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7.4. THE HEISENBERG UNCERTAINTY PRINCIPLE

111

7.4The Heisenberg Uncertainty Principle

The more precisely the position is determined, the less precisely the momentum is known in this instant, and vice versa. Werner Heisenberg, 1927.

In §7.3 we proved the Heisenberg Uncertainty Principle for the special case of the harmonic oscillator. Here we show this is a general feature of quantum mechanics. First we recall some basic facts about complex vector spaces.

1.If Ψ and Φ are any two states in our Hilbert space of states H, we have an inner product defined (Ψ, Φ) that satisfies the properties

(a), Φ) = (Φ, Ψ) where z denotes the complex conjugate of z.

(b)(c1Ψ1 + c2Ψ2, Φ) = c1 1, Φ) + c2 2, Φ) for all states Ψ1, Ψ2 and all complex numbers c1, c2.

(c)The length or norm of the state Ψ is defined to be kΨk2 = (Ψ, Ψ) 0 with kΨk = 0 if and only if Ψ = 0, the zero vector in H.

2.An operator A is called Hermitian (or self-adjoint) if

(AΨ, Φ) = (Ψ, AΦ)

for all states Ψ, Φ. In quantum mechanics observables are assumed to be Hermitian. Note this makes the expected value of the observable A in state Ψ a real number

hAi := (AΨ, Ψ) = (Ψ, AΨ)

= (AΨ, Ψ)

= hAi.

Sometimes one writes hAiΨ to denote the state in which the expected value is computed.

3. Just as in linear algebra, we have the Cauchy-Schwarz inequality

 

|, Φ)|2 ≤ kΨk2 kΦk2

(7.28)

for all states Ψ, Φ H.

 

We now assume we have observables A and B that satisfy the commutation relation

AB BA = i id

(7.29)

where id is the identity operator and i is the imaginary number, i2 = 1. We showed earlier that in units where ~ = 1 the position and momentum operators satisfy such a commutation relation. For a given state Ψ and observable A we define21

r

q

DE

A = (A − hAi)2 = hA2i − hAi2 0.

21In classical probability A is called the standard deviation of A. The quantity A is a measure of the deviation of A from its expected value hAi.

112

CHAPTER 7. QUANTUM HARMONIC OSCILLATOR

We now prove that if observables A and B satisfy (7.29) then

 

 

1

 

 

 

A · B

 

.

(7.30)

 

2

Proof: Let Ψ denote any normalized state vector, i.e. kΨk = 1. Apply (7.29) to Ψ to obtain

ABΨ BAΨ = iΨ

 

Now take the inner product of each side with the state Ψ to obtain

 

(ABΨ, Ψ) (BAΨ, Ψ) = i , Ψ)

 

which simplifies to22

 

(BΨ, AΨ) (AΨ, BΨ) = i

(7.31)

Let t denote any real number, then by the Cauchy-Schwarz inequality (7.28)

 

|, AΨ + itBΨ)|2 ≤ kAΨ + itBΨk2

(7.32)

since kΨk = 1. Let’s simplify the left-hand side of (7.32)

 

, AΨ + itBΨ) = (Ψ, AΨ) it, BΨ) = hAi − ithBi.

 

The absolute value squared of this is

 

hAi2 + t2hBi2.

 

We now examine the right-hand side of (7.32)

 

kAΨ + itBΨk2 = (AΨ + itBΨ, AΨ + itBΨ)

 

= kAΨk2 + it {(BΨ, AΨ) (AΨ, BΨ)} + t2kBΨk2

 

= kAΨk2 t + t2kBΨk2 by use of (7.31).

(7.33)

Thus the inequality (7.32) becomes

 

hAi2 + t2hBi2 ≤ kAΨk2 t + t2kBΨk2.

Using the fact that kAΨk2 = (AΨ, AΨ) = (A2Ψ, Ψ) = hA2i (and similarly for B) and the definition of A (and similarly for B), the above inequality can be rewritten as

t2 B)2 t + (ΔA)2 0.

This holds for all real t. The above is a quadratic polynomial in t that is always nonnegative. This means that the discriminant of the quadratic polynomial must be nonpositive, i.e.

b2 4ac 0. That is,

1 4 (ΔA)2 B)2 0

which implies that

A · B 12

which is what we want to prove.

When A is the position operator and B is the momentum operator we get the Heisenberg

Uncertainty Principle which states

x · p ~2

where we have returned to physical units. The appearance of Planck’s constant ~ in the right hand side shows that ~ sets the scale for quantum mechanical phenomena.

22Note (ABΨ, Ψ) = (BΨ, AΨ) and (BAΨ, Ψ) = (AΨ, BΨ) since A and B are Hermitian. Also note on the right hand side we used the fact that (Ψ, Ψ) = kΨk2 = 1.

7.5. EXERCISES

113

7.5Exercises

#1.

Using (7.17) and (7.19), prove (7.20) and (7.21).

#2. Averages hxˆ4i and hpˆ4i:

For the state ψn, compute the averages hxˆ4i and hpˆ4i.

#3.

Prove (7.25). (The proof is similar to the proof of (7.24).)

#4.

 

 

 

 

 

 

 

Define the operators

 

 

 

x + dx

 

a

=

2

 

 

 

1

 

d

 

a

 

 

 

x dx

 

=

2

 

 

 

1

 

d

 

That is, if ψ = ψ(x), then

 

 

 

 

 

 

 

()(x) = 2 (x) + dx

 

 

1

 

 

 

and similarly for a . Using (7.20) and (7.21) show that for the harmonic oscillator wave functions ψn

 

 

 

 

ψn−1, n 1, aψ0 = 0,

n

=

 

n

 

ψn

 

 

 

 

 

a

= n + 1 ψn+1, n = 0, 1, . . . ,

a aψn

= n ψn, n = 0, 1, . . . ,

(aa a a) ψn

= ψn, n = 0, 1, . . . .

Explain why this last equation implies the operator equation

[a, a ] = id.

In quantum mechanics the operator a is called an annihilation operator and the operator a is called a creation operator. On the basis of this exercise, why do you think they have these names?

114

CHAPTER 7. QUANTUM HARMONIC OSCILLATOR

#5. Hermite Polynomials

We obtained the Hermite polynomials from the recurrence relation (7.13). Alternatively, we have a generating formula for the Hermite polynomials. Starting with this (which many books take as the definition of the Hermite polynomials), we may obtain the Schr¨odinger equation.

1.Verify that the first three Hermite polynomials H0(ξ), H1(ξ) and H2(ξ) are given using the generating formula

 

2

 

dn

2

 

Hn(ξ) = (1)neξ

 

 

 

e−ξ

 

(7.34)

 

n

2.The generating function (7.34) can be used to give an alternative generating function for Hermite polynomials. Show that

e−z2+2zξ =

zn

Hn(ξ) .

(7.35)

 

n=0 n!

 

 

X

 

Hint: Let F (z) = e−z2 and consider the Taylor expansion of eξ2 F (z ξ) about the point z = 0.

3.Derive (7.17) from (7.35). Now derive (7.19) using (7.34) and the newly derived (7.17).

4.Use (7.17) and (7.19) to show that the Hermite polynomials are solutions of the Her-

mite equation

 

 

 

 

 

d2

 

d

 

 

 

Hn(ξ) 2ξ

 

Hn(ξ) + 2nHn(ξ) = 0.

(7.36)

 

2

5. We know that the Hermite polynomials satisfy

 

 

 

 

 

 

Nn2

Z−∞ Hn(ξ)Hm(ξ)e−ξ2 = δnm.

(7.37)

Here by setting ψn(ξ) = NnHn(ξ)e−ξ2 /2 we see that ψn(ξ) are orthonormal in L2(R). Use (7.36) to obtain the di erential equation that ψn(ξ) satisfy. You should obtain (7.9) with ε = 2n + 1. This implies that ψn(ξ) = NnHn(ξ)e−ξ2/2 is the eigenfunction corresponding to the eigenvalue of the Hamiltonian operator.

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