为何PowerShell中需对Join-Path返回值显式转字符串赋值给XmlNode？

根据官方文档，Join-Path函数返回字符串类型，理论上可直接用作字符串值，但实际操作中，若不对Join-Path的结果进行显式转换，PowerShell会抛出如下错误：

Cannot set "path" because only strings can be used as values to set XmlNode properties.
At C:\Users\Rayman\Downloads\example.ps1:6 char:1
+ $node.path = (Join-Path 'C:\Users\Rayman\Downloads' 'VSCode-win32-x64 ...
+ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    + CategoryInfo          : NotSpecified: (:) [], SetValueException
    + FullyQualifiedErrorId : XmlNodeSetShouldBeAString

为何需要对一个本已是字符串类型的结果进行显式字符串转换？以下是最小复现示例：

脚本代码 #

$xmlPath = 'C:\Users\Rayman\Downloads\example.xml'
[xml]$xml = Get-Content $xmlPath
$node = $xml.SelectSingleNode('//Software')
$node.SetAttribute('path', '')
# 不清楚为何必须用[string]转换字符串类型变量
$node.path = [string](Join-Path 'C:\Users\Rayman\Downloads' 'VSCode-win32-x64-1.73.1')

XML文件 #

<?xml version="1.0" encoding="UTF-8"?>
<Configuration>
    <Software id="VSCode" />
</Configuration>

使用的PowerShell版本为5.1.19041.5965。

内容的提问来源于stack exchange，提问作者Rayman