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Put two non-interacting particles of mass 𝑀 in a one-dimensional harmonic oscillator potential of frequency 𝜔. The
two particle wave function is:
Ψ(𝑥1 , 𝑥2 ) = 𝜓𝑛 (𝑥1 ) ∗ 𝜓𝑚 (𝑥2 )
Where 𝜓𝑛 and 𝜓𝑚 are the n’th and m’th Harmonic oscillator eigenfunction and 𝑥1 and 𝑥2 are positions for the two
particles. We can write the wave functions as the following.
1
4
1
𝑀𝜔𝑥1
𝑀𝜔
−
𝜓𝑛 (𝑥1 ) =
∗(
) ∗ 𝑒 2ℏ ∗ 𝐻𝑛 (𝑦)
𝑛
𝜋ℏ
√2 ∗ 𝑛!
1
4
1
𝑀𝜔𝑥2
𝑀𝜔
−
𝜓𝑚 (𝑥2 ) =
∗(
) ∗ 𝑒 2ℏ ∗ 𝐻𝑚 (𝑦)
𝑚
𝜋ℏ
√2 ∗ 𝑚!
Where 𝑦 = √
𝑀𝜔
𝜋
∗ 𝑥 for 𝑥 = 𝑥1 , 𝑥2
We need to first find the expectation value
〈𝑥1 − 𝑥2 〉:
〈𝑥1 − 𝑥2 〉 = 〈𝑥1 〉 − 〈𝑥2 〉
〈𝑥1 〉 𝑎𝑛𝑑 〈𝑥2 〉 can be represented as:
∞
〈𝑥1 〉 = ∫ (𝜓𝑛∗ (𝑥1 ) ∗ 𝑥1 ∗ 𝜓𝑛 (𝑥1 ))𝑑𝑥1
−∞
∞
∗ ( )
〈𝑥2 〉 = ∫ (𝜓𝑚
𝑥2 ∗ 𝑥2 ∗ 𝜓𝑚 (𝑥2 ))𝑑𝑥2
−∞
∞
∞
∗ ( )
〈𝑥1 − 𝑥2 〉 = ∫ (𝜓𝑛∗ (𝑥1 ) ∗ 𝑥1 ∗ 𝜓𝑛 (𝑥1 ))𝑑𝑥1 − ∫ (𝜓𝑚
𝑥2 ∗ 𝑥2 ∗ 𝜓𝑚 (𝑥2 ))𝑑𝑥2
−∞
−∞
Putting in the value we find out that these integrals become 0, so:
〈𝑥1 − 𝑥2 〉 = 0
Since the expectation value is finding the probability of the particle around a point, and in this case, the probability of
finding a particle to the right of the other particle is equal to the probability of finding it to the left, so it becomes zero at the
exact point.
Now we will need to find
〈(𝑥1 − 𝑥2 )2 〉:
〈(𝑥1 − 𝑥2 )2 〉 = 〈𝑥12 − 2𝑥1 𝑥2 + 𝑥22 〉 = 〈𝑥12 〉 − 〈2𝑥1 𝑥2 〉 + 〈𝑥22 〉 = 〈𝑥12 〉 − 〈2𝑥1 〉〈𝑥2 〉 + 〈𝑥22 〉
So we can do this in steps. For not crowding this document we can take:
(𝜓𝑛∗ (𝑥1 ) ∗ 𝑥1 ∗ 𝜓𝑛 (𝑥1 )) = 𝑁
∗ ( )
(𝜓𝑚
𝑥2 ∗ 𝑥2 ∗ 𝜓𝑚 (𝑥2 )) = 𝑀
First let’s find
〈𝑥12 〉:
∞
〈𝑥12 〉
=∫
−∞
(𝜓𝑛∗ (𝑥1 )
2
∞
∗ 𝑥1 ∗ 𝜓𝑛 (𝑥1 )) 𝑑𝑥1 = ∫ 𝑁 2 𝑑𝑥1
−∞
〈𝑥12 〉
=
5
− −2∗𝑚
2 2
∗ 𝑀𝜔 ∗ 𝐻𝑚 (𝑦)4
3
𝑀𝜔 2
√𝜋 ∗ ( ℏ ) ∗ ℏ ∗ (𝑚!)2
Now for 〈2𝑥1 〉〈𝑥2 〉:
∞
∞
〈2𝑥1 〉〈𝑥2 〉 = ∫ (2𝑁)𝑑𝑥1 ∗ ∫ (𝑀)𝑑𝑥2 = 0
−∞
−∞
2
And finally for 〈𝑥2 〉:
∞
〈𝑥22 〉
=∫
∞
∗ ( )
(𝜓𝑚
𝑥2
∗ 𝑥2 ∗ 𝜓𝑚 (𝑥2 ))𝑑𝑥2 = ∫ (𝑀)𝑑𝑥2
−∞
−∞
5
〈𝑥22 〉 =
2−2−2∗𝑛 ∗ 𝑀𝜔 ∗ 𝐻𝑛 (𝑦)4
3
2
𝑀𝜔
√𝜋 ∗ ( ℏ ) ∗ ℏ ∗ (𝑛!)2
When we add all these together then we get:
5
〈𝑥12 〉 − 〈2𝑥1 〉〈𝑥2 〉 + 〈𝑥22 〉 =
2−2−2∗𝑚 ∗ 𝑀𝜔 ∗ 𝐻𝑚 (𝑦)4
3
2
5
−0+
𝑀𝜔
√𝜋 ∗ ( ℏ ) ∗ ℏ ∗ (𝑚!)2
2−2−2∗𝑛 ∗ 𝑀𝜔 ∗ 𝐻𝑛 (𝑦)4
3
𝑀𝜔 2
√𝜋 ∗ ( ℏ ) ∗ ℏ ∗ (𝑛!)2
Or simply:
〈(𝑥1 − 𝑥2 )2 〉 = 〈𝑥12 〉 + 〈𝑥22 〉 =
2
−5−2∗𝑚
4
2
∗ 𝑀𝜔 ∗ 𝐻𝑚 (𝑦)
3
2
2
𝑀𝜔
) ∗ ℏ ∗ (𝑚!)
√𝜋 ∗ (
+
ℏ
〈(𝑥1 − 𝑥2 )2 〉 = 〈𝑥12 〉 + 〈𝑥22 〉 = 𝑀𝜔(
2
−5−2∗𝑚
4
2
∗ 𝐻𝑚 (𝑦)
3
2
𝑀𝜔) ∗ ℏ ∗ (𝑚!)2
ℏ
√𝜋 ∗ (
2
−5−2∗𝑛
4
2
∗ 𝑀𝜔 ∗ 𝐻𝑛(𝑦)
3
𝑀𝜔)2 ∗ ℏ ∗ (𝑛!)2
√𝜋 ∗ (
ℏ
+
2
−5−2∗𝑛
4
2
∗ 𝐻𝑛 (𝑦)
3
2
𝑀𝜔) ∗ ℏ ∗ (𝑛!)2
ℏ
√𝜋 ∗ (
)