Calculate for the first-excited state $ψ_{1} (x)$ $psi_1(x)$ of the simple harmonic oscillator the value of the Heisenberg Uncertainty relation?

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Calculate for the first-excited state $ψ_{1} (x)$ $psi_1(x)$ of the simple harmonic oscillator the value of the Heisenberg Uncertainty relation?

$psi_1(x) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)xe^(-momegax^2//2ℏ)$

This was a question on my Physics exam. I know how to do it now, so I plan to post it. :)

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Answer 1 · 2018-01-05T09:32:17

From the fruits of our labor,

$DeltaxDeltap = 3ℏ//2$

CAUTION: EXTREMELY LONG ANSWER!

First of all, the uncertainty principle is

$DeltaxDeltap >= ℏ//2$ ,

so we ought to get a value $>= ℏ//2$ .

Next, the uncertainties are defined as follows:

$DeltaA = sqrt(<< A^2 >> - << A >>^2)$ , $" "bb((1))$

where $<< A >>$ is the expectation value, or average value, of the observable $A$ .

An expectation value in one dimension is given by:

$<< A >> = << psi | A | psi >> = int_("allspace") psi^"*" hatA psi dx$ $" "bb((2))$

where $hatA$ is the operator corresponding to the observable $A$ .

So, we need to calculate $<< x^2 >>$ , $<< x >>^2$ , $<< p^2 >>$ , and $<>^2$ , given that

$psi_1(x) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)xe^(-momegax^2//2ℏ)$ ,

a real-valued function.

POSITION UNCERTAINTY

We have that $psi^"*" = psi$ and so, from $(2)$ :

$<< x^2 >> = int_(-oo)^(oo) psi^"*" x^2 psi dx$

$= ((momega)/(piℏ))^(1//2)((2momega)/ℏ)int_(-oo)^(oo) x^4 e^(-momegax^2//ℏ)dx$

$= 2/(sqrtpi)((momega)/(ℏ))^(3//2)int_(-oo)^(oo) x^4 e^(-momegax^2//ℏ)dx$

A nice way to solve this is to start from this one (which is the only one I know of these kinds of integrals):

$int_(-oo)^(oo) e^(-momegax^2//ℏ)dx$

From Leibniz's integral rule for constant integration bounds (which was not taught in the class, by the way),

$d/(dalpha)int_(-oo)^(oo) e^(-alphax^2)dx = int_(-oo)^(oo) (del)/(delalpha)[e^(-alphax^2)]dx$

$= -int_(-oo)^(oo) x^2e^(-alphax^2)dx$

This integral can be evaluated in polar coordinates to be $-(pi/alpha)^(1//2)$ . As a result, with $alpha = momega//ℏ$ ,

$color(green)(int_(-oo)^(oo) x^2 e^(-momegax^2//ℏ)dx) = -d/(dalpha)[-(pi/alpha)^(1//2)]$

$= sqrtpi cdot -1/2alpha^(-3//2)$

$= color(green)((sqrtpi)/(2alpha^(3//2)))$ $" "bb((3))$

Following the sequence once more,

$d/(dalpha)int_(-oo)^(oo) x^2e^(-alphax^2)dx = int_(-oo)^(oo) (del)/(delalpha)[x^2e^(-alphax^2)]dx$

$= -int_(-oo)^(oo) x^4e^(-alphax^2)dx$

Therefore,

$color(green)(int_(-oo)^(oo) x^4e^(-alphax^2)dx) = -d/(dalpha)[(sqrtpi)/(2alpha^(3//2))]$

$= -(sqrt(pi)/2 cdot -3/2 alpha^(-5//2))$

$= color(green)((3sqrtpi)/(4alpha^(5//2)))$ $" "bb((4))$

And so, using result $(4)$ ,

$color(green)(<< x^2 >>) = 2/(sqrtpi)((momega)/(ℏ))^(3//2)int_(-oo)^(oo) x^4 e^(-momegax^2//ℏ)dx$

$= 2/(sqrtpi)((momega)/(ℏ))^(3//2) cdot (3sqrtpi)/(4((momega)/(ℏ))^(5//2))$

$= 2/cancel(sqrtpi)cancel(((momega)/(ℏ))^(3//2)) cdot (3cancel(sqrtpi))/(4((momega)/(ℏ))^(cancel(5//2)))$

$= color(green)(3/2 (ℏ)/(momega))$ $" "bb((5))$

Next, the average position is:

$<< x >> = int_(-oo)^(oo) psi^"*"xpsidx$

$= 2/(sqrtpi)((momega)/(ℏ))^(3//2)int_(-oo)^(oo) x^3 e^(-momegax^2//ℏ)dx$

The integrand is an odd function times an even function, which is then odd. The integral of an odd function over a symmetric interval is zero, so $<< x >> = 0$ .

Therefore, using relation $(1)$ and result $(5)$ :

$barul|stackrel(" ")(" "color(green)(Deltax) = sqrt(<< x^2 >> - << x >>^2) = sqrt(<< x^2 >>) = color(green)(sqrt(3/2 (ℏ)/(momega)))" ")|$ $" "bb((6))$

MOMENTUM UNCERTAINTY

Similarly, we now need the analogous values for the momentum. The operator for momentum needed, which is $hatp = -iℏ(del)/(delx)$ . So, $hatp^2 = -ℏ^2(del^2)/(delx^2)$ .

From $(2)$ :

$<< p^2 >> = -ℏ^2int_(-oo)^(oo) psi^"*"(del^2psi)/(delx^2)dx$

The next thing we should evaluate then, is the second derivative of $psi$ .

$(del^2psi)/(delx^2) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)d^2/(dx^2)[xe^(-momegax^2//2ℏ)]$

The derivative of $e^(-alphax^2//2)$ for simplicity is $-alphaxe^(-alphax^2//2)$ , and so, [from a lot of product rule...](http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2+(xe%5E(-ax%5E2%2F2))

$=> (alpha/pi)^(1//4)sqrt(2alpha) cdot d/(dx)[-alphax^2e^(-alphax^2//2) + e^(-alphax^2//2)]$

$= (alpha/pi)^(1//4)sqrt(2alpha) cdot [-alpha(-alphax^3e^(-alphax^2//2) + 2xe^(-alphax^2//2)) - alphaxe^(-alphax^2//2)]$

$= (alpha/pi)^(1//4)sqrt(2alpha) cdot (alpha^2x^2 - 3alpha)xe^(-alphax^2//2)$

Let's not put the constants back in yet. Continuing, we got:

$(del^2psi)/(delx^2) = (alpha/pi)^(1//4)sqrt(2alpha) cdot (alpha^2x^2 - 3alpha)xe^(-alphax^2//2)$ $" "bb((7))$

From this result $(7)$ , the integral becomes, after some simplification:

$<< p^2 >> = -ℏ^2(alpha/pi)^(1//2)(2alpha) int_(-oo)^(oo) xe^(-alphax^2//2) cdot (alpha^2x^2 - 3alpha)e^(-alphax^2//2)dx$

$= -(2ℏ^2)/sqrtpi alpha^(3//2) int_(-oo)^(oo) (alpha^2x^2 - 3alpha)x^2e^(-alphax^2)dx$

$= -(2ℏ^2)/sqrtpi alpha^(3//2) {alpha^2int_(-oo)^(oo) x^4e^(-alphax^2)dx - 3alphaint_(-oo)^(oo)x^2e^(-alphax^2)dx}$

Fortunately we have basically already done these integrals. From results $(3)$ and $(4)$ , we then get:

$= -(2ℏ^2)/sqrtpi alpha^(3//2) {alpha^2((3sqrtpi)/(4alpha^(5//2))) - 3alpha((sqrtpi)/(2alpha^(3//2)))}$

$= -(2ℏ^2)/sqrtpi alpha^(cancel(3//2)) {((3sqrtpi)/(4cancel(alpha^(1//2)))) - 3((sqrtpi)/(2cancel(alpha^(1//2))))}$

$= -(2ℏ^2)/cancelsqrtpi alpha (-(3cancelsqrtpi)/4)$

$= ℏ^2 cdot alpha (3/2)$

Recalling that $alpha = momega//ℏ$ , we get:

$color(green)(<< p^2 >> = 3/2momegaℏ)$ $" "bb((8))$

Last one!! The average momentum from $(2)$ is:

$<> = int_(-oo)^(oo) psi^"*"hatppsidx$

where $hatp = -iℏ(del)/(delx)$ as stated before.

$<> = -iℏ((momega)/(piℏ))^(1//2)((2momega)/ℏ)int_(-oo)^(oo) xe^(-momegax^2//2ℏ) d/dx[xe^(-momegax^2//2ℏ)]dx$

But there is a nice trick here. The derivative of $psi$ , which in this case is an odd function (the odd $x$ times the even $e^(-alphax^2//2)$ equals an odd function), becomes even.

The result is that the integrand is an odd function times an even function, giving another odd function to integrate over a symmetric interval.

So, $<> = 0$ . As a result, using $(8)$ in $(1)$ , we get:

$barul|stackrel(" ")(" "color(green)(Deltap) = sqrt(<< p^2 >> - <>^2) = sqrt(<< p^2 >>) = color(green)(sqrt(3/2 momegaℏ))" ")|$ $" "bb((9))$

FINAL RESULT!

FINALLY, we can then evaluate the uncertainty relation! Using results $(6)$ and $(9)$ :

$color(blue)ul(DeltaxDeltap) = sqrt(3/2 ℏ/cancel(momega)) cdot sqrt(3/2 cancel(momega)ℏ)$

$= color(blue)ul(3ℏ//2)$

And this is indeed $>= ℏ//2$ , as the uncertainty principle requires!

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Calculate for the first-excited state $ψ_{1} (x)$ $psi_1(x)$ of the simple harmonic oscillator the value of the Heisenberg Uncertainty relation?

$psi_1(x) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)xe^(-momegax^2//2ℏ)$

This was a question on my Physics exam. I know how to do it now, so I plan to post it. :)

1 Answer

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Calculate for the first-excited state psi_1(x)ψ1(x)psi_1(x) of the simple harmonic oscillator the value of the Heisenberg Uncertainty relation?

psi_1(x) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)xe^(-momegax^2//2ℏ)psi_1(x) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)xe^(-momegax^2//2ℏ) This was a question on my Physics exam. I know how to do it now, so I plan to post it. :)

1 Answer

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Impact of this question

Calculate for the first-excited state $ψ_{1} (x)$ $psi_1(x)$ of the simple harmonic oscillator the value of the Heisenberg Uncertainty relation?

$psi_1(x) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)xe^(-momegax^2//2ℏ)$

This was a question on my Physics exam. I know how to do it now, so I plan to post it. :)