/22-02-introduction-to-appliednuclear-physics-spring-2012/lecture-notes/MIT22_02S12_lec_ch6.pdf [2] Operators in Quantum Mechanics, http://www.phys.spbu.ru/content/File/Library/studentlectures/schlippe/qm07-03.pdf [3] Probability, Expectation Values, and Uncertainties, http://physics.mq.edu.au/∼jcresser/Phys201/LectureNotes/ProbabilitiesExpectationValues.pdf [4] The basis of a Quantum harmonic oscillator, https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator [5] The basis of Hermite polynomials, https://en.wikipedia.org/wiki/Hermite_polynomials [6] Harmonic Oscillator Physics, https://www.reed.edu/physics/courses/P342.S10/Physics342/page1/files/Lecture.9.pdf [7] Quantum Mechanics: Commutation, https://www1.udel.edu/pchem/C444/spLectures2010/03252010b.pdf [8] Introduction to Quantum Mechanics - David J. Griffiths, http://www.fisica.net/mecanica-quantica/Griffiths%20 -%20Introduction%20to%20quantum%20mechanics.pdf [9] The mathematics of square integrable functions, https://en.wikipedia.org/wiki/Square-integrable_function
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