Bending The Heisenberg Uncertainty Principle

                      Anwar Mohiuddin𝑎, Abhijeet K. Jha𝑏 and Prasanta K. Panigrahi𝑎 

     IISER −Kolkata, Mohanpur Campus, Nadia,

𝑎 West Bengal −741252
IISER −Pune, b Sutarwadi Road, Pashan, Pune, Maharastra −411021

The celebrated Heisenberg Uncertainty Principle ∆x.∆p≥ ħ/2 can allow measurement accuracies less than ∆x or ∆p. Classical analog of this is known as sub- Fourier sensitivity. We illustrate this phenomenon in a step by step process using the example of compass state, as suggested by Zurek.

  A number of canonically conjugate variables appear in mechanics, like co-ordinate-

momentum and time-energy. The fact that, they are related through Fourier transform, restricts their measurement accuracies. For example, it is well-known from the theory of Fourier transform that, ∆𝑥. ∆𝑘~1, where the Fourier transform of a function 𝐹(𝑥) of the co-ordinate variable is related to its Fourier counterpart 𝐹 (𝑘) in the form, 𝐹 𝑥 =
𝑑3𝑘 (2𝜋)3 𝑒𝑖𝑘𝑥𝐹 (𝑘). In quantum mechanics, the above uncertainty product leads to the Heisenberg uncertainty relation, ∆𝑥. ∆𝑝≥ ħ 2, where 𝑝= ħ𝑘.

  For a Gaussian state of the type, 
                  𝜓 𝑥 = (

𝑚𝜔 2ħ ) 1 4𝑒−𝑚𝜔𝑥2 2ħ , familiar from the harmonic oscillator problem, the uncertainty relation leads to an equality, ∆𝑥. ∆𝑝= ħ 2.
Explicit calculation yields, ∆𝑥= < 𝑥2 > −< 𝑥>2≡ < 𝑥2 >=
ħ 2𝑚𝜔, and, ∆𝑝= < 𝑝2 > −< 𝑝>2≡ < 𝑝2 >= ħ𝑚𝜔 2 .

Here, < 𝑥2 >=
𝜓∗ 𝑥 𝑥2𝜓(𝑥)𝑑𝑥 ∞ −∞ and < 𝑝2 >=
𝜓∗ 𝑥 (−ħ2 𝜕2 𝜕𝑥2)𝜓(𝑥)𝑑𝑥 ∞ −∞ . By use of more general Gaussian states, like squeezed states, one can reduce one of the uncertainties: ∆𝑥→ ∆𝑥 𝜆 , and ∆𝑝→𝜆∆𝑝, maintaining ∆𝑥. ∆𝑝= ħ 2. It is then natural to ask, if such states exist for which it is possible to measure variation in 𝑥 (or 𝑝), which is less than ∆𝑥 (or ∆𝑝). In Fourier transform, this is known as sub-Fourier sensitivity and has been experimentally demonstrated recently, through appropriate combination of laser beams [1]. In the quantum domain, it was demonstrated by Zurek [2], that the above can be achieved through special states like cat and compass states. These states are superposition of familiar Gaussian states and hence the reason behind this sensitivity can be appreciated without tedious effort. The following problem illustrates this, in a step by step process.

Q1) Show that the displaced Gaussian function 𝑒− 𝑥−𝛼 2/2 ≡ < 𝑥 ∣ 𝛼> is an Eigen state of
a = 𝑥+ 𝜕 𝜕𝑥 , with Eigen value𝛼.

Proof: Since a < 𝑥 ∣ 𝛼> = 𝑥+ 𝜕 𝜕𝑥 𝑒− 𝑥−𝛼 2/2 = 𝑥𝑒− 𝑥−𝛼 2/2 + 𝜕 𝜕𝑥𝑒− 𝑥−𝛼 2/2 = 𝛼𝑒− 𝑥−𝛼 2/2, the displaced Gaussian function. 𝑒− 𝑥−𝛼 2/2 is an Eigen state of a. It is worth noting that < 𝑥 ∣ 𝛼> is known as the coherent state in literature, which describes laser. A discerning reader will recognize that modulo constant factors, a is the annihilation operator of the harmonic oscillator problem.

Q2) Given that𝜓= 𝑁(𝑒− 𝑥−𝛼 2 2

• 𝑒− 𝑥+𝛼 2 2 ), find out the normalization constant N from the square integrability condition:
  𝜓∗𝜓𝑑𝑥= 1 ∞ −∞ . Hint: one can take 𝛼 to be real and use the formula
  𝑒−𝑎𝑥2𝑑𝑥=
  𝜋 𝑎 . ∞ −∞

Solution: assuming N and 𝛼 to be real;

𝜓∗𝜓 𝑑𝑥=
𝑁 𝑒− 𝑥−𝛼 2 2

• 𝑒− 𝑥+𝛼 2 2

∗ 𝑁 𝑒− 𝑥−𝛼 2 2

• 𝑒− 𝑥+𝛼 2 2 𝑑𝑥= 1 ∞ −∞ ∞ −∞

= 𝑁2
𝑒 𝑥−𝛼 2 + 𝑒 𝑥+𝛼 2 + 2𝑒− 𝑥2+𝑎2 𝑑𝑥= 1 ∞ −∞

Substituting 𝑥−𝛼= 𝑦 in the first expression and carrying out similar manipulations in the last two expressions, the above integrals can be straightforwardly evaluated and one obtains 2 𝜋𝑁2 1 + 𝑒−𝛼2 = 1 , yielding 𝑁=
1 𝜋 1 4
1 2 1+𝑒−𝛼2
1 2 .

Q3) Given that 𝜙= 𝑒− 𝑥−𝛼 2 4𝜎2 + 𝑒− 𝑥+𝛼 2 4𝜎2 𝑒−𝑖𝑘𝑥 and
𝜙𝛿 = 𝑒− 𝑥−𝛼 2 4𝜎2 + 𝑒− 𝑥+𝛼 2 4𝜎2 𝑒−𝑖𝑘𝑥𝑒𝑖𝛿𝑥, Calculate the overlap integral, 𝐼=
𝜙𝛿 ∗𝜙 𝑑𝑥, ∞ −∞ and find out the points it vanishes. Give physical interpretation for this phenomenon.
Solution: Taking 𝛼 to be real, for simplicity, one finds 𝐼=
𝑒− 𝑥−𝛼 2 4𝜎2 + 𝑒− 𝑥+𝛼 2 4𝜎2 𝑒−𝑖𝑘𝑥𝑒𝑖𝛿𝑥 ∗ 𝑒− 𝑥−𝛼 2 4𝜎2 + 𝑒− 𝑥+𝛼 2 4𝜎2 𝑒−𝑖𝑘𝑥 𝑑𝑥 ∞ −∞

=

(𝑒−((𝑥2+𝛼2−2𝑥𝛼+2𝜎2𝑖𝛿𝑥)/2𝜎2) + 𝑒−((𝑥2+𝛼2+2𝑥𝛼+2𝜎2𝑖𝛿𝑥 )/2𝜎2) + 2𝑒−((𝑥2+𝛼2+2𝜎2𝑖𝛿𝑥) /2𝜎2) ) ∞ −∞ 𝑑𝑥.
We now consider each term individually: 1st term =
𝑒−((𝑥2+𝛼2−2𝑥 𝛼−𝜎2𝑖𝛿 + 𝛼−𝜎2𝑖𝛿 2− 𝛼−𝜎2𝑖𝛿 2)/2𝜎2) 𝑑𝑥 ∞ −∞

Redefining the variable as 𝑥−𝛼+ 𝜎2𝑖𝛿 / 2 𝜎= 𝑧, and using the above mentioned result we obtain 2𝜋𝜎𝑒−𝜎2𝛿2/2𝑒𝑖𝛼𝛿. Similarly we get the second and third terms as 2𝜋𝜎𝑒−𝜎2𝛿2/2𝑒−𝑖𝛼𝛿 and 2 2𝜋𝜎𝑒−𝜎2𝛿2/2𝑒−𝛼2 2𝜎2 respectively. We note that the third term is completely real as compared to the first two terms in the integral. Adding the results leads to, 𝐼= 2𝜋𝜎𝑒−𝜎2𝛿2 2 𝑒−𝑖𝛿𝑥+ 𝑒𝑖𝛿𝑥+ 2𝑒−𝛼2 2𝜎2 .
We also note that the first two terms in the above result lead to an oscillatory factorcos⁡(𝛿𝛼), where as the third term led to Gaussian factor 𝑒−𝛼2/2𝜎2. It can be ea

…(Full text truncated)…