Finding the y axis on a probability density graph

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Homework Help Overview

The discussion revolves around understanding the y-axis of a probability density graph, particularly in the context of quantum mechanics and wavefunctions. Participants are exploring the relationship between probability and probability density, and how to correctly interpret and manipulate these concepts mathematically.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to derive the correct probability density from a given wavefunction, questioning the validity of their initial assumptions about the graph's nature. There is discussion about the necessity of integrating the correct function and the implications of negative values in probability density.

Discussion Status

Some participants have identified misunderstandings regarding the nature of the graph and the definitions of probability versus probability density. There is ongoing exploration of how to properly convert probability into probability density, with some guidance offered on the need to square the wavefunction.

Contextual Notes

Participants are navigating constraints related to the definitions of probability density and the requirements for total probability, as well as the implications of negative values in their calculations.

Sorin2225
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Homework Statement
I was given a probability density graph and from it I had to figure out what B was in nm^-1/2
Relevant Equations
I think it's just from the graph.
I assumed to find it I would need to find the area under the graph. I also assumed that the part under x would cancel out so I would be left with 2b*10=1 if it was, in fact, true that it had to equal to one. So my final answer was (1/10)/2 nm^-1 but the actual answer was 0.0845 nm^-1/2 and I'm unsure of how this answer was gained.
 

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That is not a probability density. A probability density can not be negative.

Judging by the use of the symbol ##\psi## I'm guessing this is a quantum mechanical wavefunction, so the actual probability density is ##|\psi|^2##. Your method was correct, that the total probability (total area) must be 1. So make that change and see how it works out.

The units of ##b## are another clue that you're supposed to be squaring these values to get the density, because ##b^2 \times## [length] is unitless.
 
To get b however I'm doing 3b*10-2b*10+2b*10=1 and rearranging for this but I'm nowhere near the right answer
 
Sorin2225 said:
To get b however I'm doing 3b*10-2b*10+2b*10=1

You shouldn't be. You're integrating ##\psi## with respect to ##x## and I just told you that ##\psi## is not the probability density, not the thing you should be integrating. Therefore trying to get a total probability of 1 by integrating ##\psi## is not a correct thing to do.

Again, you want to integrate the probability density. ##\psi## is not the probability density. Find the probability density and then integrate that.

I know you started from this assumption, in your original post:
Sorin2225 said:
I was given a probability density graph...

but what I'm trying to say is that you were not given a probability density graph. A probability density can not be negative. ##\psi## is by inspection not a probability density.
 
Yes sorry, I realized that my initial assumption about the type of graph was wrong. I am confused on how I am meant to be changing the probability into probability density. To do this I attempted
to square the probability
 

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