Question Video: Particle Location Probability given its Wave Function

A wave function of a particle with mass ๐‘š is given by ฮจ(๐‘ฅ) = cos ๐›ผ๐‘ฅ if โˆ’๐œ‹/2๐›ผ โ‰ค ๐‘ฅ โ‰ค ๐œ‹/2๐›ผ, ฮจ(๐‘ฅ) = 0 otherwise, where ๐›ผ = 1.00 ร— 10ยนโฐ mโปยน. Find the probability the particle can be found in the interval (0.00 โ‰ค ๐‘ฅ โ‰ค 0.50 ร— 10โปยนโฐ) m.


Video Transcript

A wave function of a particle with mass ๐‘š is given by ฮจ as a function of ๐‘ฅ equals cosine ๐›ผ๐‘ฅ if ๐‘ฅ is greater than or equal to negative ๐œ‹ over 2๐›ผ and less than or equal to positive ๐œ‹ over two ๐›ผ and zero otherwise, where ๐›ผ is 1.00 times 10 to the 10th inverse meters. Find the probability the particle can be found in the interval 0.00 to 0.50 times 10 to the negative 10th meters.

So given this wave functions ฮจ of ๐‘ฅ and this constant value for ๐›ผ, we want to find the probability of a particle being located within particular distance boundaries; weโ€™ll call this probability capital ๐‘ƒ. Given this wave function, the value for ๐›ผ, and these boundaries, the first thing weโ€™d like to do is figure out how the given boundary conditions match up with the boundaries in our piecewise function.

When we know that, weโ€™ll have figured out what our wave functions ฮจ actually looks like over the interval weโ€™re interested in. So our first question is to look at the lower bounds of our piecewise function and the lower bounds of the given range in our problem statement. We would like to know is negative ๐œ‹ over two ๐›ผ less than or equal to 0.00.

Since ๐›ผ is a positive number, that means overall negative ๐œ‹ over two ๐›ผ is negative. So indeed negative ๐œ‹ over two ๐›ผ is less than 0.00. Next we want to figure out is the upper bound, 0.50 times 10 to the negative 10th meters, is that less than or equal to positive ๐œ‹ over two ๐›ผ.

If we plug in the given value for ๐›ผ, we see that the right side of this expression is ๐œ‹ over two times 10 to the negative 10th meters, whereas the left side is one-half times 10 to the negative 10th meters. ๐œ‹ over two is roughly 1.5, so that means this inequality also is true. Taken together, taken together, these two conditions being satisfied means that ฮจ of ๐‘ฅ over the range that weโ€™re interested in exploring it is always equal to the cosine of ๐›ผ๐‘ฅ, never zero.

Since ฮจ of ๐‘ฅ is a wave function, one condition it might satisfy as all wave functions do is the condition of normalization. All that means is that if we integrate the square of the function ฮจ of ๐‘ฅ for over all space, from negative infinity to positive infinity, then that integral must be equal to one. This is the mathematical statement of saying that if we search everywhere over all space for this particle, then the chance of our finding it is one.

In other words, it exists somewhere. Now when we go to normalize our function ฮจ of ๐‘ฅ, instead of using boundaries of negative infinity to positive infinity, we use boundaries of negative ๐œ‹ over two ๐›ผ to positive ๐œ‹ over two ๐›ผ. The reason we can do that is because ฮจ is zero elsewhere, so thereโ€™s no point in integrating over a place where the particle canโ€™t exist. Now hereโ€™s the wave function ฮจ of ๐‘ฅ weโ€™re going to use in this integration equation.

Weโ€™re going to use the equation ๐ถ, a constant, times the cosine of ๐›ผ๐‘ฅ. Now you may say why weโ€™re using ๐ถ; there was no ๐ถ that appeared in the original piecewise function, so why weโ€™re introducing that? And the reason weโ€™re using ๐ถ is because we donโ€™t know if this originally stated ฮจ of ๐‘ฅ is normalized. It may be; that is, it maybe we find a ๐ถ is equal to one, but it may not be. It may be that ๐ถ is a different value, in which case weโ€™ll be properly normalizing this function by solving for ๐ถ.

So plugging in for ฮจ of ๐‘ฅ with ๐ถ cosine ๐›ผ๐‘ฅ, our expression simplifies to ๐ถ squared cosine squared of ๐›ผ๐‘ฅ ๐‘‘๐‘ฅ. And remember, by the normalization condition, all of this equals one. So we want to integrate this expression and find out what capital ๐ถ is equal to. Since ๐ถ is a constant, we can remove it from inside the integral. To integrate this cosine squared function, we can split this expression up into two different simpler integrals.

We can do that by recalling the half-angle identity that the cosine squared of a variable ๐‘ฅ equals one plus the cosine of two ๐‘ฅ all divided by two. With this substitution in place, we can now make two integrals out of this one single expression. The single integral splits up into two integrals: the first one, simply, of the constant one-half and the second one of cosine of two ๐›ผ๐‘ฅ. Notice that both integrals still have the same upper and lower boundary conditions and that we havenโ€™t forgotten about the factor of one-half in the second integral.

When we evaluate the first integral in this expression, we find that itโ€™s equal to ๐ถ squared over two times ๐‘ฅ evaluated at the lower boundary condition and then to the upper boundary condition. When we enter these values in for ๐‘ฅ, we find that this is ๐œ‹ over two ๐›ผ minus minus ๐œ‹ over two ๐›ผ or ๐œ‹ divided by ๐›ผ. So thatโ€™s the first integral. For the second integral, we can recall that the integral of cosine of ๐‘ฅ is sin of ๐‘ฅ, and weโ€™ll also remember to divide our answer for the integral of cosine of two ๐›ผ๐‘ฅ by the derivative of the argument of the cosine of two ๐›ผ๐‘ฅ, to be consistent with the chain rule in differentiation.

So our integral is the sine of two ๐›ผ๐‘ฅ divided by two ๐›ผ evaluated from ๐œ‹ over two ๐›ผ down to negative ๐œ‹ over two ๐›ผ. Now as we insert our upper and our lower boundaries into this function, notice what happens in each case. Two ๐›ผ cancels in the numerator and denominator, leaving us with, in the first case, sine of ๐œ‹ and, in the second case, sine of negative ๐œ‹. As we consider the sine function though, we recall that the sine of ๐œ‹ is equal to zero and the sine of negative ๐œ‹ is also zero.

So the entire second integral evaluates to zero over the interval weโ€™re considering. All that means weโ€™re left with an expression thatโ€™s simpler than we might have expected: ๐ถ squared over two times ๐œ‹ over ๐›ผ is equal to one. When we rearrange this expression to solve for ๐ถ, we find itโ€™s equal to the square root of two ๐›ผ over ๐œ‹, so ๐ถ is not one. But the good news is now our wave function is normalized, so our final normalized wave function is ฮจ of ๐‘ฅ equals the square root of two ๐›ผ over ๐œ‹ times the cosine of ๐›ผ๐‘ฅ.

Now weโ€™re ready to solve for the probability ๐‘ƒ that a particle described by this wave function appears in the bounds that weโ€™ve been given in the problem statement. That probability is equal to our wave function ฮจ of ๐‘ฅ, the magnitude squared, integrated from our lower bound to our upper bound. Weโ€™ll solve this integral in a similar way as the one we solved to find the normalization constant ๐ถ. First, we factor out the constant, then we use the half-angle identity to rewrite our integral into more manageable separate integrals.

Before splitting them up, we can simplify by canceling out the factor of two in numerator and denominator. Then with the integrals split into two separate integrals with the same upper and lower bounds, we start by evaluating the first one. The integral of one with the respect to ๐‘ฅ is ๐‘ฅ. And evaluating this from the upper to the lower bounds, since the lower bound is zero, the first integral overall simply equals ๐›ผ over ๐œ‹ times the upper bound.

Looking at the second integral, once again weโ€™ll remember that the integral of the cosine of ๐‘ฅ is the sine of ๐‘ฅ, and weโ€™ll also remember to divide by the derivative, with respect to ๐‘ฅ, of the argument of this trigonometric function. This integral, sine of two ๐›ผ ๐‘ฅ over two ๐›ผ, means that the ๐›ผ in the denominator and the one multiplied in the constant cancel out. Factoring out the one- half from beneath the sine function, now weโ€™re ready to evaluate this trigonometric function at the given boundary conditions.

If we look at the second of these two terms, notice that the argument of the sin function comes down to zero, and the sine of zero is itself zero. So the second integral reduces to the sine of two ๐›ผ times the upper bound. Weโ€™re now ready to take our given value of ๐›ผ and plug it in where it appears in this expression. With ๐›ผ plugged in, when we look at our first term, it simplifies to one over 2.00๐œ‹.

And our second term, when we evaluate it using a calculator, evaluates to approximately 0.841 over two ๐œ‹. And when we add these two numbers together, we find that, to three significant figures, ๐‘ƒ is 0.293 or 29.3 percent. Thatโ€™s the probability of our finding a particle between the lower and upper bounds given.

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