Minimum average value of position

In summary, the conversation discusses the calculation of the average value of position <x> using the expression ϕ(x) and the difficulty in determining the limits for the integral. It is suggested to equate the derivative of ϕ(x) to 0 to find the minimum value, but it is pointed out that this approach is incorrect. The correct approach is to calculate <x> as a definite integral and find the values of a and b that result in the minimum value.
  • #1
tanaygupta2000
208
14
Homework Statement
Let ψ₀(x) and ψ₁(x) be the wave functions corresponding to the ground state and the first excited states of a one dimensional harmonic oscillator respectively. Consider the normalized state ϕ(x) = αψ₀(x) + βψ₁(x), where α and β are real numbers. The values of α and β for which <x>, the average value of the position is a minimum are:

(a) α = -β = 1/√2
(b) α = β = 1/√2
(c) α = 1/√3 and β = -√(2/3)
(d) α = 1/√3 and β = √(2/3)
Relevant Equations
For one-dimensoinal harmonic oscillator,
Ground state wavefunction, ψ₀(x) = (mw/πℏ)^(1/4) exp[-(mw/2ℏ)x^2]
First excited state wavefunction, ψ₁(x) = (mw/πℏ)^(1/4) * x√(2mw/ℏ) * exp[-(mw/2ℏ)x^2]

Average value of position, <x> = ∫xϕ(x) dx
After getting the values of ψ₀(x) and ψ₁(x), I put them in the expression of ϕ(x) to get:
ϕ(x) = (mw/πℏ)^(1/4) * exp[-(mw/2ℏ)x^2] * [α + βx√(2mw/ℏ)]

Now when attempting to find the value of <x> by ∫xϕ(x) dx, I am having trouble determining the limits, as I am getting nothing useful by integrating from -∞ to ∞.
Also, do I even need to find <x>, can't I equate the derivative of simply ϕ(x) to 0 to get the minima, as <x> ∝ ϕ(x) which is relatively easier to differentiate?
Kindly help !
 
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  • #2
tanaygupta2000 said:
Average value of position, <x> = ∫xϕ(x) dx

This is not right.
 
  • #3
tanaygupta2000 said:
Also, do I even need to find <x>, can't I equate the derivative of simply ϕ(x) to 0 to get the minima, as <x> ∝ ϕ(x) which is relatively easier to differentiate?
Kindly help !

By "minimum" value, I think they mean the greatest negative value of ##\langle x \rangle##. They do not mean ##\langle x \rangle = 0##. You could do that with ##a = 1, b = 0##, for example.

I think you have to calculate ##\langle x \rangle##, unless there is a clever trick I'm missing.
 
  • #4
tanaygupta2000 said:
Also, do I even need to find <x>, can't I equate the derivative of simply ϕ(x) to 0 to get the minima, as <x> ∝ ϕ(x) which is relatively easier to differentiate?
As PeroK pointed out, your expression for <x> is wrong.
Also note that there is a difference between differentiating <x> and differentiating either ϕ(x) or its square which is the probability density: The probability density is a function of x, whereas <x> is not. <x> is a definite integral. <x> is now dependent on a and b. The statement of the problem requires you to find the values of a and b such that <x> is minimum.
 
  • Informative
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Related to Minimum average value of position

What is the concept of minimum average value of position?

The minimum average value of position refers to the lowest average value of an object's position over a given period of time. It is a measure of the overall trend of an object's movement.

How is the minimum average value of position calculated?

The minimum average value of position is calculated by dividing the total distance traveled by an object by the total time taken. This gives the average speed of the object. The minimum average value of position is then determined by finding the lowest point on a graph of the object's position over time.

What is the significance of the minimum average value of position in physics?

In physics, the minimum average value of position is important because it can provide information about the acceleration of an object. If the minimum average value of position is decreasing, it indicates that the object is decelerating, while an increasing minimum average value of position suggests that the object is accelerating.

Can the minimum average value of position be negative?

Yes, the minimum average value of position can be negative. This occurs when an object is moving in the negative direction (e.g. left or down) and its position is decreasing over time. However, the magnitude of the minimum average value of position is always positive.

How is the minimum average value of position used in real-world applications?

The minimum average value of position is used in many real-world applications, such as calculating the fuel efficiency of a vehicle or predicting the motion of objects in sports. It is also used in physics experiments to analyze the motion of objects and determine the effects of different forces on their movement.

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