EE201/MSE207 Lecture 8 Measurement and uncertainty principle - - PowerPoint PPT Presentation

EE201/MSE207 Lecture 8

Measurement and uncertainty principle

Determinate state Theorem: If 𝐵 Ψ = 𝜇 Ψ , then measurement of 𝐵 in state |Ψ〉 will certainly give result 𝜇 (therefore, such Ψ is called determinate state) Proof: a) from postulate 5 of the previous lecture: 𝑄𝜇 = 𝑔

𝜇 Ψ 2;

b) from postulate 4: 𝐵 = 𝜇, 𝐵2 = 𝜇2  no variance. Also, from postulate 6, measurement will not change such determinate state. In contrast, if |Ψ〉 is not an eigenvector of 𝐵, then measurement of 𝐵 can give different results, and changes (collapses) |Ψ〉.

Theorem 1 If two Hermitian operators commute, 𝐵, 𝐶 = 0, then there exists a basis consisting of eigenvectors of both 𝐵 and 𝐶 simultaneously. (These states are determinate states of both 𝐵 and 𝐶, i.e. 𝜏

𝐵 = 𝜏𝐶 = 0. )

Theorem 2 (generalized uncertainty principle) For Hermitian 𝐵 and 𝐶, if 𝐵, 𝐶 ≠ 0, then Example

𝑦, 𝑞 = 𝑗ℏ



𝜏𝑦

2 𝜏𝑞 2 ≥

ℏ 2

2



𝜏𝑦𝜏𝑞 ≥ ℏ 2 Compatible and incompatible observables

Question: When a state can be a determinate state for two operators 𝐵 and 𝐶?

(why important: consider measurement sequence A, B, A)

Answer: If 𝐵 and 𝐶 commute, 𝐵, 𝐶 = 0, then this is possible (“compatible”); if 𝐵, 𝐶 ≠ 0, then this is usually impossible (“incompatible”). More rigorously, two theorems from linear algebra (without proof).

(Heisenberg uncertainty principle) (𝜏 is the standard deviation)

𝜏

𝐵 2 𝜏𝐶 2 ≥

1 2𝑗 𝐵, 𝐶

2

Energy-time uncertainty

You can often find inequality

Δ𝐹 Δ𝑢 ≥ ℏ 2

1) It is formally incorrect, but often gives correct intuition. 2) It is correct in the following sense. Theorem For any observable 𝑅 (which does not explicitly depend on time)

𝜏𝐹 𝜏𝑅 𝑒〈𝑅〉 𝑒𝑢 ≥ ℏ 2

(So, if anything changes significantly during Δ𝑢, then the energy spread should be large enough, Δ𝐹 Δ𝑢 ≥ ℏ 2. In stationary state Δ𝐹 = 0, therefore nothing changes.) Proof Straightforward from 𝜏𝐼𝜏𝑅 ≥ 1 2𝑗 𝐼, 𝑅

2

and

𝑒〈𝑅〉 𝑒𝑢 = 𝑗 ℏ 𝐼, 𝑅

𝑒〈𝑅〉 𝑒𝑢 = 𝑒 𝑒𝑢 Ψ 𝑅Ψ = 𝑒Ψ 𝑒𝑢 𝑅Ψ + Ψ 𝑅 𝑒Ψ 𝑒𝑢 + Ψ 𝜖 𝑅 𝜖𝑢 Ψ = = −𝑗 ℏ 𝐼Ψ 𝑅Ψ + Ψ 𝑅 −𝑗 ℏ 𝐼Ψ = 𝑗 ℏ Ψ 𝐼 𝑅Ψ − Ψ 𝑅 𝐼Ψ = 𝑗 ℏ 〈Ψ| 𝐼, 𝑅 Ψ〉 = ( 𝑗 ℏ) 〈 𝐼, 𝑅 〉

𝑦-representation and 𝑞-representation

It is easy to check that both

𝑦 and 𝑞 are Hermitian (as for any observable).

We need to prove 𝑔 𝑈𝑕 = 〈 𝑈𝑔|𝑕〉. For 𝑈 = 𝑦 this is very simple: 𝑔∗ 𝑦 𝑦𝑕 𝑦 𝑒𝑦 = 𝑦 𝑔 𝑦

∗𝑕 𝑦 𝑒𝑦.

For 𝑈 = 𝑞 we need integration by parts: 𝑔∗ 𝑦

−𝑗ℏ 𝑒

𝑒𝑦 𝑕 𝑦 𝑒𝑦 =

= 𝑗ℏ 𝑒𝑔∗(𝑦)

𝑒𝑦

𝑕 𝑦 𝑒𝑦 = −𝑗ℏ 𝑒𝑔 𝑦

𝑒𝑦 ∗

𝑕 𝑦 𝑒𝑦.

Find eigenstates of

𝑦 and 𝑞 𝑈𝑔 𝑦 = 𝜇 𝑔(𝑦)

Eigenstates of 𝑦:

𝑦 𝑔 𝑦 = 𝜇 𝑔(𝑦)  𝑔 𝑦 = 𝐵 𝜀(𝑦 − 𝜇)

Not normalizable, choose 𝐵 = 1.

𝑔

𝜇 𝑦 = 𝜀(𝑦 − 𝜇), then 𝑔

𝜇 𝑔 𝜈 = 𝜀 𝜈 − 𝜇

Check: 𝜀 𝑦 − 𝜇 𝜀 𝑦 − 𝜈 𝑒𝑦 = 𝜀(𝜈 − 𝜇) (since 𝑦 = 𝜈)

• rthonormal basis

Eigenstates of 𝑞:

−𝑗ℏ 𝑒 𝑒𝑦 𝑕 𝑦 = 𝜇 𝑕(𝑦) 𝑕 𝑦 = 𝐶 𝑓𝑗𝜇

ℏ𝑦



𝑕𝜇 𝑦 =

1 2𝜌ℏ 𝑓𝑗𝜇

ℏ𝑦, then 𝑕𝜇 𝑕𝜈 = 𝜀 𝜈 − 𝜇

Check: 1

2𝜌ℏ 𝑓−𝑗𝜇

ℏ𝑦𝑓𝑗𝜈 ℏ𝑦 𝑒𝑦 =

1 2𝜌ℏ 𝑓𝑗𝜈−𝜇

ℏ 𝑦 𝑒𝑦 = 𝜀(𝜈 − 𝜇)

since

−∞ ∞ 𝑓𝑗𝑙𝑦𝑒𝑦 = 2𝜌 𝜀 𝑙

and 𝜀 𝑏𝑦 = 𝜀(𝑦)

|𝑏|

Digression: proof of the formula

−∞ ∞ 𝑓𝑗𝑙𝑦𝑒𝑦 = 2𝜌 𝜀 𝑙

Fourier transform 𝑔 𝑦 =

1 2𝜌 𝑓𝑗𝑙𝑦𝐺 𝑙 𝑒𝑙, 𝐺 𝑙 = 1 2𝜌 𝑓−𝑗𝑙𝑦 𝑔 𝑦 𝑒𝑦

Therefore 𝐺 𝑙 = 1 2𝜌 𝑒𝑦 𝑓−𝑗𝑙𝑦 1 2𝜌 𝑓𝑗𝑙′𝑦𝐺 𝑙′ 𝑒𝑙′ = 1 2𝜌 𝑓𝑗 𝑙′−𝑙 𝑦𝑒𝑦 𝐺 𝑙′ 𝑒𝑙 𝜀(𝑙 − 𝑙′)

• rthonormal basis

Similarity of 𝑦-representation and 𝑞-representation

𝑔

𝜇 𝑦 = 𝜀(𝑦 − 𝜇)

𝑦-eigenstates (x-basis) 𝑞-eigenstates (p-basis)

𝑕𝜇 𝑦 = 1 2𝜌ℏ 𝑓𝑗𝜇

ℏ𝑦

We can say that Ψ 𝑦 are actually components of vector |Ψ〉 in x-basis:

Ψ 𝑦 =

−∞ ∞

Ψ 𝑦′ 𝜀 𝑦 − 𝑦′ 𝑒𝑦′

basis vectors components Similarly, we can write it in p-basis:

Ψ 𝑦 =

−∞ ∞

Φ 𝑞 1 2𝜌ℏ 𝑓𝑗𝑞

ℏ𝑦 𝑒𝑞

basis vectors components Φ 𝑞 = 1 2𝜌ℏ

−∞ ∞

𝑓−𝑗𝑞

ℏ𝑦 Ψ 𝑦 𝑒𝑦

We can regard Φ 𝑞 as a wavefunction in p-space Actually, more often people use Φ 𝑙 in 𝑙-space, where 𝑙 = 𝑞/ℏ, then eigenstates (basis vectors) are 1

2𝜌 𝑓𝑗𝑙𝑦.

Operators 𝑦 and 𝑞 in p-space

𝑞Φ 𝑞 = 𝑞 Φ(𝑞)

Expected by analogy. Formal proof:

= −𝑗ℏ Φ 𝑞 1 2𝜌ℏ 𝑗𝑞 ℏ 𝑓𝑗𝑞

ℏ𝑦𝑒𝑞 = 𝑞 Φ 𝑞

1 2𝜌ℏ 𝑓𝑗𝑞

ℏ𝑦𝑒𝑞

−𝑗ℏ 𝜖Ψ(𝑦) 𝜖𝑦 = −𝑗ℏ 𝜖 𝜖𝑦 Φ 𝑞 1 2𝜌ℏ 𝑓𝑗𝑞

ℏ𝑦𝑒𝑞 =

𝑦Φ 𝑞 = 𝑗ℏ 𝜖 𝜖𝑞 Φ(𝑞)

Proof: 𝑦Ψ(𝑦) = Φ 𝑞

1 2𝜌ℏ 𝑦 𝑓𝑗𝑞

ℏ𝑦𝑒𝑞 =

−𝑗ℏ 𝜖 𝜖𝑞 𝑓𝑗𝑞

ℏ𝑦

= by parts = 𝑗ℏ 𝑒Φ(𝑞) 𝑒𝑞 1 2𝜌ℏ 𝑓𝑗𝑞

ℏ𝑦𝑒𝑞

Commutator does not change 𝑦, 𝑞 = 𝑦, −𝑗ℏ 𝜖 𝜖𝑦 = 𝑗ℏ 𝑦, 𝑞 = 𝑗ℏ 𝜖 𝜖𝑞 , 𝑞 = 𝑗ℏ

Probabilities Ψ 𝑦

2𝑒𝑦 and Φ 𝑞 2𝑒𝑞 Φ 𝑞 is as good a wavefunction as Ψ 𝑦 For x-measurement, probability to find particle near 𝑦0 is 𝒬 𝑦0 𝑒𝑦 (postulate 5, rem. a), with 𝒬 𝑦0 = 𝑔

𝑦0 Ψ 2 = 𝜀 𝑦 − 𝑦0 Ψ 𝑦 𝑒𝑦 2 = Ψ 𝑦0 2

Similarly, for p-measurement, probability to find momentum near 𝑞0 is 𝒬 𝑞0 𝑒𝑞, with 𝒬 𝑞0 = 𝑕𝑞0 Ψ

2 = 1 2𝜌ℏ e−𝑗𝑞0

ℏ 𝑦 Ψ 𝑦 𝑒𝑦

2

= Φ 𝑞0

2

Minimum-uncertainty states

From Heisenberg uncertainty relation, we know that 𝜏𝑦𝜏𝑞 ≥

ℏ 2

Which states satisfy the lower bound, 𝜏𝑦𝜏𝑞 =

ℏ 2 ?

Answer:

Ψ 𝑦 = 𝐵 𝑓−𝑏 𝑦−𝑦0 2/2ℏ 𝑓𝑗𝑞0𝑦/ℏ

(in optics they are called “squeezed states”) It has 𝑦 = 𝑦0 , 𝑞 = 𝑞0 ,

𝜏𝑦 = ℏ 2𝑏 , 𝜏𝑞 = 𝑏ℏ 2 ,

In p-representation very similar form: 𝐵 = 𝑏 𝜌ℏ

1/4

Φ 𝑞 = 1 2𝜌ℏ 𝑓−𝑗𝑞

ℏ𝑦 Ψ 𝑦 𝑒𝑦 = 𝑓𝑗𝑞0𝑦0/ℏ 𝐵

𝑏 𝑓− 𝑞−𝑞0 2/(2ℏ𝑏) 𝑓−𝑗𝑦0𝑞/ℏ